Contents - SelfStudys

GEOMETRICAL OPTICS

ContentsParticular's Page No.

Theory 01 – 72

Exercise # 1 73 – 90Part - I : Subjective QuestionsPart - II : Objective QuestionsPart - III : Match The Column

Exercise # 2 91 – 103Part - I : Only one option correct typePart - II : Integer Type QuestionsPart - III : One or More than one option correct typePart - IV : Comprehension

Exercise # 3 104 – 118Part - I : Previous Years IIT-JEE/JEE Advance ProblemsPart - II : Previous Years AIEEE / JEE Mains Problems

Answers 119 – 121

JEE (Advance) SyllabusRectilinear propagation of light; Reflection and refraction at plane and spherical surfaces; Total internal reflection;Deviation and dispersion of light by a prism; Thin lenses; Combinations of mirrors and thin lenses; Magnification.

JEE (Main) SyllabusReflection of Light by Spherical Mirrors, Refraction, Total Internal Reflection, Refraction at Spherical Surfaces and byLenses, Refraction through a Prism, Dispersion by a Prism, Some Natural Phenomena due to Sunlight, OpticalInstruments

Note: Marked Questions can be used for Revision.

Ranker Problems 122 – 125

Answers 126

SAP (Self Assessment Paper) 127 – 130

Answers 130

JEE (Adv.)-Physics Geometrical Optics

GEOMETRICAL OPTICS

INTRODUCTIONThe branch of Physics called optics deals with the behavior of light and other electromagnetic waves. Under

many circumstances, the wavelength of light is negligible compared with the dimensions of the device as in the

case of ordinary mirrors and lenses. A light beam can then be treated as a ray whose propagation is governed

by simple geometric rules. The part of optics that deals with such phenomenon is known as geometric optics.

PROPAGATION OF LIGHT

Light travels along straight line in a medium or in vacuum. The path of light changes only when there is an

object in its path or where the medium changes. We call this rectilinear (straight–line) propagation of light.

Light that starts from a point A and passes through another point B in the same medium actually passes

through all the points on the straight line AB. Such a straight line path of light is called a ray of light. Light rays

start from each point of a source and travel along straight lines till they fall on an object or a surface separating

two media (mediums). A bundle of light rays is called a beam of light.

Apart from vacuum and gases, light can travel through some liquids and solids. A medium in which light can

travel freely over large distances is called a transparent medium. Water, glycerine, glass and clear plastics are

transparent. A medium in which light cannot travel is called opaque. Wood, metals, bricks, etc., are opaque. In

materials like oil, light can travel some distance, but its intensity reduces rapidly. Such materials are called

translucent.

REFLECTION OF LIGHTWhen light rays strike the boundary of two media such as air and glass, a part of light is bounced back into thesame medium. This is called Reflection of light.

(i) Regular / Specular reflection :

When the reflection takes place from a perfect plane surface

then after reflection rays remain parallel.

It is called Regular reflection.

(ii) Diffused reflection

When the surface is rough, light is reflected from the surface from bits

of its plane surfaces in irregular directions. This is called diffused

reflection. This process enables us to see an object from any position.


LAWS OF REFLECTION1. The incident ray, the reflected ray and the normal at the point of incidence lie in the same plane.

This plane is called the plane of incidence (or plane of reflection). This condition can be

expressed mathematically as R . ( × N ) = N . ( × R ) = . ( N × R ) = 0 where , N and Rare vectors of any magnitude along incident ray, the normal and the reflected ray respectively.

i rref

lected

rayincident ray

O

mirror

plane normal to mirror

reflecting surface

2. The angle of reflection is equal to the angle of incident i.e. i = r.

reflec

ted ra

yincident ray

O O

i rref

lected

rayincident ray

normalnormal

O

i rref

lected

rayincident ray

normal

er

n

\\\\\\\\\\\\\\

\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

\\\\\\\\

3. Normal Incidence : In case light is incident normally,

i = r = 0

= 180º

Note : We say that the ray has retraced its path.4. Grazing Incidence : In case light strikes the reflecting surface tangentially,

i = r = 90

= 0º or 360º

Note : In case of reflection speed (magnitude of velocity) of light remains unchanged but in Grazingincidence velocity remains unchanged.

OLVED XAMPLES EExample 1. Show that for a light ray incident at an angle ‘i’ on getting reflected the angle of deviation is

= 2i or 2i.

Solution :

From figure (b) it is clear that light ray bends either by 1 anticlockwise or by 2 (= 2 – 1) clockwise.From figure (a) 1 = – 2i .

2 = + 2i .


REAL AND VIRTUAL SPACESA mirror, plane or spherical divides the space into two ;

(a) Real space, a side where the reflected rays exist.

(b) Virtual space is on the other side where the reflected rays do not exist.

Real Space

VirtualSpace

Realspace

Virtualspace

\\\\\\\\\\\\\\\\ \\\\\ \\ \\\\\\\\\\\\\\\\\\\\\\\\\\\

Realspace

Virtualspace

\\\\\\

\\\\\\

\\\\\\

\\ \\\ \

\\\\\\

\\\\\\

\\\\\\

\\\\\\

\\

• OBJECT

Object is decided by incident rays only. The point object is that point from which the incident rays actuallydiverge (Real object) or towards which the incident rays appear to converge (virtual object).

Point Object(Real)

Point

Virtual(Object)

Real

Object Point

\\\\\\\\\\\\\\\\ \\\\\\\\\\\\\\\\\\\\\\\\\\

• IMAGE

Image is decided by reflected or refracted rays only. The point image is that point at which the refracted /reflected rays reflected from the mirror, actually converge (real image) or from which the refracted /reflectedrays appear to diverge (virtual image).

IO

(Real image)

\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

(Virtual image)

O I

\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ \ \\ \\\ \\\\\\\\\\\\\\\\\

PLANE MIRRORPlane mirror is formed by polishing one surface of a plane thin glass plate .It is also said to besilvered on one side.

A beam of parallel rays of light, incident on a plane mirror will get reflected as a beam of parallelreflected rays.


• Point objectCharacteristics of image due to Reflection by a Plane Mirror :

(i) Distance of object from mirror = Distance of imagefrom the mirror.

(ii) All the incident rays from a point object will meet at asingle point after reflection from a plane mirror whichis called image.

(iii) The line joining a point object and its image is normalto the reflecting surface.

(iv) For a real object the image is virtual and for a virtualobject the image is real

(v) The region in which observer's eye must be present in order to view the image is called fieldof view.

OLVED XAMPLES EExample 2. Figure shows a point object A and a plane mirror MN. Find the position

of image of object A, in mirror MN, by drawing ray diagram. Indicatethe region in which observer’s eye must be present in order to view

the image. (This region is called field of view).

Solution : See figure, consider any two rays

emanating from the object. N1 and

N2 are normals ;

i1 = r1 and i2 = r2

The meeting point of reflected rays R1 and R2 is image A’. Thoughonly two rays are considered it must be understood that all raysfrom A reflect from mirror such that their meeting point is A´. Toobtain the region in which reflected rays are present, join A´ with

the ends of mirror and extend. The following figure shows this

region as shaded. In figure there are no reflected rays beyond therays 1 and 2, therefore the observers P and Q cannot see theimage because they do not receive any reflected ray.

• Extended object :An extended object like AB shown in figure is a combination of infinitenumber of point objects from A to B. Image of every point object will beformed individually and thus infinite images will be formed. A´ will beimage of A, C´ will be image of C, B´ will be

image of B etc. All point images together form extended image.

Thus extended image is formed of an extended object.

JEE (Adv.)-Physics Geometrical Optics• Properties of image of an extended object, formed by a plane mirror :

(1) Size of extended object = size of extended image.

(2) The image is erect, if the extended object is placed parallel to the mirror.

(3) The image is inverted if the extended object lies perpendicular to the plane mirror.

(4) If an extended horizontal object is placed infront of a mirror inclined 45° with the horizontal,the image formed will be vertical. See figure.

• To see complete image in a plane mirror the minimum length of plane mirror should be half the height of aperson.

From figure. HNM and ENM are congruent

EN = HN MD = EN = 12 HE

Similarly EN' M' and LN' M' are congruent

H

N

E

N'

L

M'

M

G

H2

D

Length of the mirror MM' = MD + M'D = 12 HE +

12 EL

= 12 (HE + EL) =

12 HL

Minimum of length of mirror is just half of the person.

• This result does not depend on position of eye (height of the eye from ground).

• This result is independent of distance of person in front of mirror.


• Relation between velocity of object and image :From mirror property : x im = - xom , yim = yom and zim = zom

Here xim means ‘x’ coordinate of image with respect to mirror. Similarly others have meaning.

•vop

von vin

vip

Stationary Plane Mirror

on inv v , opv = ipv

though speed of object and image are the same

vop = component of velocity of object along parallel to mirror.

von = component of velocity of object along normal to mirror.

v ip = component of velocity of image along parallel to mirror.

v in = component of velocity of image along normal to mirror.

•

vop

vmp

von

vmn

vin

vip

If mirror is moving ip opv v and im omn nv v

in mnv v = – on mnv v in mn onv 2v v

mnv = component of velocity of mirror along normal.

opv = component of velocity of object along mirror..

onv = component of velocity of object along normal

ipv = component of velocity of image along mirror..

inv = component of velocity of image along normal.

OLVED XAMPLES EExample 3. There is a point object and a plane mirror. If the mirror is moved by 10 cm away from the

object find the distance which the image will move.

Solution : We know that x im = - xom or x i – xm = xm – xo

or x i – xm = xm – xo.

In this Q. xo = 0 ; xm = 10 cm.

Therefore x i = 2 xm – xo = 20 cm.

JEE (Adv.)-Physics Geometrical OpticsExample 4. An object moves with 5 m/s towards right while the mirror moves with 1m/s towards the left

as shown. Find the velocity of image.

Solution Take as + direction. v i - vm = vm - v0

v i - (-1) = (-1) - 5

v i= - 7m/s.

7 m/s and direction towards left.

Example 5. Find the velocity of the image.

Solution : oxv = (–10 cos 37°) i = ˆ8i and

2m/s

37°O

10m/s

oyv = (10 sin 37°) j = ˆ6 j

ix mx oxv 2v v = 2 (–2i) – ˆ8i = ˆ4i and iy oyˆv v 6 j

• Keeping the mirror fixed if the incident ray is rotated by some angle, the reflected ray is also rotated by thesame angle but in opposite sense. (See Fig. 1)

(Fig. 1)

I2N1 R2I1 R1

(Fig. 2)

R2

N1

R1

2

I N2

• Keeping the incident ray fixed, if the mirror is rotated by some angle, then the reflected ray rotates by doublethe angle in the same sense. (See Fig. 2)

• Images formed by two plane mirrors :If rays after getting reflected from one mirror strike second mirror, the image formed by first mirror willfunction as an object for second mirror, and this process will continue for every successive reflection.

OLVED XAMPLES EExample 6. Figure shows a point object placed between two parallel

mirrors. Its distance from M1 is 2 cm and that from M2 is 8cm. Find the distance of images from the two mirrorsconsidering reflection on

mirror M1 first.

Solution : To understand how images are formed see the following figure and table. You will require toknow what symbols like 121 stands for. See the following diagram.


I1 I12 I1212I121

Incident rays

Reflected by

Reflected rays

Object Image Object distance

Image distance

Rays 1 M1 Rays 2 O 1 AO = 2cm A 1 = 2 cm Rays 2 M2 Rays 3 1 12 B 1 = 12 cm B 12 = 12 cm Rays 3 M1 Rays 4 12 121 A 12 = 22cm A 121 = 22cm Rays 4 M2 Rays 5 121 1212 B 121 =32cm B 1212=32cm

Similarly images will be formed by the rays striking mirror M2 first. Total number of images = .

• Locating all the images formed by two plane mirrors• If there are two plane mirror inclined to each other at an angle the number of image (n) of a point object

formed are determined as follows.

(a) If 360

m is even then number of images n = m – 1

(b) If 360

m is odd. There will be two case.

(i) When object is not on bisector, then number of images n = m(ii) When object is at bisector, then number of images n = m – 1

(c) If 360

m is a fraction, and the object is placed symmetrically then no. of images n =nearest even

integerNu mber of images formed if object is

placed S.No. in degree

360m

asymmetrically symm etrically

1. 0 2. 30 12 11 11 3. 45 8 7 7 4. 60 6 5 5 5. 72 5 5 4 6. 75 4.8 – 4 7. 90 4 3 3 8. 112.5 3.2 – 4 9. 120 3 3 2


• If the object is placed between two plane mirrors then images are formed due to multiple reflections. At eachreflection, a part of light energy is absorbed. Therefore, distant images get fainter.

• Total deviation produced by the combination of two plane mirrors which

normal

incide

nt

reflected

O

C

AQ M1

mirrorB

2

90–

90–

1

M2

mirror

\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ \

\ \\\\\

\\\\ \\

\\\\\ \

\\\\\\

\ \\\\ \

\\ \\ \\

\\\\\ \

\\\\\\

\ \\\

are inclined at an angle from each other.

= 1 + 2 = 180 – 2 + 180 – 2 = 360 – 2 ( + ) ...(i)

From QAB, + 90 – + 90 – = 180 = + ...(ii)

Putting the value of in (i) from (ii), = 360 – 2

ELF RACTICE ROBLEMSS P P1. What should be the angle between two plane mirrors so that whatever be the angle of incidence, the incident

ray and the reflected ray from the two mirrors be parallel to each other [USS - [MP PMT 1997; DCE 2001, 03]

(1) 60° (2) 90° (3) 120° (4) 175°

2. A light bulb is placed between two plane mirrors inclined at an angle of 60°. The number of images formed are[USS - SCRA 1994; AIIMS 1997; RPMT 1999; AIEEE 2002; Orissa JEE 2003; MP PMT 2004; MP PET 2004]

(1) 6 (2) 2 (3) 5 (4) 4

3. A ray of light incidents on a plane mirror at an angle of 30°. The deviation produced in the ray is

(1) 30° (2) 60° (3) 90° (4) 120°

4. A man runs towards mirror at a speed of 15m/s. What is the speed of his image USS - [CBSE PMT 2000]

(1) 7.5 m/s (2) 15 m/s (3) 30 m/s (4) 45 m/s

Ans. 1. (2) 2. (3) 3. (4) 4. (2)

• SPHERICAL (CURVED) MIRRORCurved mirror is part of a hollow sphere. If reflection takes place from the inner surface then

The mirror is called concave and if its outer surface acts as reflector it is convex.

P

C

Msphericalsurface concave mirror convex mirror

principalaxis

sphericalmirror

M' \\\\\\\\\\ \\\ \\\ \\\\\\\\\\\\\\\\\ \\\\\ \

\ \\\\ \

\ \\\\\

\\\\\\

\\\\\\

\\\

C CF FP P

M' M'

M M

ir

• Definitions FOR THIN SPHERICAL MIRRORS(i) Pole is any point on the reflecting surface of the mirror. For convenience we take it as the midpoint P of the

mirror (as shown).

(ii) Principal–section is any section of the mirror such as MM' passing through the pole is called principal–section.

(iii) Centre of curvature is the centre C of the sphere of which the mirror is a part.


(iv) Radius of curvature is the radius R of the sphere of which the mirror is a part.

(v) Principal–axis is the line CP, joining the pole and centre of curvature of the mirror.

(vi) Principal– focus is an image point F on principal axis for which object is at infinity.

focus

parallel to axis

centre of curvature

C F

normal

M

P

M'

\\\\\\\\\ \\ \\\ \\\\ \\ \\\\ \\\\\\\\\\\\\\\\\\\\\\\\\

parallel to axis

centre of curvature

M

P

M'

ir

focusF C

\\\\\\\\\\\\\\\\\\\\\\\\\\\\ \ \\ \\\ \ \\\\\ \\\\\\ \\\\(vii) Focal–length is the distance PF between pole P and focus F along principal axis.

(viii) Aperture, in reference to a mirror, means the effective diameter of the light reflecting area of the mirror.

(ix) Focal Plane is the plane passing through focus and perpendicular to principal axis.

C

A

P

B

////////////////// //// //////////////////////

Focalplane

F/

/////////////////// ////////////////////////////

C

A

F P

BFocalplane

(x) Paraxial Rays Those rays which make small angle with normal at point of incidence and hence are closeto principal axis.

n

PC

\\\\\\\\\\\ \\\\\\\\ \\ \ \\ \\\\\\\\\\\\\\\\\\\\\\\\\\\

n

////////////////// //// / ////////////////////////////

( is very small) C

//////////////// // / ///////////////////////

(xi) Marginal rays :

Rays having a large angle of incidence

n

PC

//////////// //////// //// // //////////////////////////////

( is large)

• Paraxial rays (not parallel to principal axis)

PC

F'

F

////////////////////////////////////////////// /// ////// / // ////////////////////////////////////////

Such rays after reflection meet at a point in the focal plane(F'), such that

FF 'tan

FPFF 'f

FF ' = f


OLVED XAMPLES EExample : 7 Find the angle of incidence of ray for which it passes

through the pole, given that M || CP.

Solution : MIC = C P =

M || CP MIC = CP =

C = CP

C P = CP =

In C P all angle are equal

3 = 180º = 60º

Example : 8 Find the distance CQ if incident light ray parallel to principalaxis is incident at an angle i. Also find the distance CQ

if i 0.

Solution : cos i = CQ2R

CQ = icos2R

As i increases cos i decreases. If i is a small angle cos i 1

Hence CQ increases CQ = R/2

So, paraxial rays meet at a distance equal to R / 2 from center of curvature, which is called focus.

SIGN– CONVENTION

C F P

negative

light

positive

nega

tive

posi

tive

CF

positive

negative

nega

tive

posi

tive

light

positivenegative

nega

tive

posi

tive

positive

O

y

xP

//////// /////// //////////// /////////////////////////////////////// //////

//////

//////

//////

//////

//// //

/ /// //

//// //

// ////

//////

// ////

• Along principal axis, distances are measured from the pole ( pole is taken as the origin).

• Distance in the direction of light are taken to be positive while opposite to be negative.

• The distances above principal axis are taken to be positive while below it negative.

• Whenever and wherever possible the ray of light is taken to travel from left to right.

• Formulae for Reflection from spherical mirrors :

(a) Mirror formula : 1v +

1u =

2R =

1f

X-coordinate of centre of Curvature and focus of Concave mirror are negative and those forConvex mirror are positive. In case of mirrors since light rays reflect back in X-direction,therefore -ve sign of v indicates real image and +ve sign of v indicates virtual image.


OLVED XAMPLES EExample 9. Figure shows a spherical concave mirror with its pole at (0, 0)

and principle axis along x axis. There is a point object at(–40 cm, 1cm), find the position of image.

Solution : According to sign convention,u = –40 cm h1 = +1 cm f = – 5 cm.

f1

u1

v1

51

401

v1

; v = 740

cm. ; uv

hh

1

2

h2 = – uv

× h1 = 40

17

40

= 71

cm.

The position of image is cm71,cm

740

Example 10. Converging rays are incident on a convex spherical mirror so that their extensions intersect30 cm behind the mirror on the optical axis. The reflected rays form a diverging beam sothat their extensions intersect the optical axis 1.2 m from the mirror . Determine the focallength of the mirror.

Solution : In this case, u = + 30v = + 120

f1

= v1

+ u1

= 1201

+ 301

f = 24 cm

MAGNIFICATIONTransverse or lateral magnificationIf one dimensional object is placed perpendicular to the principal axis then linear magnification is called transverse

or lateral magnification. i

o

h vm

h u , m =f

vfuf

f

Magnification Image Magnification Image|m|> 1 enlarged |m| < 1 diminishedm < 0 inverted m > 0 erect

• Longitudinal magnificationIf one dimensional object is placed with its length along the principal axis then linear magnification is calledlongitudinal magnification.

Longitudinal magnification 2 1L

2 1

v vlength of imagem

length of object u u u2v1

u1

objectimage

v2

\\\\\\\\\\\\\\\\ \\\\ \\ \\\\\\\\\\\\\\\\\\\\\\\\\For small objects only : L

dvm

dudifferentiation of

1 1 1v u f

gives us 2 2

dv du0

v u

2dv vdu u

so 2

2L

dv vm m

du u

Nature of Object Nature of Image Inverted or erectReal Real InvertedReal Virtual ErectVirtual Real ErectVirtual Virtual Inverted


IMAGE FORMED BY THE CONCAVE MIRRORS.No. Position of object Ray diagram Position Nature of Size of

of image image image

1. At infinity at focus real and inverted very small

2. Between infinity and between focus real and inverted small

centre of cruvature

C Fho

hi

/////////// /// ////// / //// ////////////////////////////

M

P

M'

and centre of

curvature 3. At centre of curvature at centre of real and inverted equal to

curvature object size

4. Between focus andC F

Ohi

h0

M

P

M'

/////////// /// /// // / //// /////////////////////////////

between centre real and inverted enlarged

centre of curvature of curvatureand infinity

5. At focus at infinity real and inverted very large

6. Between pole and focusC

F

M

P

M'

h0

hi

// ///// // //////// /// /////// /////////////////////////////////////

between poles virtual and enlarged

and focus erect

7. At infinity

M

P

M'

F C

I

at F virtual, erect and very small (m<<+1)

Object placed at infinity

///////////////////////////// / // / /// //// / // //// /// /

at focus virtual and erect very small

8. Between pole and infinity

M

P

M'

F CIO

Virtual, erect, diminished(m<+1), between P and F

Object placed in front of mirror

////////////////////////// // /// // / // ///// /// // // ///

between focus virtual and erect small

& pole


OLVED XAMPLES EExample 11. An extended object is placed perpendicular to the principle axis of a concave mirror of radius

of curvature 20 cm at a distance of 15 cm from pole. Find the lateral magnification produced.

Solution : u = – 15 cm f = – 10 cm

Using f1

u1

v1

we get, v = – 30 cm

m = uv

= – 2.

Aliter : m = uff

= )15(1010

= – 2

Example 12. A person looks into a spherical mirror. The size of image of his face is twice the actual sizeof his face. If the face is at a distance 20 cm then find the nature of radius of curvature of themirror.

Solution : Person will see his face only when the image is virtual. Virtual image of real object is erect.

Hence m = 2

uv

= 2 v = 40 cm

Applying f1

u1

v1

; f = – 40 cm or R = 80 cm.

Aliter : m = uff

2 = )20(ff

f = – 40 cm or R = 80 cm

Example 13. The focal length of a concave mirror is 30cm. Find the position of the object in front of the mirror,so that the image is three times the size of the object.

Solution : As the object is in front of the mirror it is real and for real object the magnified image formed byconcave mirror can be inverted (i.e.,real) or erect (i.e.,virtual), so there are two possibilities.

(a) If the image is inverted (i.e., real)

fm

f u30

330 u

u = – 40 cm

Object must be at a distance of 40 cm in

CO

FI

M

P

M'120cm

40cm

////// /////////// // // // // ////////////////////////

front of the mirror (in between C and F).

(b) If the image is erect (i.e., virtual)

f 30m 3 u = 20 cm

f u 30 uObject must be at a distance of 20 cm in

CF

M

P

M'

O

I

/ //////// ////// ///// // ///////////////////////////

front of the mirror (in between F and P).


Example 14. A thin rod of length f3

is placed along the principal axis of a concave mirror of focal length f such

that its image which is real and elongated, just touches the rod. What is magnification ?

Solution : Image is real and enlarged, the object must be between C and F. One end A' of the imagecoincides with the end A of rod itself, so

vA = uA ,A A

1 1 1v v f

i.e., vA = uA = – 2f

F

2f

53 f

f3

52 f

B' A'

A BP

M

M'

C

/// // ////// /// // /// / // //////////////////////

so it clear that the end A is at C. the length of rod is f3

distance of the other end B from P is uB = f 5

2f f3 3

if the distance of image of end B from P is vB then BB

1 1 1 5v f

5v f 2f3

the length of the image B A

5 1| v | | v | f 2f f

2 2 and magnification B A

B A

1fv v 32m

1u u 2f3

Negative sign implies that image is inverted with respect to object and so it is real.

Example 15. An image of a candle on a screen is found to be double its size. When the candle is shiftedby a distance 5 cm then the image become triple its size. Find the nature and ROC of themirror.

Solution : Since the images formed on screen it is real. Real object and real image implies concavemirror.

Applying, m = uf

for – 2 = )u(f

f....(1)

After shifting – 3 = )5u(ff

....(2)

[Why u + 5 ? , why not u – 5 : In a concave mirror are size of real image will increase, onlywhen the real object is brought closer to the mirror. In doing so, its x coordinate willincrease]

From (1) & (2) we get,

f = – 30 cm or R = 60 cm


VELOCITY OF IMAGE OF MOVING OBJECT (SPHERICAL MIRROR)(a) Velocity component along axis (Longitudinal velocity)

When an object is coming from infinite towards the focus of concave mirror

1 1 1v u f 2 2

1 dv 1 duv dt u dt

= 0 2

ix ox2

vv v

u 2

oxm v

Mj

iO

M'

// /// //////// // //// / // / // /////////////////////////////////

ix ox

dv duv velocity of image along principal-axis; v velocity of object along principal-axis

dt dt

(b) Velocity component perpendicular to axis (Transverse velocity)

m = I

0

h v fh u f u

hI = 0

fh

f u

Mj

iO

M'

// /// //////// // //// / // / // /////////////////////////////////

Idhdt

= 0dhf

f u dt ; oyiyˆv mv j

rI

ro

dhvelocity of image to principal-axis

dtdh

velocity of object to principal-axisdt

Note : Here principal axis has been taken to be along x–axis.

POWER OF A MIRROR

The power of a mirror is defined as 1 100

Pf(m) f(cm)

OLVED XAMPLES EExample 16. A point object is placed 60 cm from pole of a concave mirror of focal length 10 cm on the

principle axis. Find

(a) the position of image

(b) If object is shifted 1 mm towards the mirror along principle axis find the shift inimage. Explain the result.

Solution : (a) u = – 60 cm

f = – 10cm

fufuv = )10(60

)60(10= 50

600 = –12 cm.


(b) f1

u1

v1

Differentiating , we get dv = – 2

2

uv

du = –2

6012

[1 mm ] = – 251

mm

[ du = 1mm; sign of du is +ve because it is shifted in +ve direction defined by signconvention.]

• –ve sign of dv indicates that the image will shift towards negative direction.

• The sign of v is negative. Which implies the image is formed on negative side of pole. (A) and(B) together imply that the image will shift away from pole.

Note that differentials dv and du denote small changes only.

Newton's Formula: XY = f 2

X and Y are the distances ( along the principal axis ) of the object and image respectivelyfrom the principal focus. This formula can be used when the distances are mentioned orasked from the focus.

In case of spherical mirrors if object distance (x) and image distance (y) are measured from

focus instead of pole, u = –(f + x) and v = – (f + y), by v1

+ u1

= f1

we can write

– )yf(

1 – )xf(1

= f1– on solving xy = f2

This is Newton's formula.

ELF RACTICE ROBLEMSS P P5. A dimenished virtual image can be formed only in

(1) Plane mirror (2) A concave mirror (3) A convex mirror (4) Concave-parabolic mirror

6. An object 5 cm tall is placed 1m from a concave spherical mirror which has a radius of curvature of 20 cm.The size of the image is

(1) 0.11 cm (2) 0.50 cm (3) 0.55 cm (4) 0.60 cm

7. In a concave mirror experiment, an object is placed at a distance x1 from the focus and the image is formedat a distance x2 from the focus. The focal length of the mirror would be

(1) x1x2 (2) 21xx (3) 2

xx 21 (4) 2

1xx

8. Given a point source of light, which of the following can produce a parallel beam of light

(1) Convex mirror (2) Concave mirror

(3) Concave lens (4)Two plane mirrors inclined at an angle of 90°

Ans. 5. (3) 6. (3) 7. (2) 8. (2)

(c) In case of successive reflection from mirrors, the overall lateral magnification is given bym1 × m2 × m3 ......, where m1 , m2 etc. are lateral magnifications produced by individual mirrors.h1 and h2 denote the y coordinate of object and image respectively.


OLVED XAMPLES EExample 17. Find the position of final image after three successive

reflections taking first reflection on m1.

Solution : reflection :Focus of mirror = – 10 cm u = – 15 cmApplying mirror formula :

f1

u1

v1

v = – 30 cm.

For reflection on plane mirror :u = – 10 cm v = 10 cm

For reflection on curved mirror again :u = – 50 cm ; f = – 10 cm

Applying mirror formula :

f1

u1

v1

; v = – 12.5 cm.

Example 18. A concave mirror of focal length 10 cm and convex mirror of focal length 15 cm are placed facingeach other 40 cm apart. A point object is placed between the mirror on their common axis and 15cm from the concave mirror. Find the position of image produced by the reflection first at concavemirror and then at convex mirror.

Solution : For M1 mirror O act as a object, let its image is I1 then,

u = – 15 cm, f = –10 cm

15cmO

M2M1

15cm

f=+15 f=–10

I2 I1

///// /////////// ////// // /// // // ///////////////////////////////////////////

//////

//////

//////

/ /////

// ////

//////

//////

//////

//////

//////

//

1 1 1v 15 10

v = – 30 cm

Image I1 will act as a object for mirror M2 its distance

from mirror M2.

u1 = –(40 – 30) cm = –10 cm

so 11 1 1

1 1 1 1 1 1v 6cm

v u f v 10 15

So final image I2 is formed at a distance 6 cm behind the convex mirror and is virtual.

Example 19. A parabolic reflecting surface given by 2y

x2

, is placed at oirign, as shown. An incidnet ray

moving along y = 1 is incident on it. The ray gets reflected by the surface twice. The deviation

suffered by the ray is n

radians. Find n.

Solution :2y

x2

dy 1dx y , at y = 1, slope = 1

///////////////////// ///////////////////

incident ray

x-axis

i = 45°, r = 45° & deviation in first reflection = 90°.

Similarly for second reflection, Net deviation = 180°.


REFRACTIONRefraction is the phenomenon in which direction of propagation of light changes at the boundary when it passesfrom one medium to the other. In case of refraction frequency does not change.

• Laws of Refraction(i) Incident ray, refracted ray and normal always lie in the same plane. e

1

2

n

r

i

rIn vector form ˆ ˆ ˆ(e n).r 0

(ii) The product of refractive index and sine of angle of incidence at apoint in a medium is constant. 1 sin i = 2 sin r (Snell's law)

In vector form 1ˆ ˆ ˆ ˆe n r n

• Absolute refractive index

It is defined as the ratio of speed of light in free space 'c' to that in a given medium v. or n = cv

Denser is the medium, lesser will be the speed of light and so greater will be the refractive index,

vglass < vwater, G > W

• Relative refractive indexWhen light passes from one medium to the other, the refractive index

v1

v2

of medium 2 relative to 1 is written as 1 2 and is defined as

2 2 11 2

1 1 2

(c / v ) v(c / v ) v

• Bending of light ray

According to Snell's law, 1sin i = 2 sin r i

r

airR

D

water

normalincident ray

refracted ray

(i) If light passes from rarer to denser medium 1 = R and 2 = D

so that D

R

sin i1

sin ri > r

In passing from rarer to denser medium, the ray bends towards the normal.

(ii) If light passes from denser to rarer medium 1= D and 2 = R

i

rR

Dwater

normal

denser medium

rare medium refracted ray

incide

nt ray

R

D

sin i1

sin r i < r

In passing from denser to rarer medium, the ray bends away from thenormal

• Special cases :Normal incidence : i = 0

from snell’s law : r = 0


• Deviation of a Ray Due to RefractionDeviation ( ) of ray incident

at i and refracted at r

is given by = |i r| .

OLVED XAMPLES EExample 20. A light passes through many parallel slabs one by one as shown in figure.

Prove that n1sini1 = n2sini2 = n3sini3 = n4sini4 =.............[Remember this]. Also prove that ifn1 = n4 then light rays in medium n1 and in medium n4 are parallel.

Solution : We have,

2

1

isinisin

= 1

2

nn

n1 sin i1 = n2 sin i2 ....(i)

Similarly n2 sin i2 = n3 sin i3so onso, n1 sin i1 = n2 sin i2 = n3 sin i3 = ..............

n1 sin i1 = n4 sin i4 sin i1 = sin i4 ( n1 = n2)

so, i1 = i4Hence, light rays in medium n1 and in medium n4 are parallel.

Example 21. A light ray is incident on a glass sphere at an angle ofincidence 600 as shown. Find the angles r, r’,e and the totaldeviation after two refractions.

Solution : Applying Snell’s law 1sin600 = 3 sinr

r = 300

From symmetry r’ = r = 300 .

Again applying snell’s law at second surface 1sin e = 3 sinr

e = 600

Deviation at first surface = i – r = 600 – 300 = 300

Deviation at second surface = e – r’ = 600 – 300 = 300

Therefore total deviation = 600 .

JEE (Adv.)-Physics Geometrical OpticsExample 22. Find the angle a made by the light ray when it gets refracted

from water to air, as shown in figure .

Solution : Snell’s Law

Wsin W = a sin a ; 153

34

sin a ;

sin a = 54

; a = sin–154

Example 23. Find the speed of light in medium ‘a’ if speed of light in medium ‘b’ is 3c

where c = speed of

light in vacuum and light refracts from medium ‘a’ to medium ‘b’ making 45º and 60ºrespectively with the normal.

Solution : Snell’s Law

a sin a = b sin b ;av

c sin a =

bvc

sin b.

avc

sin 45º = 3/cc

sin 60º. ; va = 33c2

• Principle of Reversibility of Light Rays(a) A ray travelling along the path of the reflected ray is reflected along the path of the incident

ray.

(b) A refracted ray reversed to travel back along its path will get refracted along the path ofthe incident ray. Thus the incident and refracted rays are mutually reversible.

REFRACTION THROUGH A PARALLEL SLABWhen light passes through a parallel slab, having same medium on both sides, then

(a) Emergent ray is parallel to the incident ray.

Note : Emergent ray will not be parallel to the incident ray if the medium on both the sides of slab aredifferent.

• LATERAL SHIFTThe perpendicular distance between incident and emergent ray is known as lateral shift.

Lateral shift d = BC and t = thickness of slab

i

r

A

D

D

i– r

O

C

N

Bd

In BOC BC d

sin i rOB OB

d = OB sin(i – r) ...(i)

In OBD OD t t

cos r OBOB OB cos r

...(ii)

From (i) and (ii) t

d sin(i r)cos r


OLVED XAMPLES EExample 24. Find the lateral shift of light ray while is passes through a parallel glass slab of thickness

10 cm placed in air. The angle of incidence in air is 60º and the angle of refraction in glassis 45º.

Solution : d = rcos)ri(sint

= 45cos)4560(sin10

= 45cos15sin10

= 10 2 sin 15º.

APPARENT DEPTH AND NORMAL SHIFTIf a point object in denser medium is observed from rarer medium and boundary is plane, then fromSnell's law we have D sin i = R sin r...(i)

If the rays OA and OB are close enough to reach the eye.

sin i tan i = ac

pd

and sin r tan r = ap

pd

here dac = actual depth, dap = apparent depth

So that equation (i) becomes ac D 1

D Rac ap ap R 2

dp pdd d

i

R

object

r i

D

A B

rp

O

I

(If R = 1, D = ) then acap

dd so dap < dac ...(ii)

The distance between object and its image, called normal shift (x)

x = dac – dap ac

ap

dd ;

acac ac

d 1x d d 1 ...(iii) If dac = d then

1x d 1

object

dap

dac image

image dap

dac

object

JEE (Adv.)-Physics Geometrical Optics• Object in a rarer medium is seen from a denser medium

ac 1 R

ap 2 D

d 11

d i

R

iO Bapparent

heightdap

dAC

A B

densermedium

shift

actual height

ximageI

dac

D

dap = dac i.e., dap > dac

A high flying object appears to be higher than in reality.

x = dap – dac x = [ – 1] dac

OLVED XAMPLES EExample 25. An object lies 100 cm inside water .It is viewed from air nearly normally. Find the apparent

depth of the object.

Solution : d =relativen

d=

13/4

100=75 cm

Example 26. See the figure

(i) Find apparent height of the bird

(ii) Find apparent depth of fish

(iii) At what distance will the bird appear to the fish.

(iv) At what distance will the fish appear to the bird

(v) If the velocity of bird is 12 cm/sec downward and the fish is 12 cm/sec in upwarddirection, then find out their relative velocities with respect to each other.

Solution : (i) d´B = 4/3

36

341

36 = 48 cm

(ii) d´F = 3/436

= 27 cm

(iii) For fish : dB = 36 + 48 = 84 cm(iv) For bird : dF = 27 + 36 = 63 cm.

(v) Velocity of fish with respect to bird = 11

341212

// = 21 cm/sec.

Velocity of bird with respect to fish = 11

431212

// = 28 cm/sec.

OLVED XAMPLES E

TRANSPARENT GLASS SLAB (Normal shift)When an object is placed infront of a glass slab, it shift the object

in the direction of incident light and form a image at a distance x.

1x t 1


OLVED XAMPLES EExample 27. A 20 cm thick glass slab of refractive index 1.5 is kept infront of a plane mirror. An object is kept

in air at a distance 40 cm from the mirror. Find the position of image w.r.t an observer near theobject. What is effect of separation between glass slab and the mirror on image.

Solution : Shifting in object due to glass slab x = d 1 1 20

1 20 1 cm1.5 3

Distance of object from mirror (as seen by mirror) 20 10040 cm

3 3

Image will be formed at a distance 100

3cm from mirror M.

Shifting in image due to glass slab = 203

cm

So distance of image from mirror = 100 20 80

3 3 3cm

Distance of image from the actual plane mirror is independent of separation b between glass slaband the mirror. If the distance is more then brightness of image will be less.

Example 28. An object is placed 21 cm infront of a concave mirror of radius of curvature 20 cm. A glass slabof thickness 3 cm and refractive index 1.5 is placed closed to the mirror in space between theobject and the mirror. Find the position of final image formed if distance of nearer surface of theslab from the mirror is 10 cm.

Solution : shift by slab x = d 1 1

1 3 1 1cm1.5

for image formed by mirror u = – (21 – 1) cm = – 20 cm.

1 1 1 1 1 1u v f 20 v 10

v = – 20 cm

shift in the direction of light v = – (20 + 1) = – 21 cm.


Example 29. A particle is dropped along the axis from a height f2

on a concave mirror of focal length f as shown in

figure. Find the maximum speed of image.

///////////////////////////////////////////

//////

t=0

h=f/2 g

Solution : 2 2IM OMv m v m gt

where 2 2

12 2 22 2

f f 2f 2f 4f gtm v gt

f u f gt f gtf gt f gtf2 2

For maximum speed IImax

dv f 30 t v 3fg

dt 3g 4

• Refraction through a composite slab (or refraction through a number of parallelmedia, as seen from a medium of R. I. n0)Apparent depth (distance of final image from final surface)

= rel1

1n

t +

rel2

2nt +

rel3

3nt

+......... + reln

n

nt

Apparent shift

= t1 rel1n11 + t2

rel2n11 +........+

relnnn1 tn

Where ' t ' represents thickness and ' n ' represents the R.I. of the respective media, relative to themedium of observer. (i.e. n1rel = n1/n0 , n2 rel = n2/n0 etc.)

OLVED XAMPLES EExample 30. See figure. Find the apparent

depth of object seen below

surface AB.

Solution : Dapp = d

=

8.12

20 +

8.15.1

15 = 18 + 18 = 36 cm.

ELF RACTICE ROBLEMSS P P9. Monochromatic light is refracted from air into the glass of refractive index . The ratio of the wavelength of

incident and refracted waves is [USS - JIPMER 2000; MP PMT 1996, 2003]

(1) 1 : (2) 1 : 2 (3) : 1 (4) 1 : 1

10. The index of refraction of diomond is 2.0, velocity of light in diamond in cm/second is approximately [USS - CPMT 1975; MNR 1987; UPSEAT 2000]

(1) 6×1010 (2) 3.0×1010 (3) 2×1010 (4) 1.5×1010

JEE (Adv.)-Physics Geometrical Optics11. A beam of light is converging towards a point on a screen. A plane glass plate whose thickness in the

direction of the beam = t, refractive index = , is introducted in the path of the beam. The convergence pointis shifted by

(1) away11t (2) away11t (3) nearer11t (4) nearer11t

12. When light travels from air to water and from water to glass again from glass to CO2 gas and finally through air.The relation between their refractive indices will be given by

(1) an × ngl × glngas × gasna = 1 (2) an × ngl × gasngl × glna = 1

(3) an × ngl × glngas = 1 (4) There is no such relation

13. When light enters from air to water, then its

(1) Frequency increases and speed decreases

(2) Frequency is same but the wavelength is smaller in water than in air

(3) Frequency is same but the wavelength in water is greater than in air

(4) Frequency decreases and wavelength is smaller in water than in air

14. A mark at the bottom of a liquid appears to rise by 0.1 m. The depth of the liquid is 1m. The refractive indexof the liquid is

(1) 1.33 (2) 109

(3) 910

(4) 1.5

Ans. 9. (3) 10. (4) 11. (1) 12. (1) 13. (2) 14. (3)

TOTAL INTERNAL REFLECTIONWhen light ray travel from denser to rarer medium it bend away from the normal if the angle of incident isincreased, angle of refraction will also increased. At a particular value of angle the refracted ray subtend 900

angle with the normal, this angle of incident is known as critical angle ( C). If angle of incident further increasethe ray come back in the same medium this phenomenon is known as total internal reflection.

CONDITIONS

• Angle of incident > critical angle [i > c]

• Light should travel from denser to rare medium Glass to air, water to air, Glass to water

Snell's Law at boundary xx', D sin C = R sin 90° RC

D

sin


Graph between angle of deviation ( ) and angle of incidence (i) as rays goes from denser to raremedium

• If i < c µDsini = µR sin r; 1 D

R

r sin sin i so 1 D

R

r i sin sin i i

• If i > c ; = – 2i

• A point object is situated at the bottom of tank filled with a liquid

of refractive index upto height h. It is found light from the sourcecome out of liquid surface through a circular portion above the object

2

C C 2 2 22 2 2 2

r 1 1 r 1 rsin & sin

r hr h r h

2 2 2 2 2 2 2r r h ( 1)r h radius of circular portion

2

hr

1 and area = r2

OLVED XAMPLES EExample 31. Find the radius of circle of illuminance, If a luminous object is placed at a distance h from the

interface in denser medium.

Solution : tan c = hr

.

µd

µr

1

2

3 TOI

circle of illuminance

r = h tan c.

But C = sin–1

rd /1

so, r = h tan rd

1

/1sin

= 2r

2d

r.h


SOME ILLUSTRATIONS OF TOTAL INTERNAL REFLECTION• Sparkling of diamond

The sparkling of diamond is due to total internal reflection inside it. As refractive index for diamond is 2.5 so

c = 24°. Now the cutting of diamond are such that i > C. So TIR will take place again and again inside it. Thelight which beams out from a few places in some specific directions makes it sparkle.

• Optical FibreIn it light through multiple total internal reflections is propagated along the axis of a glass fibre of radius of few microns in which index of refraction ofcore is greater than that of surroundings.

• Mirage and loomingMirage is caused by total internal reflection in deserts where due to heating of the earth, refractive index of airnear the surface of earth becomes lesser than above it. Light from distant objects reaches the surface of earthwith i > C so that TIR will take place and we see the image of an object along with the object as shown infigure.

Similar to 'mirage' in deserts, in polar regions 'looming' takes place due to TIR. Here decreases with heightand so the image of an object is formed in air if (i> C ) as shown in figure.

KEY POINTS• A diver in water at a depth d sees the world outside through a horizontal circle of radius. r = d tan c.• In case of total internal reflection, as all (i.e. 100%) incident light is reflected back into the same medium

there is no loss of intensity while in case of reflection from mirror or refraction from lenses there is some lossof intensity as all light can never be reflected or refracted. This is why images formed by TIR are much brighterthan formed by mirrors or lenses.

OLVED XAMPLES EExample 32. Find the max. angle that can be made in glass medium ( = 1.5) if a light ray is refracted

from glass to vacuum.Solution : 1.5 sin C = 1 sin 90º, where C = critical angle.

sin C = 2/3C = sin–1 2/3

Example 33. Find the angle of refraction in a medium (µ = 2) if light is incident in vacuum, making angleequal to twice the critical angle.

Solution : Since the incident light is in rarer medium. Total Internal Reflection can not take place.

C = sin–1 1

= 30º i = 2C = 60º

Applying Snell’s Law. 1 sin 60º = 2 sin r

sin r = 43

r = sin–1 43

.

JEE (Adv.)-Physics Geometrical OpticsExample 34. A rectangular block of glass is placed on a printed page laying on a horizontal surface. Find the

minimum value of the refractive index of glass for which the letters on the page are not visible fromany of the vertical faces of the block.

Solution : The situation is depicted in figure. Light will not emerge out from the vertical face BC if at it

i > C or sin i > sin C sin i > C

1 1as sin ... (i)

But from Snell's law at O 1 × sin = sin r

And in OPR, r + 90 + i = 180 r + i = 90° r = 90 – i

So sin = sin (90 – i) = cos i sin

cos i

so 2

2 sinsin i 1 cos i 1 ... (ii)

so substituting the value of sin i from equation (ii) in (i),

2

2

sin 11 i.e., 2 > 1 + sin2 (sin2 )max = 1 2 > 2 > 2 min= 2

Example 35. What should be the value of angle so that light enteringnormally through the surface AC of a prism (n=3/2) does not

cross the second refracting surface AB.

Solution : Light ray will pass the surface AC without bending since it is incident normally. Suppose itstrikes the surface AB at an angle of incidence i.

i = 90-

For the required condition:

90° – > C

or sin (90° ) > sinC

or cos > sinC = 231/ = 3

2or < cos-1

32

.

ELF RACTICE ROBLEMSS P P15. A cut diamond sparkles because of its [USS - NCERT 1974; RPET 1996; AFMC 2005]

(1) Hardness (2) High refractive index

(3) Emission of light by the diamond (4) Absorption of light by the diamond

16. Critical angle of light passing from glass to air is minimum for [USS - [NCERT 1975; RPMT 1999; MP PMT 2002]

(1) Red (2) Green (3) Yellow (4) Violet

JEE (Adv.)-Physics Geometrical Optics17. For total internal reflection to take place, the angle of incidence i and the refractive index of the medium

must satisfy the inequality [USS - MP PMT 1994]

(1) isin

1(2)

isin1

(3) sin i < (4) sin i >

18. Critical angle is that angle of incidence in the denser medium for which the angle of refraction in rarer mediumis [USS - MP PMT 1996]

(1) 0° (2) 57° (3) 90° (4) 180°

Ans. 15. (2) 16. (4) 17. (1) 18. (3)

CHARACTERISTICS OF A PRISM(a) A homogeneous solid transparent and refracting medium bounded by twoplane surfaces inclined at an angle is called a prism :

3-D view

DEVIATIONPQ = incident ray

QR = Refracted ray

RS = emergent ray

A = Prism angle

i1 = incident angle on face AB

i2 = emergent angle on face AC

r1 = refracted angle on face AB

r2 = incident angle on face AC

Angle of deviation on face AB. i1 – r1

Angle of deviation on face AC i2 – r2

Total angle of deviation

i1 – r1) + (i2 – r2) i1 + i2 – (r1 + r2) .....(i)

In QOR r1 + r2 + = 180° ...(ii)

In AQOR A + = 180° ...(iii)

from (ii) and (iii) r1 + r2 = A ...(iv)

from (i) and (iv) Total angle of deviation = i1 + i2 –A

from Snell's law at surface AB sin i1 = sin r1

and at surface AC sin r2 = sin i2

JEE (Adv.)-Physics Geometrical OpticsCONDITION OF MINIMUM DEVIATION

For minimum deviation

In this condition i1 = i2 = i r1 = r2 = r and since r1 + r2 = A r + r = A 2r = A

Ar

2

Minimum deviation min = 2i – A; minA Ai , r

2 2

if prism is placed in air 1; 1 × sin i = sin r

min

min

Asin

A A 2sin sin

A2 2 sin2

if angle of prism is small A < 10° then sin

min

min

AA2

A A2

A min = A min = ( –1)A

CONDITION FOR MAXIMUM DEVIATION/GRAZING EMERGENCE• Angle of incidence (ig)for grazing emergence

For ig, e = 90°

Applying Snell's law at face AC

µsinr2 = 1 × 1 sinr2 =1µ ; r2 =

1 1sin

µ = c

r1 + r2 = A r1 = A – c

Again, Applying Snell's law at face AB

1 × sin ig = µsinr1; 1 × sin ig = µsin(A – c)

sinig = µ[sinAcos c – cosAsin c]

1 2gi sin µ 1 sin A cos A

2

c c

µ 11as sin , cos

µ µ

If i increases beyond ig, r1 increases thus r2 decreases and becomes less than c and ray emerges.

Thus i ig ray emerges, otherwise TIR. max = ig + 90° – A


NO EMERGENCE CONDITIONLet maximum incident angle on the face AB imax = 90°

1 × sin 90° = sin r1; 1 C

1sinr sin ; r1 = C ...(i)

if TIR occur at face AC then r2 > C ...(ii)

r1 + r2 = A ...(iii)

from (i) and (ii) r1 + r2 > C + C r1 + r2 > 2 C ...(iv)

from (iii) and (iv) C C C

A A A 1 1A 2 sin sin sin

A2 2 2 sin2

OLVED XAMPLES EExample 36. Refracting angle of a prism A = 60º and its refractive index is, n = 3/2, what is the angle of

incidence i to get minimum deviation. Also find the minimum deviation. Assume the surroundingmedium to be air (n = 1).

Solution : For minimum deviation,

r1 = r2 = 2A

= 30º.

applying snell’s law at I surface

1 × sin i = 23

sin 30º i = 43sin 1

min = 2sin–1 43

– 3

Example 37. See the figure.

Find the deviation caused by a prism having refracting angle 4º and refractive index 23

.

Solution : = (23

– 1) × 40 = 20

Example 38. A ray of light passes through an equilateral prism such that angle of incidence is equal of emergenceand the later is equal to 3/4th of the angle of prism. Calculate the angle of deviation. Refractive indexof prism is 1.5.

Solution : A = 60°, = 1.5 ; i1 = i2 =34

A = 45°, = ?

A + = i1 + i2 60° + = 45° + 45° = 90° – 60° = 30°

JEE (Adv.)-Physics Geometrical OpticsExample 39. A prism of refractive index 1.53 is placed in water of refractive index 1.33. If the angle of prism is 60°,

calculate the angle of minimum deviation in water. (sin 35.1° = 0.575).

Solution : Here, ag = 1.33, a

w = 1.53, A = 60°, m = ? ma

gw wg ga

w

Asin1.53 21.15

A1.33 sin2

wmg

sin(A ) A 60sin 1.15sin 0.575

2 2 2 mA

2= sin–1 (0.575) = 35.1°

m = 35.1 × 2 – 60 = 10.2°

Example 40. Find r, r´, e, for the case shown in figure.

Solution : Here = 180° – 75o = 105°

sin 45o = 2 sin r

r = sin–1

21

= 30°.

r'r ' = 180° – (r + )

= 180° – 30° – 105° = 45o

sin e = 2 sin r '

sin e = 2 × sin 45o = 1

So, = i + e – A = 45o + 90o – 75o = 60o.

Example 41. From the graph of angle of deviation versus angle of incidence i, find the prism angle


Solution :

From the graph ;

= i + e – A.

30° = 30° + 60° – A

use the result : If i and e are interchanged then we get same value of

REFRACTION AT SPHERICAL SURFACESFor paraxial rays incident on a spherical surface separating two media:

vn2

un1 =

Rnn 12 .................... (A)

where light moves from the medium of refractive index n1 to the medium of refractive index n2.

Transverse magnification (m) (of dimension perpendicular to principal axis) due to refraction at

spherical surface is given by m = RuRv

= 1

2n/un/v

OLVED XAMPLES EExample 42. Find the position, size and nature of image, for the

situation shown in figure. Draw ray diagram .

Solution : For refraction near point A, u = – 30 ; R = – 20; n1 = 2 ; n2 = 1.

Applying refraction formula

vn2

un1 =

Rnn 12

v1

– 302

= 2021

v = – 60 cm

m = 1

2

hh

= unvn

2

1 = )30(1

)60(2 = 4 h2 = 4 mm.

JEE (Adv.)-Physics Geometrical OpticsExample : 43 An air bubble in glass ( = 1.5) is situated at a distance 3 cm from a spherical surface of diameter 10

cm as shown in Figure. At what distance from the surface will the bubble appear if the surface is (a)convex (b) concave.

Solution : In case of refraction from curved surface 2 1 2 1( )

v u R

(a) 1 = 1.5 , 2 = 1, R = – 5 cm and u = –3 cm 1 (1.5) 1 1.5v ( 3) ( 5) v = –2.5 cm

the bubble will appear at a distance 2.5 cm from the convex curved surface inside the glass.

(b) 1 = 1.5 , 2 = 1 , R = 5 cm and u = –3 cm 1 (1.5) 1 1.5v ( 3) (5) v 1.66 cm

the bubble will appear at a distance 1.66 cm from the concave curved surface inside theglass.

Note : If the surface is plane then R

case (a) or (b) would yield 1 (1.5) (1 1.5)v ( 3) v = – 2cm

Example 44. In a thin spherical fish bowl of radius 10 cm filled with water ofrefractive index (4/3), there is a small fish at a distance 4 cm fromthe centre C as shown in Figure. Where will the fish appear to be,

if seen from (a) E and (b) F (neglect the thickness of glass) ?

Solution : In the case of refraction from curved surface 2 1 2 1( )v u R

(a) Seen from E 1

43

, 2 = 1, R = – 10 cm & u = –(10 – 4) = –6 cm

4 411 903 3 v 5.3cm

v 6 10 17

i.e., fish will appear at a distance 5.3 cm from E towards F

(lesser than actual distance, i.e., 6 cm)


(b) Seen from F 1

43

, 2 = 1, R = – 10 cm and u= –(10 + 4) = –14 cm

4 411 3 3

v 14 10

210v 16.154

13 cm

so fish will appear at a distance 16.154 cm from F toward E

(more than actual distance, i.e., 14 cm)

Special case:Refraction at plane Surfaces

Putting R = in the formula v

n2 un1 =

Rnn 12 , we get;

v =1

2n

un

The same sign of v and u implies that the object and the image are always on the same sideof the interface separating the two media. If we write the above formula as

v = unrel

,

it gives the relation between the apparent depth and real depth, as we have seen before.

OLVED XAMPLES EExample 45. Using formula of spherical surface or otherwise, find the apparent depth of an object placed

10 cm below the water surface, if seen near normally from air.

Solution : Put R = in the formula of the Refraction at Spherical Surfaces we get,

v = 1

2

nun

u = – 10 cm

+ direction [ of incident light]direction

airwater

n1 = 34

n2 = 1

v = – 3/4110

= – 7.5 cm

negative sign implies that the image is formed in water.


Aliter:

dapp = rel

reald

= 3/410

= 4

30

= 7.5 cm.

THIN LENSA thin lens is called convex if it is thicker at the middle and it is called concave if it is thicker at theends.

One surface of a convex lens is always convex . Depending on the other surface a convex lens iscategorized as

(a) biconvex or convexo convex , if the other surface is also convex,

(b) Plano convex if the other surface is plane and

(c) Concavo convex if the other surface is concave.

Similarly concave lens is categorized as concavo-concave or biconcave, plano-concave and convexo-concave.

For a spherical, thin lens having the same

medium on both sides:

v1

u1

= (nrel 1) 21 R

1R1

.........(a),

where nrel = medium

lens

nn and R1 and R2 are x coordinates of the centre of curvature of the 1st surface

and 2nd surface respectively.

1v

1u

= 1f Lens Maker's Formula................(b)

Lens has two Focii:

If u = , then f11

v1

v = f

If incident rays are parallel to principal axis then its refracted ray will cut the principal axis at ‘f’.

It is called 2nd focus.

JEE (Adv.)-Physics Geometrical OpticsIn case of converging lens it is positive and in case of diverging lens it is negative.

If v = that meansf1

u11

u = – f

If incident rays cuts principal axis at – f then its refracted ray will become parallel to the principalaxis. It is called 1st focus. In case of converging lens it is negative ( f is positive) and in the caseof diverging lens it positive ( f is negative)

use of – f & + f is in drawing the ray diagrams.

Notice that the point B, its image B and the pole P of the lens are collinear. It is due to parallel slabnature of the lens at the middle. This ray goes straight. (Remember this)

From the relation 1f =(nrel 1)

21 R1

R1

it can be seen that the second focal length depends on

two factors.

(A) The factor 21 R

1R1

is

(i) Positive for all types of convex lenses and

(ii) Negative for all types of concave lenses.

(B) The factor (nrel 1) is

(i) Positive when surrounding medium is rarer than the medium of lens.(ii) Negative when surrounding medium is denser than the medium of lens.

(C) So a lens is converging if f is positive which happens when both the factors (A) and (B)are of same sign.

(D) And a lens is diverging if f is negative which happens when the factors (A) and (B) are ofopposite signs.

(E) Focal length of the lens depends on medium of lens as well as surrounding.

(F) It also depends on wavelength of incident light. Incapability of lens to focus light rays ofvarious wavelengths at single point is known as chromatic aberration.


OLVED XAMPLES EExample 46. Find the behavior of a concave lens placed in a rarer medium.

Solution : Factor (A) is negative , because the lens is concave.

Factor (B) is positive , because the lens is placed in a rarer medium.

Therefore the focal length of the lens, which depends on the product of these factors, isnegative and hence the lens will behave as diverging lens.

Example 47. Show that the factor 21 R

1R1

(and therefore focal length) does not depend on which surface

of the lens light strike first.

Solution : Consider a convex lens of radii of curvature p and q as shown.

CASE 1: Suppose light is incident from left side and strikes the surface with radius of curvature p, first.

Then R1 = +p ; R2 = -q and 21 R

1R1

= q1

p1

CASE 2: Suppose light is incident from right side and strikes the surface with radius of curvature q, first.

Then R1 = +q ; R2 = -p and 21 R

1R1

= p1

q1

Though we have shown the result for biconvex lens , it is true for every lens.

Example 48. Find the focal length of the lens

shown in the figure.

Solution :f1

= (nrel – 1) 21 R

1R1

f1

= (3/2 – 1) )10(1

101

f1

= 21

× 102

f = + 10 cm.

Example 49. Find the focal length of the

lens shown in figure

Solution :f1

= (nrel – 1) 21 R

1R1

= 123

101

101

f = – 10 cm


Example 50. Find the focal length of the

lens shown in figureROC = 60 cm

ROC = 20 cm

(a) If the light is incident from left side.

(b) If the light is incident from right side.

Solution : (a) f1

= (nrel – 1) 21 R

1R1

= 123

201

601

f = 60 cm

(b) f1

= (nrel – 1) 21 R

1R1

= 123

601

201

f = 60 cm

Example 51. Point object is placed on the principal axis of a thin lens with parallel curved boundaries i.e.,having same radii of curvature. Discuss about the position of the image formed .

Solution :1f = (nrel 1)

21 R1

R1

= 0 [ R1 = R2]

1v

1u

= 0 or v = u i.e. rays pass without appreciable bending.

Example 52. Focal length of a thin lens in air, is 10 cm. Now medium onone side of the lens is replaced by a medium of refractiveindex =2. The radius of curvature of surface of lens, in contact

with the medium, is 20 cm. Find the new focal length

Solution : Let radius of surface be R1 and refractive index of lens be .Let parallel rays be incident on the lens. Applying refractionformula at first surface

1V1

= 1R1

...(1)

At surfaceV2

– 1V = 20

2...(2)

Adding (1) and (2)

1V1

+ V2

– 1V =

1R1

+ 202

= ( – 1) 201

R1

1 – 20

1 – 20

2 =

f1

(in air) + 201

– 202

v = 40 cm f = 40 cm


Example 53. Figure shown a point object and a converging lens.

Find the final image formed.

Solution :v1

– u1

= f1

v1

– 151

= 101

v1

= 101

– 151

= 301

v = + 30 cm

Example 54. See the figure

Find the position of final image formed.

Solution : For converging lens

u = –15 cm, f = 10 cm v = uffu

= 30 cm

For diverging lens

u = 5 cm

f = –10 cm v = uffu

= 10 cm

Example 55. Figure shows two converging lenses. Incident rays are parallelto principal axis. What should be the value of d so that final

rays are also parallel.

Solution : Final rays should be parallel. For this the focus of L1

must coincide with focus of L2.

d = 10 + 20

= 30 cm

Here the diameter of ray beam becomes wider.


Transverse magnification (m)Transverse magnification (m) of (of dimension perpendicular to principal axis) is given by

Magnification 1

0

height of image hm

height of object hv f f vu f u f

If the lens is thick or/and the medium on both sides is different, then we have to apply the formulagiven for refraction at spherical surfaces step by step.

OLVED XAMPLES EExample 56. An extended real object of size 2 cm is placed perpendicular to the principal axis of a

converging lens of focal length 20 cm. The distance between the object and the lens is30cm.

(i) Find the lateral magnification produced by the lens.

(ii) Find the height of the image.

(iii) Find the change in lateral magnification, if the object is brought closer to the lens by 1mm along the principal axis.

Solution : Usingv1

– u1

= f1

and m = uv

we get m = uff

.........(A) m = )30(2020

= 1020

= – 2

–ve sign implies that the image is inverted.

(ii)1

2

hh

= m h2 = mh1 = (–2) (2) = – 4 cm

(iii) Differentiating (A) we get

dm = 2)uf(f

du = 2)10()20(

(0.1) = 1002

= –.02

Note that the method of differential is valid only when changes are small.

Alternate method :u (after displacing the object)

= –(30 + 0.1) = – 29.9 cm

Applying the formula

m = uff

; m = )9.29(2020

= – 2.02

change in ‘m’ = – 0.02.

Since in this method differential is not used, this method can be used for changes, small orlarge.


Displacement method to find focal length of converging lens :Fix an object of small height H and a screen at a distance D fromobject (as shown in figure). Move a converging lens from the objecttowards the screen. Let a sharp image forms on the screen when thedistance between the object and the lens is ‘a’.

From lens formula we have

aD1

– a1

= f1

or a2 – D a + f D = 0 ...(A)

This is quadratic equation and hence two values of ‘a’ are possible. Call them a1 and a2. Thus a, anda2 are the roots of the equation. From the properties of roots of a quadratic equation,

a1 + a2 = D a1a2 = f D

Also (a1 – a2) = 212

21 aa4)aa( = fD4D2 = d (suppose).

‘d’ physically means the separation between the two position of lens.

The focal length of lens in terms of D and d.

so, a1 – a2 = 212

21 aa4)aa(

fD4D2 = d

condition, d = 0, i.e. the two position coincide

= D4

D2

Roots of the equation a2 – Da + D = 0, become imaginary if

b2 – 4ac < 0.

= D2 – 4 D < 0 = D (D – 4 ) < 0 =

for real value of a in equation a2 – Da + f D = 0

b2 – 4ac 0. = D2 – 4f D 0.

so, D 4f Dmin = 4f

Lateral magnification in displacement method:if m1 and m2 be two magnifications in two positions (In the displacement method)

m1 = 1

1

uv

= 1

1

a)aD(

m2 = 2

2

uv

= 2

2

aaD

= )aD(a

1

1


So m1 m2 = 1

1

aaD

× )aD(a

1

1 = 1.

If image length are h1 and h2 in the two cases,

then m1 = –Hh1 m2 = –

Hh2 m1 m2 = 1

221

Hhh

= 1 h1h2 = H2 H = 21 hh

ELF RACTICE ROBLEMSS P P19. The radius of curvature for a convex lens is 40 cm, for each surface. Its refractive index is 1.5. The focal length

will be [USS - MP PMT 1989]

(1) 40 cm (2) 20 cm (3) 80 cm (4) 30 cm

20. A thin lens focal length f1 and its aperture has diameter d . It forms an image of intensity I. Now the central partof the aperture upto diameter d/2 is blocked by an opaque paper. The focal length and image intensity willchange to [USS - CPMT 1989; MP PET 1997; KCET 1998]

(1) 2

and2f

(2) 4

andf (3) 2

and4f3

(4) 43andf

21. A lens of power +2 diopters is placed in contact with a lens of power –1 diopter. The combination will behavelike [USS - MP PET/PMT 1988]

(1) A convergent lens of focal length 50 cm (2) A divergent lens of focal length 100 cm

(3) A convergent lens of focal length 100 cm (4) A convergent lens of focal length 200 cm

22. If in a plano - canvex lens, the radius of curvature of the convex surface is 10 cm and the focal length of thelens is 30 cm, then the refractive index of the material of lens will be

[USS - CPMT 1986; MNR 1988; MP PMT 2002; UPSEAT 2000]

(1) 1.5 (2) 1.66 (3) 1.33 (4) 3

23. The silt of a collimator is illuminated by a source as shown in the adjoining figures. The distance between thesilt S and the collimating lens L is equal to the focal length of the lens. The correct direction of the emergentbeam will be as shown in figure [USS - CPMT 1986]

(1) 1 (2) 3 (3) 2 (4) None of the figures

JEE (Adv.)-Physics Geometrical Optics24. A convex lens makes a real image 4 cm long on a screen. When the lens is shifted to a new position without

disturbing the object, we again get a real image on the screen which is 16 cm tall. The length of the objectmust be USS - [MP PET 1991]

(1) 1/4 cm (2) 8 cm (3) 12 cm (4) 20 cm

25. A thin convex lens of refractive index 1.5 has a focal length of 15 cm in air. When the lens is placed in liquidof refractive index 4/3, its focal length will be USS - [CPMT 1974; 77; MP PMT 1992]

(1) 15 cm (2) 10 cm (3) 30 cm (4) 60 cm

Ans. 19. (1) 20. (4) 21. (3) 22. (3) 23. (3) 24. (2)

25. (4)

IMAGE FORMATION BY A LENSPosition of Object Details of Image Figure

At infinity Real, Inverted Diminished (m <<–1) At F

Between and 2F Real inverted Diminished

(m < –1) between F and 2F

At 2F Real inverted equal m = –1 At 2F

Between 2F and F Real, inverted Enlarged

(m > – 1) between 2F and

At F Real inverted Enlarged

(m >> –1) At inifinity

Between Focus & Virtual, erect enlarged

Pole (m > +1) between

and Object on same side

JEE (Adv.)-Physics Geometrical Optics(b) For real extended object, if the image formed by a single lens in inverted (i.e. m is negative) it is always

real and the lens is convergent i.e., convex. In this situation if the size of image is -

Smaller than object Equal to object Larger than object

Object between and 2F Object is at 2F Object between 2F and F

Image is between F and 2F Image is at 2F Image is between 2F and

OLVED XAMPLES EExample 57. In the displacement method the distance between the object and the screen is 70 cm and the focal

length of the lens is 16 cm, find the separations of the magnified and diminished image position of thelens.

Solution : 2 2d D 4fd (70) 4 16 70 420 20.5cm

Example 58. An object 25 cm high is placed in front of a convex lens of focal length 30 cm. If the height of imageformed is 50 cm, find the distance between the object and the image (real and virtual) ?

Solution : As object is in front of the lens, it is real and as h1 = 25 cm, f = 30 cm, h2 = – 50 cm ;

2

1

h 50m 2h 25

f 30m 2f u 30 u u = – 45 cm

v vm 2

u 45 v = 90 cm

As in this situation object and image are on opposite sides of lens, the distance between object and imaged1 = u + v = 45 + 90 = 135 cm. If the image is erect (i.e., virtual)

f 30m 2

f u 30 uu = – 15 cm

v vm 2

u 15v = 30 cm

As in the situation both image and object are in front of the lens, the distance between object and image

d2 = v – u = 30 – 15 = 15 cm.


COMBINATION OF LENSESThe equivalent focal length of thin lenses in contact is given by

1 1 1 1

1 2 3F f f f...

where f`1 , f2 , f3 are focal lengths of individual lenses.

If two converging lenses are separated by a distance d and the incident light rays are parallel to thecommon principal axis ,then the combination behaves like a single lens of focal length given by therelation

1 1 1

1 2 1 2F f fd

f f

and the position of equivalent lens is 1fFd

with respect to 2nd lens

(iii) If an equiconvex lens of focal length f is cut into equal parts by a horizontal

plane AB then the focal length of each part will be equal to that of initiallens. Because , R1 and R2 will remain unchanged. Only intensity will bereduced.

intensity I (apertures)2

intensity through a single part will be reduced

(iv) If the same lens is cut into equal parts by a vertical plane CD the focal length

of each part will be double of initial value but intensity will remain unchanged.

For equiconvex lens 1 ( 1)2f R

For plano convex lens 1

1 1f R

So 1

1 2f f f1 = 2f Focal length of each part = 2

(focal length of original lens)

(v) If a lens is made of number of layers of different refractive index for a given

wavelength then no. of images is equal to number of refractive index,

as 1

( 1)f

.

In figure number of images = 2

POWER OF LENSReciprocal of focal length in meter is known as power of lens.

SI UNIT : dioptre (D) Power of lens : 1 100

Pf(m) f(cm) dioptre [in air]


OLVED XAMPLES EExample 59. Find the lateral magnification produced by the

combination of lenses shown in the figure.

Solution :f1

= 1f1

+ 2f1

= 101

– 201

= 201

f = + 20

v1

– 101

= 201

v1

= 201

– 101

= 201

= – 20 cm m = 1020

= 2

Example 60. Find the focal length of equivalent system.

Solution :1f1

= 123

101

101

= 21

× 102

= 101

2f1

= 156

201

101

= 51

× 201030

= 1003

3f1

= 158

201

201

= 503

f1

= 1f1

+ 2f1

+ 3f1

= 101

+ 1003

+ 503

f = 13100

Ans.

ELF RACTICE ROBLEMSS P P26. Two similar plano-convex lenses are combined together in three different

ways as shown in the adjoining figure. The ratio of the focal lengths in three

cases will be

(1) 2 : 2 : 1 (2) 1 : 1 : 1

(3) 1 : 2 : 2 (4) 2 : 1 : 1

27. A concave and convex lens have the same focal length of 20 cm and are put into contact to form a lenscombination. The combination is used to view an object of 5 cm length kept at 20 cm from the lens combination.As compared to the object, the image will be USS - [CPMT 1986; RPMT 1997]

(1) Magnified and inverted (2) Reduced and erect

(3) Of the same size as the object and erect (4) Of the same size as the object but inverted

Ans. 26. (2) 27. (3)


COMBINATION OF LENS AND MIRRORThe combination of lens and mirror behaves like a

mirror of focal length ‘f’ given by

f1

= mF1

– F2

If lenses are more then one, ‘f’ is given by

f1

= mF1

– f12

For the following figure

‘f’ is given byf1

= mF1

– 21 f1

f12

OLVED XAMPLES EExample 61. Find the position of final image formed.

(The gap shown in figure is of negligible width )

Solution:eqf1

= 101

– 102

= 101

feq = – 10 cm

v1

+ 201

= 101

v = – 20 cm

Hence image will be formed on the object itself

Example : 62 See the figure

Find the equivalent focal length of the

combination shown in the figure and

position of image.

Solution : For the woreme lens

f1

= (n – 1) 21 R

1R1

= 123

101

101

= 21

× 102

= 101

Ago, Fm = R2

= 2

10 = 5 cm

eqf1

= mF1

– 2 f1

= 51

+ 2 × 101

= 52

Ans.


ELF RACTICE ROBLEMSS P P28. A plano - convex lens of focal length 20 cm silvered at the plane surface will behave as convergent mirror of

focal length. GRB - Pramod Agarwal Q.54 [Manipal 2006; Orissa (JEE) 2006]

(1) 20 cm (2) 30 cm (3) 40 cm (4) 10 cm

Ans. 28. (4)

DISPERSION OF LIGHTThe angular splitting of a ray of white light into a number of components and spreading in differentdirections is called Dispersion of Light. [It is for whole Electro Magnetic Wave in totality]. Thisphenomenon is because waves of different wavelength move with same speed in vacuum but withdifferent speeds in a medium.

Therefore, the refractive index of a medium depends slightly on wavelength also. This variation ofrefractive index with wavelength is given by Cauchy’s formula.

Cauchy's formula n ( ) = 2ba where a and b are positive constants of a medium.

Note :Such phenomenon is not exhibited by sound waves.

Angle between the rays of the extreme colours in the refracted (dispersed) light is calledangle of dispersion. = v

r (Fig. (a))

Fig (a) and (c) represents dispersion, whereas in fig. (b) there is no dispersion.

For prism of small ‘A’ and with small ‘i’ :

= v – r = (nv – nr)A

Deviation of beam (also called mean deviation)

= y = (ny – 1)A

nv, nr and ny are R. . of material for violet, red and yellow colours respectively.

OLVED XAMPLES EExample 63. The refractive indices of flint glass for red and violet light are 1.613 and 1.632 respectively.

Find the angular dispersion produced by a thin prism of flint glass having refracting angle 50.

Solution : Deviation of the red light is r = ( r – 1)A and deviation of the violet light is v = ( v – 1)A.

The dispersion = v – r = ( v – r)A = (1.632 – 1.613) × 50 = 0.0950 .

JEE (Adv.)-Physics Geometrical OpticsNote :

Numerical data reveals that if the average value of is small v – r is also small and if the averagevalue of is large v – r is also large. Thus, larger the mean deviation, larger will be the angulardispersion.

Dispersive power ( ) of the medium of the material of prism is given by: = 1nnn

y

rv

is the property of a medium.

For small angled prism ( A 10o ) with light incident at small angle i :

1nnn

y

rv =

y

rv = y

= )yellow(raymeanofdeviationdispersionangular

[ ny = 2

nn rv if ny is not given in the problem ]

n 1 = refractivity of the medium for the corresponding colour.

Example 64. Refractive index of glass for red and violet colours are 1.50 and 1.60 respectively. Find

(a) the ref. index for yellow colour, approximately

(b) Dispersive power of the medium.

Solution : (a) r ~ 2

Rv = 2

60.150.1 = 1.55

(b) = 1r

Rv = 155.1

50.160.1 = 0.18.

COMBINATION OF PRISMDeviation without dispersion ( = 0°)Two or more than two thin prism are combined in such a way that deviation occurs i.e. emergent light raymakes angle with incident light ray but dispersion does not occur i.e., light is not splitted into seven colours.

Total dispersion = = = ( V – R)A + ( 'V – 'R)A'

For no dispersion = 0 ; ( V – R)A + ( 'V – 'R)A' = 0

Therefore, V R

' 'V R

( )AA '

–ve sign indicates that prism angles are in opposite direction.

Dispersion without deviation Two or more than two prisms combine in such a way that dispersion occurs i.e., light is splitted into sevencolours but deviation do not occur i.e., emergent light ray becomes parallel to incident light ray.

Total deviation


; – A + ( '–1)A' = 0 ( 1)A

A '' 1

–ve sign indicates that prism angles are in opposite direction.

KEY POINTS• Dispersive power like refractive index has no units and dimensions and depends on the material of the prism

and is always positive.

• As for a given prism dispersive power is constant, i.e., dispersion of different wavelengths will be different andwill be maximum for violet and minimum for red (as deviation is maximum for violet and minimum for red).

• As for a given prism a single prism produces both deviation and dispersion simultaneously, i.e., asingle prism cannot give deviation without dispersion or dispersion without deviation.

OLVED XAMPLES EExample : 65 White light is passed through a prism of angle 5°. If the refractive indices for red and blue colours are

1.641 and 1.659 respectively, calculate the angle of dispersion between them.

Solution : As for small angle of prism = ( – 1)A,

B = ( 1.659 – 1) × 5° = 3.295° and R = (1.641 – 1) × 5° =3.205°

so = B – R = 3.295° – 3.205° = 0.090°

Example : 66 Prism angle of a prism is 10o. Their refractive index for red and violet color is 1.51 and 1.52 respectively.

Then find the dispersive power.

Solution : Dispersive power of prism v r

y 1 but v ry

1.52 1.511.515

2 2

Therefore 1.52 1.51 0.01

0.0191.515 1 1.515

Example : 67 The refractive indices of flint glass for red and violet colours are 1.644 and 1.664. Calculate itsdispersive power.

Solution : Here, r = 1.644, v = 1.664, = ?

Now v ry

1.664 1.6441.654

2 2 v r

y

1.664 1.6440.0305

1 1.654 1

Example : 68 In a certain spectrum produced by a glass prism of dispersive power 0.031, it was found that

r = 1.645 and v = 1.665. What is the refractive index for yellow colour ?

Solution : Here, = 0.031, r = 1.645, v = 1.665, y = ?

v r

y 1 v ry

1.665 1.645 0.0201 0.645

0.031 0.31 y = 0.645 + 1 = 1.645

JEE (Adv.)-Physics Geometrical OpticsExample : 69 A combination of two prisms, one of flint and other of crown glass produces dispersion without

deviation. The angle of flint glass prism is 15°. Calculate the angle of crown glass prism and angulardispersion of red and violet. ( for crown glass = 1.52, for flint glass = 1.65, for crown glass 0.20,

for flint glass = 0.03).

Solution : Here, A = 15°, A' = ?, = 0.03, ' = 0.02, = 1.65, ' = 1.52,

For no deviation, + ' = 0

( – 1)A + ( ' – 1)A' = 0 (1.65 – 1)15° + (1.52 –1)A' = 0 A' = 0.65 15

0.52= –18.75°

Negative sign indicates that two prisms must be joined in opposition. Net angular dispersion

( v – r)A + ( 'v – 'r)A' = ( – 1)A + ' ( ' –1)A' = 0.03 (1.65 – 1)15° + 0.02 (1.52 – 1) (–18.75°)

= 0.2925 – 0.195 = 0.0975°

Example : 70 If two prisms are combined, as shown in figure,

find the total angular dispersion and angle of

deviation suffered by a white ray of light incident

on the combination.

Solution : Both prisms will turn the light rays towards their bases and hence in same direction.Therefore turnings caused by both prisms are additive.

Total angular dispersion

= + ’ = ( V – R) A + ( ’V – ’R) A’

= (1.5 – 1.4) 4º + (1.7 – 1.5)2º = 0.8°

Total deviation

= + ´

= 12

RV A + 12

'' RV A’A’ = 12

4.15.1 0.4º + 1

25.17.1

0.2º

= (1.45 – 1) 0.4º + (1.6 – 1) 0.2º

= 0.45 × 0.4º + 0.6 × 0.2º

= 1.80 + 1.2 = 3.0º Ans.

Example : 71 Two thin prisms are combined to form an achromatic combination. For I prism A = 4º,

R = 1.35, Y = 1.40, v = 1.42. for II prism ’R= 1.7, ’Y = 1.8 and ’R = 1.9 find the prismangle of II prism and the net mean deviation.

Solution : Condition for achromatic combination.

= ’

( V – R)A = ( ’V – ’R)A’

A’ = 4.17.19.1

º4)35.142.1(

Net = ~ ’ = ( Y – 1)A ~ ( ’Y – 1)A’ = (1.40 – 1) 4º ~ (1.8 – 1) 1.4° = 0.48º.

JEE (Adv.)-Physics Geometrical OpticsExample : 72 A crown glass prism of angle 50 is to be combined with a flint prism in such a way that the

mean ray passes undeviated. Find (a) the angle of the flint glass prism needed and (b) theangular dispersion produced by the combination when white light goes through it. Refractiveindices for red, yellow and violet light are 1.5, 1.6 and 1.7 respectively for crown glass and1.8,2.0 and 2.2 for flint glass.

Solution : The deviation produced by the crown prism is

= ( – 1)A

and by the flint prism is :

' = ( ' – 1)A'.

The prisms are placed with their angles inverted with respect to each other. The deviations are alsoin opposite directions. Thus, the net deviation is :

D = – ' = ( – 1)A – ( ' – 1)A'. .................(1)

(a) If the net deviation for the mean ray is zero,

( – 1)A = ( ' – 1)A'.

or, A' = )1'()1(

A = 05102161

.

.=30

(b) The angular dispersion produced by the crown prism is :

v – r = ( v – r)A

and that by the flint prism is,

'v – 'r = ( 'v – 'r)A

The net angular dispersion is,

( v – r)A – ( 'v – 'r)A

= (1.7 – 1.5) × 50 – (2.2 – 1.8) × 30

= – 0.20 .

The angular dispersion has magnitude 0.20 .

ELF RACTICE ROBLEMSS P P29. Formula for dispersive power is (where symbols have their usual meanings)

or

If the refractive indices of crown glass for red, yellow and violet colours are respectively r , y , and ,

then the dispersive power of this glass would be

(A) 1r

y(B) 1y

r(C)

ry

y(D) 1

yr

30. Angle of minimum deviation for a prism of refractive index 1.5 is equal to the angle of prism. The angle of prismis (cos 41° = 0.75)

(A) 62° (B) 41° (C) 82° (D) 31°

JEE (Adv.)-Physics Geometrical Optics31. In the formation of primary rainbow, the sunlight rays emerge at minimum deviation from rain-drop after

(A) One internal reflection and one refraction

(B) One internal reflections and two refractions

(C) Two internal reflections and one refraction

(D) Two internal reflections and two refractions

32. Dispersive power depends upon(A) The shape of prism (B) Material of prism (C) Angle of prism (D) Height of the prism

Ans. 29. (B) 30. (C) 31. (B) 32. (B)

CHROMATIC ABERRATION

The image of a object in white light formed by a lens is usually colored and blurred. This defect of image is calledchromatic aberration and arises due to the fact that focallength of a lens is different for different colors. For a

single lens 1 2

1 1 11

f R R and as of lens is

maximum for violet while minimum for red, violet isfocused nearest to the lens while red farthest from it. Itis defect of lens.Longitudinal or Axial Chromatic AberrationWhen a white object O is situated on the axis of a lens, then images of different colors are formed at differentpoints along the axis. The formation of images of different colors at different positions is called 'axial' orlongitudinal chromatic aberration. The axial distance between the red and the violet images IR – IV is known aslongitudinal aberration. When white light is incident on lens, image is obtained at different point on the axisbecause focal length of lens depend on wavelength. f fR > fV

fR – fV = fy Axial or longitudinal chromatic aberrationIf the object is at infinity, then the longitudinal chromatic aberration is equal to the difference infocal–lengths (fRfV ) for the red and the violet rays.LATERAL CHROMATIC ABERRATIONAs the focal–length of the lens varies from color

to color, the magnification m = f

u f produced

by the lens also varies from color to color.Therefore, for a finite–size white object AB, theimages of different colors formed by the lens areof different sizes.The formation of images of different colors indifferent sizes is called lateral chromaticaberration. The difference in the height of the red

image BR AR and the violet image BV AV is known

as lateral chromatic aberration. LCA = hR – hV

JEE (Adv.)-Physics Geometrical OpticsACHROMATISM

If two or more lens combined together in such a way that this combination produce image at a same pointthen this combination is known as achromatic combination of lenses.

1 2 1

y y 1 2 2

f'+ =0 0

f f' f f f1

2

For combination of lens. 1 2

1 1 1F f f (Apply sign convention in numerical)

STRUCTURE OF EYE :Light enters the eye through a curved front surface, the corner. It passes through the pupil which is thecentral hole in the iris. The size of the pupil can change under control of muscles. The light is further focussedby the eye-lens on the retina. The retina is a film of nerve fibres covering the curved back surface of theeye. The retina contains rods and cones which sense light intensity and colour, respectively, and transmitelectrical signals via the optic nerve to the brain which finally processes this information. The shape (curvature)and therefore the focal length of the lens can be modified somewhat by the ciliary muscles. For example,when the muscle is released, the focal length is about 2.5 cm and (for a normal eye) objects at infinityare in sharp focus on the retinas. When the object is brought closer to the eye, in order to maintain thesame image-lens distance ( 2.5 cm), the focal length of the eye-lens becomes shorter by the actionof the ciliary muscles. This property of the eye in called accommodation. If the object is too close to theeye, the lens cannot curve enough to focus the image on to the retina, and the image is blurred.

The closest distance for which the lens can focus light on the retina is called the least distance of distinctvision, or the near point. The standard value (for normal vision) taken here is 25 cm. (Often the near pointis given the symbol D.)

DEFECTS OF VISIONRegarding eye it is nothing that:

(1) In eye convex eye-lens forms real inverted and diminished image at the retina by changing its convexity(the distance between eye lens and retina is fixed)

(2) The human eye is most sensitive to yellow green light having wavelength 5550 A0 and least to violet(4000 A0) and red (7000 A0)

(3) The size of an object as perceived by eye depends on its visual-angle when object is distant its visualangle and hence image I1 at retina is small (it will appear small) and as it is brought near to the

eye its visual angle 0 and hence size of image I2 will increase.


(A) (B)(4) The far and near point for normal eye are usually taken to be infinity and 25 cm respectively ie., normal

eye can see very distant object clearly but near objects only if they are at distance greater than 25cm from the eye. The ability of eye to see objects from infinite distance to 25 cm from it is calledPower of accommodation.

(5) If object is at infinity i.e., parallel beam of light enters the eye is least strained and said to be relaxedor unstrained. However, if the object is at least distance of distint vision (L.D.D.V] i.e., D (=25 cm)eye is under maximum strain and visual angle is maximum.

(6) The limit of resolution of eye is one minute ie., two object will not be visible distintely to the eye ifthe angle substanded by them on the eye is lesser than one minute.

(7) The persistance of vision is (1/10) sec i.e., If time interval between two consecutive light pulses is lesserthan 0.1 sec eye cannot distinguish them separately. This fact is taken into account in motion pictures.

In case of eye following are the common defects of vision.

DEFECTS OF EYESMYOPIA [or Short–sightedness or Near– sightedness]

(i) Distant object are not clearly visible, but near object are clearly visible because image is formed before theretina.

(ii) To remove the defect concave lens is used.

The maximum distance. Which a person can

see without help of spectacles is known as

far point.

JEE (Adv.)-Physics Geometrical OpticsIf the reference of object is not given then it is

taken as infinity.

In this case image of the object is formed at the far point of person.

1 1 1 1 1 1P P

v u f distance of far point (inm) distance of object (inm) f

100 100

Pdistance of far point (in cm) distance of object (incm)

LONG–SIGHTEDNESS OR HYPERMETROPIA

(i) Near object are not clearly visible but far object are clearly visible.

(ii) The image of near object is formed behind the retina.

(iii) To remove this defect convex lens is used.

Near Point :–The minimum distance which a person can see without help of spectacles.

• In this case image of the object is formed at the near point.

• If reference of object is not given it is taken as 25 cm.

1 1 1 1 1 1P P

v u f distance of near point (inm) distance of object (inm) f

distance of near point = –ve, distance of object = –ve, P = +ve

PRESBYOPIAIn this case both near and far object are not clearly visible. To remove this defect two separate spectacles onefor myopia and other for hypermetropia are used or bifocal lenses are used.

JEE (Adv.)-Physics Geometrical OpticsASTIGMATISMIn this defect eye can not see object in two orthogonal direction clearly. It can be removed by using cylindricallens in particular direction.

OLVED XAMPLES EExample : 73 A person can not see clearly an object kept at a distance beyond of 100 cm. Find the nature and the

power of lens to be used for seeing clearly the object at infinity.

Solution : For lens u = – and and v = – 100 cm

1 1 11 1f v 100cm concave

v u f v f Power of lens

1 1P 1D

f 1

Example : 74 A far sighted person has a near point of 60 cm. What power lens should be used for eye glassessuch that the person can read this book at a distance of 25 cm.

Solution : Here v = – 60 cm, u = –25 cm

1 1 1 1 1 300f cm

f v u 60 25 7

1 1Power = 2.33D

f in m 3/7

ELF RACTICE ROBLEMSS P P33. For a normal eye , the least distance of distinct vision is

(1) 0.25 m (2) 0.50 m (3) 25 m (4) Infinite

34. For the myopic eye, the defect is cured by

(1) Convex lens (2) Concave lens (3) Cylindrical lens (4) Toric lens

35. Lens used to remove long sightedness (hypermetropia ) is

or

A person suffering from hypermetropia requires which type of spectacle lenses

(1) Concave lens (2) Plano- concave lens

(3) Convexo- concave lens (4) Convex lens

36. Image formed on the retina is

(1) Real and inverted (2) Virtual and erect (3) Real and erect (4) Virtual and inverted

Answers:

33. (1) 34. (2) 35. (4) 36. (1)


OPTICAL INSTRUMENTS

SIMPLE MICROSCOPEWhen object is placed between focus and optical centre a virtual, magnified and erect image is formed

Magnifying power (MP) = visual angle with instrument ( )

maximum visual angle for unaided eye ( )

hDuMP

h uD

(i) When the image is formed at infinity :

by lens equation 1 1 1 1 1 1

u fv u f u f

So MPDu

Df

(ii) If the image is at minimum distance of clear vision D :

1 1 1 1 1 1D u f u D f

[v = –D and u = –ve]

Multiplying by D both the sides D D D D

1 MP 1u f u f

.

OLVED XAMPLES EExample : 75 A man with normal near point 25 cm reads a book with small print using a magnifying glass, a thin

convex lens of focal length 5 cm.

(a) What is the closest and farthest distance at which he can read the book when viewing through themagnifying glass ?

(b) What is the maximum and minimum MP possible using the above simple microscope ?

Solution : (a) As for normal eye far and near point are and 25 cm respectively, so for magnifier vmax= – and

vmin = –25 cm. However, for a lens as 1 1 1 f

uv u f / v 1


So u will be minimum when v = minimum = –25 cm i.e. min5 25

u 4.17cm5 / 25 1 6

Ans u will be maximum when v = maximum = i.e., umax = 5

51

= – 5 cm

So the closest and farthest distance of the book from the magnifier (or eye) for clear viewing

are 4.17 cm and 5 cm respectively.

(b) As in case of simple magnifier MP = (D/u). So MP will be minimum when u = max = 5 cm

min25 D

MP 55 and MP will be maximum when u = min = (25/6) cm

max25 D

MP 6 125/6

COMPOUND MICROSCOPE

Compound microscope is used to get more magnified image. Object is placed infront of objective lens andimage is seen through eye piece. The aperture of objective lens is less as compare to eye piece becauseobject is very near so collection of more light is not required. Generally object is placed between F – 2F dueto this a real inverted and magnified image is formed between 2F – . It is known as intermediate image A'B'.The intermediate image act as a object for eye piece. Now the distance between both the lens are adjustedin such a way that intermediate image falls between the optical centre of eye piece and its focus. In thiscondition, the final image is virtual, inverted and magnified.

Total magnifying power = Linear magnification × angular magnification MP = m0me = 0

0 e

v Du u

(i) When final image is formed at minimum, distance of distinct vision.

0 0 0 0 2

0 e 0 0 e 0 e 1 e

v f f v hD D D DMP 1 1 1 1

u f f u f f f h f

Length of the tube = v0 + |ue|


(ii) When final image is formed at infinity e ee e e e e

1 1 1 1 1 1u f

v u f u f

0 0 0 0 2

0 e 0 0 e 0 e 1 e

v f f v hD D D DMP

u f f u f f f h f. Length of the tube L = v0 + fe

Sign convention for solving numerical u0 = –ve, v0 = +ve, f0 = +ve,

ue = –ve, ve = –ve, fe = +ve, m0 = –ve, me = +ve, M = –ve.

Example 76. A thin convex lens of focal length 5 cm is used as a simple microscope by a person with normal nearpoint (25 cm). What is the magnifying power of the microscope ?

Solution : Here, f = 5 cm; D = 25 cm, M = ? D 25

MP 1 1 6f 5

Example 77. A compound microscope consists of an objective lens of focal length 2.0 cm and an eye piece offocal length 6.25 cm, separated by a distance of 15 cm. How far from the objective should an objectbe placed in order to obtain the final image at (a) the least distance of distinct vision (25 cm) (b)infinity?

Solution : Here, f0 = 2.0 cm; fe = 6.25 cm, u0 = ?

(a) ve = – 25 cm e e e

1 1 1v u f e e e

1 1 1 1 1 1 4 5u v f 25 6.25 25 25 ue = – 5 cm

As distance between objective and eye piece = 15 cm; v0 = 15 – 5 = 10 cm

00 0 0 0 0 0

1 1 1 1 1 1 1 1 1 5 10u 2.5 cm

v u f u v f 10 2 10 4

Magnifying power = 0

0 e

v D 10 251 1 20

|u | f 2.5 6.25

(b) ve = , ue = fe = 6.25 cm v0 = 15 – 6.25 = 8.75 cm.

00 0 0 0 0 0

1 1 1 1 1 1 1 1 2 8.75 17.5u 2.59cm

v u f u v f 8.75 20 17.5 6.75

Magnifying power = 0 0

0 e 0 e

v vD D 8.75 251 13.51

|u | f u u 2.59 6.25

ELF RACTICE ROBLEMSS P P1. The focal lengths of the objective and eye lens of a microscope are 1 cm and 5cm respectively If the magnifying

power for the relaxed eye is 45, then the length of the tube is

(1) 30 cm (2) 25 cm (3) 15cm (4) 12 cm

JEE (Adv.)-Physics Geometrical Optics2. In a compound microsocope magnification will be large ,if the focal length of the eye piece is:

(1) Large (2) Smaller

(3) Equal to that of objective (4) Less than that of objective

3. Magnifying power of a simple microscope is ( when final image is formed at D= 25 cm from eye)

(1) fD

(2) fD1 (3)

Df1 (4)

fD1

4. If in compound microscope m1 and m2 be the linear magnification of the objective lens and eye lens respectivelythen magnifying power of the compound microscope will be

(1) m1 – m2 (2) 21 mm (3) ( m1 + m2) /2 (4) m1 ×m2

Answers: 1. (3) 2. (2) 3. (2) 4. (4)

ASTRONOMICAL TELESCOPE

A telescope is used to see distant object, objective lens forms the image A'B' at its focus. This image A'B'acts as a object for eyepiece and it forms final image A"B".

0 0

e

e

h 'f fvisual angle with instrument ( )

MP MP A B hh 'visual angle for unaided eye ( ) uu

(i) If the final image is at infinity ve = – , ue = –ve

e ee e

1 1 1u f

u f. So 0

e

fMP

f and length of the tube L = f0 + fe

(ii) If the final image is at D : ve = –D ue = –ve

e

e e e e e

f1 1 1 1 1 1 11

D u f u f D f D So 0 0 e

e e

f f fMP 1

u f D

Length of the tube is L f ue0


S. No.

Compound – Microscope S. No.

Astronomical – Telescope

1. It is used to increase visual angle of near tiny object.

1. It is used to increase visual angle of distant large objects.

2. In it field and eye lens both are convergent, of short focal length and aperture.

2. In it objective lens is of large focal length and aperture while eye lens of short focal length and aperture and both are convergent.

3. Final image is inverted, virtual and enlarged and at a distance D to from the eye.

3. Final image is inverted, virtual and enlarged at a distance D to from the eye.

4. MP does not change appreciably if objective and eye lens are interchanged as [MP ~ (LD / f0 fe)]

4. MP becomes (1/m2) times of its initial value if objective and eye–lenses are interchanged as MP ~ [f0 / fe]

5. MP is increased by decreasing the focal length of both the lenses.

5. MP is increased by increasing the focal length of objective lens and by decreasing the focal length of eyepiece

6. RP is increased by decreasing the wavelength of light used.

2 sinRP

6. RP is increased by increasing the aperture of objective.

DRP

1.22

OLVED XAMPLES E

Example : 78 A small telescope has an objective lens of focal length 144 cm and an eyepiece of focal length 6.0cm. What is the magnifying power of the telescope ? What is the separation between the objectiveand the eyepiece? When final image is formed at infinity.

Solution : Here, f0 = 144 cm; fe = 6.0 cm, MP = ?, L = ?

MP = 0

e

f 14424

f 6.0 and L = f0 + fe = 144 + 6.0 = 150.0 cm

Example : 79 Diameter of the moon is 3.5 × 103 km and its distance from earth is 3.8 × 105 km. It is seen by atelescope whose objective and eyepiece have focal lengths 4m and 10cm respectively. What will theangular diameter of the image of the moon.

Solution : 0

e

f 400MP 40

f 10 . Angle subtended by the moon at the objective =3

5

3.5 103.8 10

= 0.009 radian.

Thus angular diameter of the image = MP × visual angle = 40 × 0.009 = 0.36 radian

= 0.36 180

213.14

Example : 80 A telescope consisting of an objective of focal length 60 cm and a single–lens eyepiece of focallength 5 cm is focussed at a distant object in such a way that parallel rays emerge from the eyepiece. If the object subtends an angle of 2° at the objective, then find the angular width of the image.

Solution : 0 0

e e

f f 60MP 2 24

f f 5

Example : 81 The focal lengths of the objective and the eye piece of an astronomical telescope are 60 cm and 5cm respectively. Calculate the magnifying power and the length of the telescope when the final imageis formed at (i) infinity, (ii) least distance of distinct vision (25 cm)

JEE (Adv.)-Physics Geometrical OpticsSolution : (i) When the final image is at infinity, then :

MP = – 0

e

ff = –

605

= – 12 and length of the telescope is L = f0 + fe = 60 + 5 = 65 cm

(ii) For least distance of distinct vision, the magnifying power is :

0 e

e

f f 60 5 12 6MP 1 1 14.4

f D 5 25 5

Now e e e e e

1 1 1 1 1 1 1 1 1f v u 5 25 u u 25 5 ue = – 4.17 cm |ue| = 4.17 cm

The length of telescope in this position is L = f0 + |ue| = 60 + 4.17 = 64.17 cm

GALILEO’S TELESCOPEConvex lens as objective.

Concave lens as eyepiece.

Field of v iew is much smaller

eyepiece lens in concave.

(i) e

e

e

0

vf1

ffM

(ii) e

0

ff

M

Final image is at L = f0 – fe

(iii) Df1

ffM e

e

0Final image is at D. L = f0 – ue

OLVED XAMPLES EExample 82. A telescope consists of two convex lens of focal length 16 cm and 2 cm. What is angular magnification

of telescope for relased eye? What is the separation between the lenses?

If object subtends an angle of 0.5º on the eye, what will be angle subtended by its image ?

Solution : Angular magnification

M = 82

16fF cm

Separation between lenses

= F + f = 16 + 2 = 18 cm

Here = 0.5º

Angular subtended by image

= M = 8 × 0.5º = 4º

JEE (Adv.)-Physics Geometrical OpticsExample 83. The magnifying power of the telescope if found to be 9 and the separation between the lenses is 20

cm for relased eye. What are the focal lengths of component lenses ?

Solution : Magnification M = fF

Separaton between lenses

d = F + f

Given fF

= 9 i.e., F = 9f ......(1)

and F + f = 20 ......(2)

Putting value of F from (1) in (2), we get

9f + f = 20 10 f = 20

cm21020

F = 9f = 9 × 2 = 18 cm

F = 18 cm, f = 2 cm

ELF RACTICE ROBLEMSS P P1. The magnifying power of a telescope can be increased by

(1) Increasing focal length of the system (2) Fitting eye piece of high power

(3) Fitting eye piece of low power (4) Increasing the distance of objects

2. A simple telescope, consisting of an objective of focal length 60 cm and a single eye lens of focal length 5 cmis focussed on a distant object is such a way that parallel rays comes out from the eye lens. If the objectsubtends an angle 2° at the objective, the angular width of the image

(1) 10° (2) 24° (3) 50° (4) 1/6°

3. If the telescope is reversed i.e., seen from the objective side

(1) Object will appear very small

(2) Object will appear very large

(3) There will be no effect on the image formed by the telescope

(4) Image will be slightly greater than the earlier one

4. The aperture of a telescope is made large, because

(1) To increase the intensity of image (2) To decrease the intensity of image

(3) To have greater magnification (4) To have lesser resolution

5. The magnifying power of a telescope is 9. When it is adjusted for parallel rays, the distance between theobjective and the eye-piece is found to be 20 cm. The focal length of the two lenses are

(1) 18 cm, 2 cm (2) 11 cm, 9 cm (3) 10 cm, 10 cm (4) 15 cm, 5 cm

6. A reflecting telescope utilizes

(1) A concave mirror (2) A convex mirror (3) A prism (4) A plano-convex lens

Answers: 1. (2) 2. (2) 3. (1) 4. (1) 5. (1) 6. (1)


SCATTERING OF LIGHT(i) When light from some source (ga, sun, stars) enters the earth atmosphere then it gets reflected in variousdirection by the particles of dust, smoke and gas molecules. The phenomenon of this diffuse reflection isknown an scattering This was initially suggested by Tindal.

(ii) According to Rayleigh, the intensity (I) of scattered light is inversely proportional to the fourth power ofwavelength of light. i.e.

i.e. I 41

That is the reason why red light ( more) gets scattered minimum and violet light ( less) gets scatteredmaximum

(IR = 16IV)

(iii) Consequences of scattering of light -

(a) Appearance of blue colour of sky.

(b) The danger signals are made red.

(c) Appearance of black colour of sky in the absence of atmosphere.

(d) Appearance of red colour of sun at sun rise and sun set.

Luminous bodies –– The bodies which emit light themselves are known as luminous bodies.


Problem 1. An object is kept fixed in front of a plane mirror which is moved by 10 m/s away from theobject, find the velocity of the image.

Solution : OMIM VV

G,MG,OG,MG, VVVV

2VV

V G,OG,G,M

2V G, ( G,OV = 0 )

2V G, = 10 i m/s

G,V = 20 i m/s

Problem 2. Find the position of final image after three successive reflections taking first reflection on m1.

Solution : 1st reflection at mirror m1 : u = –15 cm, f = – 10cm

v1

= f1

– u1

v = fuuf

= 10)15()10()15(

= 5150

cm = –30 cm.

Thus, image is formed at a point 5 cm right of m2 which will act as an object for the reflection at m2

For 2nd reflection at m2u = 5 cm, f = 10 cm

v = fuuf

= 105105

= 550

= – 10 cm.

3rd reflection at m1 again.u = –15 cm f = – 10 cm

v = fuuf

= 10)15()10(15

= –30 cm. Ans.

JEE (Adv.)-Physics Geometrical OpticsProblem 3. A coin is placed 10 cm in front of a concave mirror . The mirror produces a real image that

has diameter 4 times that of the coin. What is the image distance.

Solution : m = 1

2

dd

= – uv

We have, u = 10 cm (virtual object) as real image is formedv = – mu= –4 ×10 cm= – 40 cm Ans.

Problem 4. A small statue has a height of 1 cm and is placed in front of a spherical mirror . The imageof the statue is inverted and is 0.5cm tall and located 10 cm in front of the mirror. Find thefocal length and nature of the mirror.

Solution : We have m = 1

2

hh

= – 15.0

= – 0.5

v = – 10 cm (real image)

But m = f

vf– 0.5 =

f10f

f = 320

cm

so, concave mirror. Ans.Problem 5. A light ray deviates by 300 (which is one third of the angle of incidence) when it gets refracted

from vacuum to a medium. Find the refractive index of the medium.Solution :

= i – r

3i

= i – r = 30º.

i = 90º2i = 3r

r = 3i2 = 60º

So, µ = º60sinº90sin

= 2/31

= 32

Ans.

Problem 6. A coin lies on the bottom of a lake 2m deep at a horizontal distance x from the spotlight (asource of thin parallel beam of light) situated 1 m above the surface of a liquid of refractive

index = 2 and height 2m. Find x.

1m

2m

450

xcoin

eye


Solution : 2 = rsin

º45sin

sin r = 21

r = 30º

x = RQ + QP

= 1m + 2tan30° m

= 321 m Ans.

Problem 7. A ray of light falls at an angle of 30º onto a plane-parallel glass plate and leaves it parallel tothe initial ray. The refractive index of the glass is 1.5. What is the thickness d of the plate if

the distance between the rays is 3.82 cm? [Given : sin–1 31

= 19.5º ; cos 19.5º = 0.94 ;

sin 10.5º = 0.18]

Solution :

µ = 1.5Using s = rcos

)ri(sind

d = )rº30sin(rcos82.3

..................... (1)

Also, 1.5 = rsinº30sin

sin r = 31

so, r = 19.5º

So, d = )º5.19º30sin(º5.19cos82.3

= º5.10sin94.082.3

= 18.094.082.3

= 19.948 cm

0.2 cm

Problem 8. Find the focal length of a plano-convex lens with R1 = 15 cm and R2 = . The refractive indexof the lens material n = 1.5.

Solution :f1

= (n – 1) 21 R

1R1

= (1.5 – 1) 1

151

. = 0.5 × 151

= 30 cm .

JEE (Adv.)-Physics Geometrical OpticsProblem 9. Find the focal length of a concavo-convex lens (positive meniscus) with R1 = 15 cm and R2 =

25 cm. The refractive index of the lens material n = 1.5.

Solution :f1

= (1.5 – 1) 251

151

= 0.5 150610

. = 4

300 = 75 cm

Problem 10. Figure shows a point object and a diverging lens.

Find the final image formed.

Solution :v1

– u1 =

f1

v1

= 101

+ )10(1

= – 102

. = V = – 5 cm

Problem 11. A pin is placed 10 cm in front of a convex lens of focal length 20 cm, made of material having refractiveindex 1.5. The surface of the lens farther away from the pin is silvered and has a radius of curvature22 cm. Determine the position of the final image. Is the image real or virtual ?

Solution : As radius of curvature of silvered surface is 22 cm, so

M = R2

= 222

= –11 cm = – 0.11 m and hence,

PM = – 1

M= –

1011.

=1

011.D

Further as the focal length of lens is 20 cm, i.e., 0.20 m its power will be given by :

PL = 1

L

= 1

020. D. Now as in image formation, light after passing through the lens will be reflected back

by the curved mirror through the lens again P= PL + PM + PL = 2PL + PM i.e. 2 1 210

P D0.20 0.11 11

.

So the focal length of equivalent mirror 1 11 110

F m cmP 210 21

i.e., the silvered lens behave

as a concave mirror of focal length (110/21) cm. So for object at a distance 10 cm in front of

it,1 1 21v 10 110

i.e., v = – 11cm i.e., image will be 11 cm in front of the silvered lens and will be real as shown inFigure.

JEE (Adv.)-Physics Geometrical OpticsProblem 12. The focal length of the objective and eyepiece of a microscope are 2 cm and 5 cm respectively

and the distance between them is 20 cm. Find the distance of object from the objective, whenthe final image seen by the eye is 25 cm from the eyepiece. Also find the magnifying power.

Solution : Given f0 = 2 cm, fe = 5 cm

| vo | + | ue | = 20 cm

ve = – 25 cm

From lens formula eoe u1

v1

f1

51

251

f1

v1

u1

ee

ue = – 625

cm

Distance of real image from objective

vo = 20 – | ue | = 20 – 625

695

625120

cm

Nowo o o

1 1 1f v u

given 21

)6/95(1

f1

v1

u1

ooo

i.e., 19083

1909512

21

956

u1

o

uo = – 83190

= –2.3 cm

Magnifying power M = – o

o

uv

ef

D1)83/190(

6/95 3

251 = – 41.5


PART-I : SUBJECTIVE QUESTIONSSECTION (A) : PLANE MIRRORA-1. A plane mirror of length 8 cm is moving with speed 3 m/s towards a wall in situation as shown in figure. Size

of spot formed on the wall is 32/x cm. Find the value of x.

S = source of light

wall

3 m

/s

8 cm

4cm

\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

A-2. The angle between the velocity vector of object and image is . Velocity are shown in the figure. Then the

value of is.

60°

10m/s

5m/s

object

A-3. Find the angle of deviation (both clockwise and anticlockwise) suffered by a ray incident on a plane mirror,(as shown in figure) at an angle of incidence 30º.

A-4. Figure shows a plane mirror on which a light ray is incident. If the incident light ray is turned by 10º and themirror by 20º, as shown, find the angle turned by the reflected ray.

30º

20º

Reflected ray10º

A-5. A point object is placed at (0, 0) and a plane mirror 'M' is placed,

(1, 0)Object(0, 0)

·

////////////////////////////////////////////

M

30º x axis()

y axis ()

inclined 30º with the x axis.(a) Find the position of image.

(b) If the object starts moving with velocity 1 i m/s and themirror is fixed find the velocity of image.

JEE (Adv.)-Physics Geometrical OpticsA-6. Two plane mirrors are placed as shown in the figure and a point object 'O' is placed at the origin

(a) How many images will be formed.

(b) Find the position(s) of image(s).(c) Will the incident ray passing through a point 'P' (1, 1.25) take part in image formation.

SECTION (B) : SPHERICAL MIRRORB-1. A cube of side length 1mm is placed on the axis of a concave mirror at a distance of 45 cm from the pole as

shown in the figure. One edge of the cube is parallel to the axis. The focal length of the mirror is 30 cm. Findapproximate volume of the image.

B-2. A thin rod of length d/3 is placed along the principal axis of a concave mirror of focal length = d such that itsimage, which is real and elongated, just touches the rod. Find the length of the image ?

B 3. Find the diameter of the image of the moon formed by a spherical concave mirror of focal length 11.4 m. Thediameter of the moon is 3450 km and the distance between the earth and the moon is 3.8 ×105 km.

B-4. The radius of curvature of a convex spherical mirror is 1.2 m. How far away from the mirror is an object ofheight 1.2 cm if the distance between its virtual image and the mirror is 0.35 m? What is the height of theimage? [Apply formula for paraxial rays]

B-5. A converging beam of light rays is incident on a concave spherical mirror whose radius of curvature is 0.8 m.Determine the position of the point on the optical axis of the mirror where the reflected rays intersect, if theextensions of the incident rays intersect the optical axis 40 cm from the mirror’s pole.

B-6. A man uses a concave mirror for shaving. He keeps his face at a distance of 20 cm from the mirror and getsan image which is 1.5 times enlarged. Find the focal length of the mirror.

B-7. A point object is placed on the principal axis at 60 cm in front of a concave mirror of focal length 40 cm onthe principal axis. If the object is moved with a velocity of 10 cm/s (a) along the principal axis, find thevelocity of image (b) perpendicular to the principal axis, find the velocity of image at that moment.

B-8. A fluorescent lamp of length 1 m is placed horizontally at a depth of 1.2 m below a ceiling. A plane mirror oflength 0.6 m is placed below the lamp parallel to and symmetric to the lamp at a distance 2.4 m from it asshown in figure. Find the length (distance between the extreme points of the visible region along x-axis)of the reflected patch of light on the ceiling.

1m

1.2m

2.4m

0.6m

x

JEE (Adv.)-Physics Geometrical OpticsSECTION (C) : REFRACTION IN GENERAL , REFRACTION AT PLANE SURFACE AND T.I.R.

C-1. A light ray is incident at 45° on a glass slab. The slab is 3 cm thick, and the refractive index of the glass is1.5. What will the lateral displacement of the ray be as a result of its passage through the slab? At whatangle will the ray emerge from the slab ?

C-2. A light ray falling at an angle of 60° with the surface of a clean slab of ice of thickness 1.00 m is refractedinto it at an angle of 15°. Calculate the time taken by the light rays to cross the slab. Speed of light invacuum = 3 × 108 m/s.

C-3. In the given figure an observer in air (n = 1) sees the bottom of a beaker filled with water (n = 4/3) upto aheight of 40 cm. What will be the depth felt by this observer.

40cm

observer ( )çs{kd

C-4. Find the apparent distance between the observer and the object shown in the figure and shift in the positionof object.

10cm 20cm10cm

object ( )oLrq

C

D

A

B

observer( )çs{kd

C-5. A fish is rising up vertically inside a pond with velocity 4 cm/s, and notices a bird, which is diving verticallydownward along the same vertical line as that of fish and its velocity appears to be 16 cm/s (to the fish).What is the real velocity of the diving bird, if refractive index of water is 4/3 ?

C-6. In the given figure rays incident on an interface would converge 10 cm below the interface if they continuedto move in straight lines without bending. But due to refraction, the rays will bend and meet some whereelse. Find the distance of meeting point of refracted rays below the interface, assuming the rays to bemaking small angles with the normal to the interface.

C-7. Find the apparent depth of the object seen by observer A (in the figure shown)

C-8. A small object is placed at the centre of the bottom of a cylindrical vessel of radius 3 cm and height 3 3 cm

filled completely with a liquid. Consider the ray leaving the vessel through a corner. Suppose this ray and theray along the axis of the vessel are used to trace the image. Find the apparent depth of the image. Refractive

index of liquid = 3 .

JEE (Adv.)-Physics Geometrical OpticsC-9. A point source is placed at a depth h below the surface of water (refractive index = µ).The medium above the

surface of water is air (µ = 1).Find the area on the surface of water through which light comes in air from

water.C-10. Locate the image of the point P as seen by the eye in the figure.

C-11. At what values of the refractive index of a rectangular prism can a ray travel as shown in figure. The sectionof the prism is an isosceles triangle and the ray is normally incident onto the face AC.

C-12. Light is incident from glass (µ = 23

) side on interface of glass and air. Find the angle of incidence for which

the angle of deviation is 90º.

SECTION D : REFRACTION BY PRISMD-1. The cross section of a glass prism has the form of an equilateral triangle. A ray is incident onto one of the

faces perpendicular to it. Find the angle between the incident ray and the ray that leaves the prism. The

refractive index of glass is µ = 1.5.

D-2. A prism (n = 2) of apex angle 90° is placed in air (n = 1). What should be the angle of incidence so that light

ray strikes the second surface at an angle of incidence 60º.

D-3. Find the angle of deviation suffered by the light ray shown in figure forfollowing two conditions The refractive index for the prism material is µ = 3/2.(i) When the prism is placed in air ( = 1)(ii) When the prism is placed in water ( = 4/3)

D-4. The refractive index of a prism is . Find the maximum angle of the prism for which a ray incident on it willbe transmitted through other face without total internal reflection.

SECTION (E) : REFRACTION BY SPHERICAL SURFACEE-1. An object is placed 10 cm away from a glass piece (n = 1.5) of length 20 cm bounded by spherical surfaces

of radii of curvature 10 cm. Find the position of final image formed after two refractions at the sphericalsurfaces.

JEE (Adv.)-Physics Geometrical OpticsE-2. An extended object of size 2 cm is placed at a distance of 10 cm in air (n = 1) from pole, on the principal axis

of a spherical curved surface. The medium on the other side of refracting surface has refractive indexn = 2. Find the position, nature and size of image formed after single refraction through the curved surface.

E-3. A point object lies inside a transparent solid sphere of radius 20 cm and of refractive index n = 2. When theobject is viewed from air through the nearest surface it is seen at a distance 5 cm from the surface. Find theapparent distance of object when it is seen through the farthest curved surface.

E-4. There is a small air bubble inside a glass sphere (µ = 1.5) of radius 5 cm. The bubble is at 'O' at 7.5 cm below

the surface of the glass. The sphere is placed inside water (µ = 34

) such that the top surface of glass is 10

cm below the surface of water. The bubble is viewed normally from air. Find the apparent depth of the bubble.

water

observer

10cm

Cglass

O

E-5. A small object Q of length 1 mm lies along the principal axis of a spherical glass of radius R = 10 cm andrefractive index is 3/2. The object is seen from air along the principal axis from left. The distance of objectfrom the centre P is 5 cm. Find the size of the image. Is it real, inverted?

E-6. A narrow parallel beam of light is incident paraxially on a solid transparent sphere of radius r kept in air.What should be the refractive index if the beam is to be focused (a) at the farther surface of the sphere, (b)at the centre of the sphere.

SECTION (F) : LENSF-1. Given an optical axis MN and the positions of a real object AB and its image A 'B', determine diagrammatically

the position of the lens (its optical centre O) and its foci. Is it a converging or diverging lens? Is the imagereal or virtual ?

A'

NB'

B

A

M

JEE (Adv.)-Physics Geometrical OpticsF-2. Lenses are constructed by a material of refractive index 2. The magnitude of the radii of curvature are 20 cm

and 30 cm. Find the focal lengths of the possible lenses with the above specifications.F-3. Find the focal length of lens shown in the figure. Solve for three cases ns = 1.5, ns = 2.0, ns = 2.5.

ns ns

n=2

R.O.C.=40cmR.O.C.=60cm

F-4. Two glasses with refractive indices of 1.5 & 1.7 are used to make two identical double-convex lenses.(i) Find the ratio of their focal lengths.(ii) How will each of these lenses act on a ray parallel to its optical axis if the lenses are submerged

into a transparent liquid with a refractive index of 1.6?F-5. A thin lens made of a material of refractive index µ2 has a medium of refractive index µ1 on one side and a

medium of refractive index µ3 on the other side. The lens is biconvex and the two radii of curvature has equalmagnitude R. A beam of light travelling parallel to the principal axis is incident on the lens. Where will theimage be formed if the beam is incident from (a) the medium µ1 and (b) from the medium µ3 ?

F-6. An object of height 1 cm is set at right angles to the optical axis of a double convex lens of optical power 5D and 25 cm away from the lens. Determine the focal length of the lens, the position of the image, the linearmagnification of the lens, and the height of the image formed by it.

F-7. A 2.5 dioptre lens forms a virtual image which is 4 times the object placed perpendicularly on the principalaxis of the lens. Find the required distance of the object from the lens.

F-8. A pin of length 1 cm lies along the principal axis of a converging lens, the centre being at a distance of 5.5cm from the lens. The focal length of the lens is 3 cm. Find the size of the image.

F-9. The radius of the sun is 0.75 × 109 m and its distance from the earth is 1.5 × 1011 m. Find the diameter of theimage of the sun formed by a lens of focal length 40 cm.

F-10. A diverging lens of focal length 20 cm is placed coaxially 5 cm towards left of a converging mirror of focallength 10 cm .Where should an object be placed towards left of the lens so that a real image is formed at theobject itself ?

F -11. A point object is placed on the principal axis of a converging lens of focal length 15 cm at a distance of30 cm from it. A glass plate (µ = 1.50) of thickness 3 cm is placed on the other side of the lens perpendicularto the axis. Find the position of the image of the point object.

F-12. A convex lens and a convex mirror are placed at a separation of 15 cm. The focal length of the lens is 25 cmand radius of curvature of the mirror is 80 cm. Where should a point source be placed between the lens andthe mirror so that the light, after getting reflected by the mirror and then getting refracted by the lens, comesout parallel to the principal axis ?

F-13. A converging lens of focal length 10 cm and a diverging lens of focal length 5 cm are placed 5 cm apart withtheir principal axis coinciding. A beam of light travelling parallel to the principal axis and having a beamdiameter 5.0 mm, is incident on the combination. Show that the emergent beam is parallel to the incident

one. Find the beam diameter of the emergent beam. Also find out the ratio of emergent and incident intensities.

SECTION (G) : COMBINATION OF LENSES/LENS AND MIRRORSG-1. A point object is placed at a distance of 15 cm from a convex lens. The image is formed on the other side at

a distance of 30 cm from the lens. When a concave lens is placed in contact with the convex lens, the imageshifts away further by 30 cm. Calculate the focal lengths of the two lenses.

G-2. Two identical thin converging lenses brought in contact so that their axes coincide are placed 12.5 cm froman object. What is the optical power of the system and each lens, if the real image formed by the systemof lenses is four times as large as the object ?

JEE (Adv.)-Physics Geometrical OpticsG-3. The convex surface of a thin concavo-convex lens of glass of refractive index 1.5 has a radius of curvature

20 cm. the concave surface has a radius of curvature 60 cm. The convex side is silvered and placed on ahorizontal surface as shown in figure. (a) Where should a pin be placed on the axis so that its image isformed at the same place? (b) If the concave part is filled with water (µ = 4/3), find the distance through whichthe pin should be moved so that the image of the pin again coincides with the pin.

SECTION (H) : DISPERSION OF LIGHTH-1. Three thin prisms are combined as shown in figure. The refractive indices of the crown glass for red, yellow

and violet rays are µr, µy and µv respectively and those for the flint glass are µr , µy and µv respectively. Findthe ratio A /A for which (a) system produces deviation without dispersion(achromatic combination)and (b)system produces dispersion without deviation(direct vision arrangement).

H-2. A certain material has refractive indices 1.53, 1.60 and 1.68 for red, yellow and violet light respectively. (a)Calculate the dispersive power. (b) Find the angular dispersion produced by a thin prism of angle 6° made ofthis material.

H-3. A flint glass prism and a crown glass prism are to be combined in such a way that the deviation of the meanray is zero. The refractive index of flint and crown glasses for the mean ray are 1.6 and 1.9 respectively. If therefracting angle of the flint prism is 6°, what would be the refracting angle of crown prism ?

H-4. The focal lengths of a convex lens for red, yellow and violet rays are 100 cm, 99 cm and 98 cm respectively.Find the dispersive power of the material of the lens.

H-5. A thin prism of angle 5.0°, = 0.07 and µy = 1.30 is combined with another thin prism having' = 0.08 and µ'y = 1.50. The combination produces no deviation in the mean ray. (a) Find the angle of the

second prism. (b) Find the net angular dispersion produced by the combination when a beam of white lightpasses through it. (c) If the prisms are similarly directed, what will be the deviation in the mean ray? (d) Findthe angular dispersion in the situation described in (c).

SECTION (I) : FOR JEE MAINI-1. An angular magnification ( magnifying power ) of 30 X is desired using an objective of focal length 1.25cm

and an eye-piece of focal length 5 cm. How will you set up the compound microscope for normal adjustment?I -2. A small telescope has an objective lens of focal length 144 cm and an eye-piece of focal length 6.0 cm.

What is the magnifying power of the telescope? What is the separation between the objective and the eye-piece ?

I-3. A compound microscope consists of an objective lens of focal 2.0 cm and an eye-piece of focal length 6.25cm separated by a distance of 15 cm. How far from the objective should an object be placed in order toobtain the final image at (a) least distance of distinct vision ( 25 cm), (b) infinity? What is the magnifyingpower of the microscope in each case ?


PART-II : OBJECTIVE QUESTIONS* Marked Questions may have more than one correct option.

SECTION (A) : PLANE MIRROR-1. If two mirrors are kept at 60° to each other, then the number of images formed by them is :-

(A) 5 (B) 6 (C) 7 (D) 8-2. If = 110° then total number of images formed by the mirror system will be :-

(A) 2 (B) 3 (C) 4 (D) 5-3. A point object approaches a plane mirror with a speed of 10 ms–1, while the image recedes away from the

mirror with a speed of 6 ms–1 (as seen by stationary observer). The direction and magnitude of the velocity ofmirror is :-(A) towards the object, 8 ms–1 (B) towards the image, 6 ms–1

(C) away from the object, 8 ms–1 (D) away from the object, 2 ms–1

-4. A boy of Reliable PA1 batch is 1.8m tall and can see his image in a plane mirror fixed on a wall. His eyes are1.6m from the floor level. The minimum length of the mirror to see his full image is :-

(A) 0.9 m (B) 0.85 m (C) 0.8 m (D) Can't be determined-5. Two mirrors are inclined at an angle as shown in the figure. Light ray is incident parallel to one of the

mirrors. Light will start retracing its path after third reflection if :

(A) = 45° (B) = 30° (C) = 60° (D) All three-6. An unnumbered wall clock shows time 04: 25: 37, where 1st term represents hours, 2nd represents minutes

and the last term represents seconds. What time will its image in a plane mirror show.(A) 08: 35: 23 (B) 07: 35: 23 (C) 07: 34: 23 (D) none of these

-7. An object and a plane mirror are as shown in figure. Mirror is moved with velocity V as shown. The velocityof image is :

//////////////////////////

V

Mirror

Object (fixed)

(A) 2 V sin (B) 2 V (C) 2V cos (D) None of these-8. Two plane mirrors are parallel to each other and spaced 20 cm apart. An object is kept in between them at

15 cm from A. Out of the following at which point(s) image(s) is/are not formed in mirror A (distance measuredfrom mirror A):(A) 15 cm (B) 25 cm (C) 45 cm (D) 55 cm

-9. Two plane mirrors are inclined to each other at an angle 600. If a ray of light incident on the first mirror isparallel to the second mirror, it is reflected from the second mirror(A) Perpendicular to the first mirror (B) Parallel to the first mirror

(C) Parallel to the second mirror (D) Perpendicular to the second mirror


-10. A plane mirror is moving with velocity k8j5i4 . A point object in front of the mirror moves with a

velocity k5j4i3 . Here k is along the normal to the plane mirror and facing towards the object. The

velocity of the image is :

(A) k5j4i3 (B) k11j4i3 (C) k11j4i3 (D) k11j9i7

SECTION (B) : SPHERICAL MIRROR-1. An object is placed in front of a spherical mirror whose 2 times magnified image is formed on screen. Then

choose CORRECT option :-(A) Mirror is concave m = +2 (B) Mirror is concave m = –2(C) Mirror is convex m = +2 (D) Mirror is convex m = –2

-2. A particle approaches from very large distance towards concave mirror along the principal axis. By the timethe particle reaches the mirror the distance between the particle and its image(A) first decreases then increases(B) first increases then decreases(C) first increases then decreases and then again increases(D) first decreases then increases and then again decreases

-3. A luminous point object is moving along the principal axis of a concave mirror of focal length 12 cm towardsit. When its distance from the mirror is 20 cm its velocity is 4 cm/s. The velocity of the image in cm/s at thatinstant is :(A) 6, towards the mirror (B) 6, away from the mirror(C) 9, away from the mirror (D) 9, towards the mirror

-4. A point object on the principal axis at a distance 15 cm in front of a concave mirror of radius of curvature 20cm has velocity 2 mm/s perpendicular to the principal axis. The magnitude of velocity of image at that instantwill be:(A) 2 mm/s (B) 4 mm/s (C) 8 mm/s (D) 16 mm/s

-5. A particle is moving towards a fixed spherical mirror. The image:(A) must move away from the mirror(B) must move towards the mirror(C) may move towards the mirror

(D) will move towards the mirror, only if the mirror is convex.-6. In the figure M1 and M2 are two fixed mirrors as shown. If the object ' O ' moves towards the plane mirror, then

the image I (which is formed after two successive reflections from M1 & M2 respectively) will move

(A) towards right (B) towards left (C) with zero velocity (D) cannot be determined-7. The distance of an object from the focus of a convex mirror of radius of curvature ' a ' is ' b '. Then the distance

of the image from the focus is :(A) b2 / 4a (B) a / b2 (C) a2 / 4b (D) 4b / a2

-8. A point object at 15 cm from a concave mirror of radius of curvature 20 cm is made to oscillate along theprincipal axis with amplitude 2 mm. The amplitude of its image will be(A) 2 mm (B) 4 mm (C) 8 mm (D) 16 mm

JEE (Adv.)-Physics Geometrical Optics-9. The largest distance of the image of a real object from a convex mirror of focal length 20 cm can be :

(A) 20 cm (B) infinite(C) 10 cm (D) depends on the position of the object

-10.Which of the following can form erect, virtual, diminished image ?(A) plane mirror (B) concave mirror (C) convex mirror (D) none of these

-11.A real inverted image in a concave mirror is represented by graph (u, v, f are coordinates)

(A) (B) (C) (D)

-12.A concave mirror of radius of curvature 20 cm forms image of the sun. The diameter of the sun subtendsan angle 1º on the earth. Then the diameter of the image is (in cm) :(A) 2 /9 (B) /9 (C) 20 (D) /18

-13 An object is placed at a distance u from a concave mirror and its real image is received on a screenplaced at a distance of v from the mirror. If f is the focal length of the mirror, then the graph between1/v versus 1/u is

(A) (B) (C) (D)

-14. The position of a real point object and its real point image are as shown in the figure. AB is the principalaxis. This can be achieved by using :

(A) convex mirror (B) concave mirror (C) concave mirror only (D) plane mirror

SECTION (C) : LAWS OF REFRACTION, REFRACTION AT PLANE SURFACE AND T.I.R.C-1. How much water should be filled in a container of 21 cm in height, so that it appears half filled (of actual height

of the container) when viewed from the top of the container ? (Assume near normal incidence and w=4/3)(A) 8.0 cm (B) 10.5 cm (C) 12.0 cm (D) 14.0 cm

C-2. When a wave is refracted :(A) its path must change (B) its wavelength will not change(C) its velocity must change (D) its frequency must change

C-3. AB is a boundary separating two media of different refractive indices. A rayis incident on the boundary is partially reflected and partially transmitted.Choose the CORRECT statement.

(A) 3 is incident ray and 1 is refracted ray

(1) (2)

B

(3)

A

(B) 2 is incident ray and 1 is partially reflected ray

(C) 1 is incident ray and 3 is refracted ray

(D) 3 is incident ray and 2 is partially reflected ray

JEE (Adv.)-Physics Geometrical OpticsC-4. A ray of light passes from vacuum into a medium of refractive index n. If the angle of incidence is twice the

angle of refraction, then the angle of incidence is :

(A) cos–1 (n/2) (B) sin–1

(n/2) (C) 2 cos–1 (n/2) (D) 2 sin–1

(n/2)

C-5. The wavelength of light in vacuum is 6000 Aº and in a medium it is 4000 Aº. The refractive index of themedium is:

(A) 2.4 (B) 1.5 (C) 1.2 (D) 0.67

C-6. A ray of light is incident on a parallel slab of thickness t and refractive index n. If the angle of incidence issmall, then the displacement in the incident and emergent ray will be:

(A) n

)1n(t(B)

nt

(C) 1nnt

(D) None of these

C-7. What is the length of the image of the rod in mirror, according to

the observer in air? ( refractive index of the liquid is )

(A) L + L (B) L + L

Eye

L

L

L

(C) L +L

(D) None of these

C-8. A ray of light travelling in air is incident at grazing incidence on a slab with variable refractive index,n (y) = [k y

3/2 + 1]1/2 where k = 1 m 3/2 and follows path as shown in the figure. What is the total deviationproduced by slab when the ray comes out.

(A) 60º (B) 53º (C) sin 1 (4/9) (D) no deviation at all

C-9. The critical angle of light going from medium A to medium B is . The speed of light in medium A is v. Thespeed of light in medium B is:

(A) v

sin(B) v sin (C) v cot (D) v tan

C-10. A beam of light is converging towards a point. A plane parallel plate of glass of thickness t, refractive index is introduced in the path of the beam as shown in the figure. The convergent point is shifted by (assume nearnormal incidence) :

(A) t 11

away (B) t 11

away (C) t 11

nearer (D) t 11

nearer


SECTION (D) : REFRACTION BY PRISMD-1. The angle of minimum deviation from a prism is 60º. If the prism angle is 90º, the refractive index of

the material of the prism and the angle of incidence required for minimum deviation is : (Given thatsin75º = 0.96)

(A) º45sinº75sin

, 45º (B) º45sinº75sin

, 75º (C) 2 º45sinº75sin

, 75º (D) 2 , 75º

D-2. A prism having refractive index 2 and refracting angle 30º, has one of the refracting surfaces polished.

A beam of light incident on the other refracting surface will retrace its path if the angle of incidence is:

(A) 0º (B) 30º (C) 45º (D) 60º

D-3. A ray of monochromatic light is incident on one refracting face of a prism of angle 750. It passes through theprism and is incident on the other face at the critical angle. If the refractive index of the material of the prism

is 2, the angle of incidence on the first face of the prism is

(A) 300 (B) 450 (C) 600 (D) 00

D-4. A ray of light is incident at angle i on a surface of a prism of small angle A and emerges normally from the

opposite surface. If the refractive index of the material of the prism is , the angle of incidence i is nearly

equal to :

(A) A/ (B) A/(2 ) (C) A (D) A/2

D-5. A prism of refractive index 2 has refracting angle 60º. Answer the following questions (a) In order that a ray

suffers minimum deviation it should be incident at an angle :

(A) 450 (B) 900 (C) 300 (D) none

(b) Angle of minimum deviation is :

(A) 450 (B) 900 (C) 300 (D) none

(c) Angle of maximum deviation is :

(A) 450 (B) sin-1 ( 2 sin15º )

(C) 30º + sin-1 ( 2 sin15º ) (D) none

SECTION (E) : REFRACTION BY SPHERICAL SURFACEE-1. One end of a glass rod of refractive index n = 1.5 is a spherical surface of radius of curvature R. The centre

of the spherical surface lies inside the glass. A point object placed in air on the axis of the rod at the point P

has its real image inside glass at the point Q (see fig.). A line joining the points P and Q cuts the surface at

O such that OP = 2OQ. The distance PO is :-

P O Q

µ=1.5

(A) 8 R (B) 7 R (C) 2 R (D) None of these

JEE (Adv.)-Physics Geometrical OpticsE-2. The image for the converging beam after refraction through the curved surface (in the given figure) is

formed at :-

n=3/2n=1

P

30

O x

R=20cm

(A) x = 40 cm (B) x = 403

cm (C) x = 403

cm (D) x =180

7cm

E-3. There is a small black dot at the centre C of a solid glass sphere of refractive index . When seen fromoutside, the dot will appear to be located :-(A) away from C for all values of (B) at C for all values of

(C) at C for = 1.5, but away from C for 1.5 (D) at C only for 2 1.5.E-4. In the given figure a plano-concave lens is placed on a paper on which a flower is drawn. How far above its

actual position does the flower appear to be ?

(A) 10 cm (B) 15 cm (C) 50 cm (D) none of these

SECTION (F) : LENSF-1. From the figure shown establish a relation among , 1, 2, 3.

(A) 3 > 2 > 1 (B) 3 < 2 < 1 (C) 2 > 3 ; 3 = 1 (D) 2 > 1 ; 3 = 2

F-2. A concave lens of glass, refractive index 1.5, has both surfaces of same radius of curvature R. On immersionin a medium of refractive index 1.75, it will behave as a :-(A) convergent lens of focal length 3.5R (B) convergent lens of focal length 3.0 R.(C) divergent lens of focal length 3.5 R (D) divergent lens of focal length 3.0 R

F-3. Two symmetric double convex lenses A and B have same focal length, but the radii of curvature differ so that,

RA = 0.9 RB. If nA = 1.63, find nB.(A) 1.7 (B) 1.6 (C) 1.5 (D) 4/3

F-4. A convexo - concave diverging lens is made of glass of refractive index 1.5 and focal length 24 cm. Radius ofcurvature for one surface is double that of the other. Then radii of curvature for the two surfaces are (in cm):(A) 6, 12 (B) 12, 24 (C) 3, 6 (D) 18, 36

F-5. When a lens of power P (in air) made of material of refractive index is immersed in liquid of refractive index

0. Then the power of lens is:

(A)1

0P (B) 0

1P (C) 0

1.

P

0(D) None of these

JEE (Adv.)-Physics Geometrical OpticsF-6. The diameter of the sun subtends an angle of 0.50 at the surface of the earth. A converging lens of focal length

100 cm is used to provide an image of the sun on to a screen. The diameter (in mm) of the image formed isnearly(A) 1 (B) 3 (C) 5 (D) 9

F-7. In the figure given below, there are two convex lens L1 and L2 having focal length of f1 and f2 respectively. Thedistance between L1 and L2 will be :-

L1 L2

(A) f1 (B) f2 (C) f1 + f2 (D) f1 - f2F-8. A thin symmetrical double convex lens of power P is cut into three parts, as shown in the figure. Power

of A is :

(A) 2 P (B) P2

(C) P3

(D) P

F-9. A virtual erect image by a diverging lens is represented by (u, v, f are coordinates)

(A) (B) (C) (D)

F-10. An object is placed at a distance u from a converging lens and its real image is received on a screenplaced at a distance of v from the lens. If f is the focal length of the lens, then the graph between 1/v versus1/u is :

(A)

1/v

1/u

(B) (C) (D)

F-11. What should be the value of distance d so that final image is formed on the object itself. (focal lengthsof the lenses are as given in the figure).


F-12. A biconvex lens is used to project a slide on screen. The slide is 2 cm high and placed at 10 cm from thelens. The image is 18 cm high. What is the focal length of the lens ?

(A) 9 cm (B) 18 cm (C) 4.5 cm (D) 20 cm

JEE (Adv.)-Physics Geometrical OpticsF-13. A thin linear object of size 1 mm is kept along the principal axis of a convex lens of focal length 10 cm. The

object is at 15 cm from the lens. The length of the image is:(A) 1 mm (B) 4 mm (C) 2 mm (D) 8 mm

F-14. The minimum distance between a real object and its real image formed by a thin converging lens of focallength f is :(A) 4f (B) 2f (C) f (D) f/2

SECTION (G) : COMBINATION OF THIN LENS/LENS AND MIRRORS.

G-1. Two plano-convex lenses each of focal length 10 cm & refractive index32

are placed as shown in the figure. In

the space left, water R I. .43

is filled. The whole arrangement is in air. The optical power of the system is (in

dioptre) :-

(A) 6.67 (B) - 6.67 (C) 33.3 (D) 20G-2. A convex lens of focal length 25 cm and a concave lens of focal length 20 cm are mounted coaxially

separated by a distance d cm. If the power of the combination is zero, d is equal to(A) 45 (B) 30 (C) 15 (D) 5

G-3. A plano-convex lens, when silvered at its plane surface is equivalent to a concave mirror of focal length 28cm. When its curved surface is silvered and the plane surface not silvered, it is equivalent to a concave mirrorof focal length 10 cm, then the refractive index of the material of the lens is :-(A) 9/14 (B) 14/9 (C) 17/9 (D) None

G-4. In the above question the radius of curvature of the curved surface of plano-convex lens is :

(A) 9280

cm (B) 7

180cm (C) 3

39cm (D)

11280

cm

G-5. The focal length of a plano-concave lens is 10 cm, then its focal length when its plane surface is polishedis (n = 3/2) :(A) 20 cm (B) 5 cm (C) 5 cm (D) none of these

SECTION (H) : DISPERSION OF LIGHTH-1. The dispersive power of the material of a lens is 0.04 and the focal length of the lens is 10 cm. Then the

difference in the focal length of the lens for violet and red colour is :

(A) 1 mm (B) 2 mm (C) 3mm (D) 4mm

H-2. Two thin prisms of flint glass, with refracting angles of 6º and 8º respectively, possess dispersive powers inthe ratio :(A) 4 : 3 (B) 3 : 4 (C) 1 : 1 (D) 9 : 16

H-3. Chromatic aberration will be absent if for two thin lenses in contact :-

(A) ( 1/F1) + ( 2/F2) = 0 (B) ( 1/F2) + ( 2/F1) = 0

(C) (F1/ 2) + (F2/ 1) = 0 (D) ( 1/ 2) + (F1+ F2) = 0

H-4. When white light passes through a glass prism, one gets spectrum on the other side of the prism. In theemergent beam, the ray which is deviating least is or deviation by a prism is lowest for(A) Violet ray (B) Green ray (C) Red ray (D) Yellow ray

JEE (Adv.)-Physics Geometrical OpticsH-5. The angle of prism is 5° and its refractive indices for red and violet colours are 1.5 and 1.6 respectively. The

angular dispersion produced by the prism is :–

(A) 7.75° (B) 5° (C) 0.5° (D) 0.17°

H-6. Critical angle of light passing from glass to air is minimum for

(A) red (B) green (C) yellow (D) violet

H-7. A plane glass slab is placed over various coloured letters. The letter which appears to be raised the least is:

(A) violet (B) yellow (C) red (D) green

H-8. The dispersion of light in a medium implies that :

(A) lights of different wavelengths travel with different speeds in the medium

(B) lights of different frequencies travel with different speeds in the medium

(C) the refractive index of medium is different for different wavelengths

(D) all of the above.

H-9. A medium has nv = 1.56, nr = 1.44. Then its dispersive power is:

(A) 3/50 (B) 6/25 (C) 0.03 (D) None of these

H-10. Light of wavelength 4000 Å is incident at small angle on a prism of apex angle 4º. The prism has nv = 1.5 &nr = 1.48. The angle of dispersion produced by the prism in this light is :

(A) 0.2º (B) 0.08º (C) 0.192º (D) None of these

H-11. All the listed things below are made of flint glass. Which one of these have greatest dispersive power ( ).

(A) prism (B) glass slab (C) biconvex lens (D) all have same

SECTION (I) : FOR JEE MAIN

I-1. In a compound microscope, the intermediate image is -

(A) virtual, erect and magnified (B) real, erect and magnified

(C) real, inverted and magnified (D) virtual, erect and reduced

I-2. A simple microscope has a focal length of 5 cm. The magnification at the least distance of distinct vision is-

(A) 1 (B) 5 (C) 4 (D) 6

I-3. The magnifying power of a simple microscope can be increased if an eyepiece of :

(A) shorter focal length is used (B) longer focal length is used

(C) shorter diameter is used (D) longer diameter is used

I-4. The focal length of the objective of a microscope is

(A) arbitrary (B) less than the focal length of eyepiece

(C) equal to the focal length of eyepiece (D) greater than the focal length of eyepiece

I-5. An astronomical telescope has an eyepiece of focal-length 5 cm. If the angular magnification in normaladjustment is 10, when final image is at least distance of distinct vision (25cm) from eye piece, then angularmagnification will be :

(A) 10 (B) 12 (C) 50 (D) 60

JEE (Adv.)-Physics Geometrical OpticsI-6. The focal lengths of the objective and eye-lens of a microscope are 1 cm and 5 cm respectively. If the

magnifying power for the relaxed eye is 45, then the length of the tube is :

(A) 30 cm (B) 25 cm (C) 15 cm (D) 12 cm

I-7. A person with a defective sight is using a lens having a power of +2D. The lens he is using is

(A) concave lens with f = 0.5 m (B) convex lens with f = 2.0 m

(C) concave lens with f = 0.2 m (D) convex lens with f = 0.5 m

I-8. If the focal length of objective and eye lens are 1.2 cm and 3 cm respectively and the object is put 1.25 cmaway from the objective lens and the final image is formed at infinity. The magnifying power of the microscopeis :

(A) 150 (B) 200 (C) 250 (D) 400

PART-III : MATCH THE COLUMN1. Angle between two mirrors ‘ ’ and location of object is given in column I and some possible number of

images are given in column II.

Column–I Column–II

(A) = 30° and object is on bisector (P) 9

(B) = 36° and object is on bisector (Q) 11

(C) = 22.5° and object is on bisector (R) 8

(D) = 50° and object is on bisector (S) 15

2. An object O is kept perpendicular to the principal axis of a spherical mirror. In column I, coordinate of, object,u with sign in cm, type of mirror and its focal length in cm is given

Column I Column II(A) u = –18, concave mirror, 12 (P) Image is real(B) u = –12, concave mirror, 18 (Q) Image is virtual(C) u = –8, convex mirror, 10 (R) Image diminished(D) u = –10, convex mirror, 8 (S) Image is enlarged

(T) Image is inverted3. Column-I Column-II

(A) An object is placed at a distance equal to focal length from (P) Maginification is ( )

pole before convex mirror

(B) An object is placed at focus before a concave mirror (Q) Magnification is (0.5)(C) An object is placed at the centre of curvature before (R) Magnification is (1/3)

a concave mirror.(D) An object is placed at a distance equal to radius of (S) Magnification is (–1)

curvature before a convex mirror.

JEE (Adv.)-Physics Geometrical Optics4. Light is incident at surface PQ of prism as shown in column-I then match the column-I with column-II (surrounding

medium is air in all cases) Column-I Column-II

(A)40°40°

i=45°

QA=100°

P R2

(P) Total internal reflection takes place at surface QR.

(B)75°75°

i=45°

QA=30°

P R2

(Q) Light emerges normally from the surface QR

(C)45°45°

i=90°

Q

A=90°

P R2

Grazing incidence

(R) Light emerges parallel to surface QR

(D)60°

30°= 2

Normal incidence

Q

P R

(S) The light ray emerges from facePR perpendicularly

(T) When light ray passes through the prismit is parallel to the base PR.

5. An equi-convex lens of refractive index µ2 and focal length f(in air) is kept in medium of refractive index µ1.y

x

µ1 µ1

µ2

y'

x'

Column A Column B(P) If lens is cut in two equal parts (1) Lens will act as a glass slab

by a plane yy'. (µ1 = 1)(Q) If lens is cut in two equal parts by a (2) Lens will be converging and focal length will

plane xx' . (µ1 = 1) change.(R) If µ1 = µ2. (3) Lens will be converging and focal length will

remain same

(S) If µ1 > µ2. (4) Lens will be diverging and focal length willchange.

Codes :P Q R S

(A) 2 4 1 3(B) 2 3 1 4(C) 3 4 2 1(D) 1 2 3 4


PART-I : ONLY ONE OPTION CORRECT TYPE QUESTIONS1. Two plane mirrors A & B are aligned parallel to each other, as shown in the

figure. A light ray is incident to an angle of 30º at a point just inside oneend of A. The plane of incidence coincides with the plane of the figure.Themaximum number of times the ray undergoes reflections (including thefirst one) before it emerges out is : (A) 28 (B) 30 (C) 32 (D) 34

2. A linear object AB is placed along the axis of a concave mirror. The object is moving towards the mirror withspeed V. The speed of the image of the point A is 4 V and the speed of the image of B is also 4V. If centre ofthe line AB is at a distance L from the mirror then length of the object AB will be :

A B

(A) 2L3

(B) 3L5

(C) L (D) 3L4

3. The view in the figure is from above a plane mirror suspended by a thread connected to the centre of themirror at point A. A scale is located 0.75 m (the distance from point A to point P) to the right of the centre ofthe mirror. Initially, the plane of the mirror is parallel to the side of the scale; and the angle of incidence of alight ray which is directed at the centre of the mirror is 30º. A small torque applied to the thread causes themirror to turn 11.5° away from its initial position. The reflected ray then intersects the scale at point Q.

Q

//////

//////

/ //// /

/ //// /

/ //// /

/ /////

//////

// // //

// //

P

11.5º

30º

0.75m

Normal to the mirror in its intial position.

Initial Position of the mirror .

A

The distance from point P to point Q on the scale is(A) 1.00 m (B) 0.56 m (C) 1.02 m (D) 0.86 m.

4. An object of height 1 cm is kept perpendicular to the principal axis of a convex mirror of radius of curvature20 cm. If the distance of the object from the mirror is 20 cm then the distance (in cm) between heads of theimage and the object will be :

(A) 6404

9(B)

6414

9(C)

403

(D) none of these

JEE (Adv.)-Physics Geometrical Optics5. A point object is kept between a plane mirror and a concave mirror facing each other. The distance between

the mirrors is 22.5 cm. Plane mirror is placed perpendicular to principal axis of concave mirror. The radius ofcurvature of the concave mirror is 20 cm. What should be the distance of the object from the concave mirrorso that after two successive reflections the final image is formed on the object itself ? (Consider firstreflection from concave mirror)(A) 5 cm (B) 15 cm (C) 10 cm (D) 7.5 cm

6. A square ABCD of side 1mm is kept at distance 15 cm infront of the concave mirror as shown in the figure.The focal length of the mirror is 10 cm. The length of the perimeter of its image will be(nearly):

(A) 8 mm (B) 2 mm (C) 12 mm (D) 6 mm7. Two plane mirrors of length L are separated by distance L and a man M2

is standing at distance L from the connecting line of mirrors as shown in figure.A man M1 is walking in a straight line at distance 2 L parallel to mirrors at speedu, then man M2 at O will be able to see image of M1 for time :

(A) uL4

(B) uL3

(C) uL6

(D) uL9

8. In the figure shown a thin parallel beam of light is incident on a plane mirror m1 at small angle ‘ ’. m2 is aconcave mirror of focal length ‘f’. After three successive reflections of this beam the x and y coordinates of theimage is

(A) x = f – d, y = f (B) x = d + f , y = f (C) x = f – d, y = – f (D) x = d – f , y = – f9. The distance between an object and its doubly magnified image by a concave mirror is:

[Assume f = focal length](A) 3 f/2 (B) 2 f/3(C) 3 f (D) depends on whether the image is real or virtual.

10. In the shown figure M1 and M2 are two concave mirrors of the same focal length 10 cm. AB and CD aretheir principal axes respectively. A point object O is kept on the line AB at a distance 15 cm from M1.The distance between the mirrors is 20 cm. Considering two successive reflections first on M1 and thenon M2. The distance of final image from the line AB is :

3cm.

15cm

AD

BC

M1

M2

O

(A) 3 cm (B) 1.5 cm (C) 4.5 cm (D) 1 cm

JEE (Adv.)-Physics Geometrical Optics11. A man observes a coin placed at the bottom of a beaker which contains two immiscible liquids of

refractive indices 1.2 and 1.4 as shown in the figure. A plane mirror is also placed on the surface ofliquid. The distance of image (from mirror) of coin in mirror as seen from medium A of refractive index 1.2by an observer just above the boundary of the two media is :

(A) 18 cm (B) 12 cm(C) 9 cm (D) none of these

12. An observer can see through a pinhole the top end of a thin rod of height h,placed as shown in the figure. The beaker height is 3h and its radius h.When the beaker is filled with a liquid up to a height 2h, he can see thelower end of the rod. Then the refractive index of the liquid is :

(A) 25

(B) 25

(C) 23

(D) 23

13. A bird is flying up at an angle sin 1 (3/5) with the horizontal. A fish in a pond looks at that bird when it isvertically above the fish. The angle at which the bird appears to fly (to the fish) is : [nwater = 4/3](A) sin 1 (3/5) (B) sin 1 (4/5) (C) 45º (D) sin 1 (9/16)

14. An object is placed 30 cm (from the reflecting surface) in front of a block of glass 10 cm thick having itsfarther side silvered. The final image is formed at 23.2 cm behind the silvered face. The refractive indexof glass is :(A) 1.41 (B) 1.46 (C) 200/ 132 (D) 1.61

15. A ray of light strikes a plane mirror at an angle of incidence 45º as shown in the figure. After reflection,the ray passes through a prism of refractive index 1.50, whose apex angle is 4º. The angle throughwhich the mirror should be rotated if the total deviation of the ray is to be 90º is :

(A) 10 clockwise (B) 10 anticlockwise (C) 20 clockwise (D) 20 anticlockwise16. In the given figure a parallel beam of light is incident on the upper part of a prism of angle 1.8º and R.I.

3/2. The light coming out of the prism falls on a concave mirror of radius of curvature 20 cm. Thedistance of the point (where the rays are focused after reflection from the mirror) from the principal axisis: [use = 3.14]

(A) 9 cm (B) 1.5 7 mm (C) 3.14 mm (D) none of these

JEE (Adv.)-Physics Geometrical Optics17. The maximum refractive index of a material, of a prism of apex angle 90º, for which light may be transmitted

is:

(A) 3 (B) 1.5 (C) 2 (D) None of these

18. A prism having an apex angle of 40 and refractive index of 1.50 is located in front of a vertical plane mirror asshown in the figure. A horizontal ray of light is incident on the prism. The total angle through which the ray isdeviated is:

M

P

(A) 40 clockwise (B) 1780 clockwise (C) 20 clockwise (D) 80 clockwise19. For a prism of apex angle 450, it is found that the angle of emergence is 450 for grazing incidence. Calculate

the refractive index of the prism.(A) (2)1/2 (B) (3)1/2 (C) 2 (D) (5)1/2

20. In the figure shown a point object O is placed in air. A spherical boundary of radius of curvature 1.0 mseparates two media. AB is principal axis. The refractive index above AB is 1.6 and below AB is 2.0. Theseparation between the images formed due to refraction at spherical surface is:

(A) 12 m (B) 20 m (C) 14 m (D) 10 m21. A beam of diameter ‘ d ‘ is incident on a glass hemisphere as shown in the figure. If the radius of curvature of

the hemisphere is very large in comparison to d, then the diameter of the beam at the base of the hemispherewill be :

(A)43

d (B) d (C) 3d

(D)32

d

22. A thin lens of focal length f and its aperture diameter d, forms a real image of intensity I. Now the central partof the aperture upto diameter (d/2) is blocked by an opaque paper. The focal length and image intensitywould change to :(A) f/2, I/2 (B) f, I/4 (C) 3f/4, I/2 (D) f, 3I/4

23. In the given figure an object ' O ' is kept in air in front of a thin plano convex lens of radius of curvature10 cm. It's refractive index is 3/2 and the medium towards right of plane surface is water of refractiveindex 4/3. What should be the distance ' x ' of the object so that the rays become parallel finally.


JEE (Adv.)-Physics Geometrical Optics24. Which one of the following spherical lenses does not exhibit dispersion? The radii of curvature

of the surface of the lenses are as given in the diagrams.

(A) (B) (C) (D)

25. A beam of white light is incident on hollow prism of glass as shown in figure. Then :

(i

white light

(A) the light emerging from prism gives no dispersion(B) the light emerging from prism gives spectrum but the bending of all colours is away from base.(C) the light emerging from prism gives spectrum, all the colours bend towards base, the violet the most

and red the least.(D) the light emerging from prism gives spectrum, all the colours bend towards base, the violet the least

and red the most.26. White light is incident on the interface of glass and air as shown in the figure. If green light is just totally

internally reflected then the emerging ray in air contains :

(A) yellow, orange, red (B) violet, indigo, blue(C) all colours (D) all colours except green

PART-II : INTEGER TYPE QUESTIONS1. Two spherical mirrors (convex and concave) having the same focal length of 36 cm are arranged as shown in

figure so that their optical axes coincide. The separation between the mirrors is 1 m. At what distance fromthe concave mirror should an object be placed so that its images formed by the concave and convex mirrorsindependently are identical in size?

2. A plane mirror 50 cm long, is hung on a vertical wall of a room, with its lower edge 50 cm above theground. A man stands infront of the mirror at a distance 2 m away from the mirror. If his eyes are at aheight 1.8 m above the ground, If the length (distance between the extreme points of the visible regionperpendicular to the mirror) of the floor visible to him due to reflection from the mirror is 45/x. Then valueof x is :

JEE (Adv.)-Physics Geometrical Optics3. A convex mirror and a concave mirror each of focal length f are placed coaxially. They are separated by

4f and their reflecting surfaces face each other. A point object is kept on the principle axis at a distancex from the concave mirror such that final image after two reflections, first on the concave mirror, is on

the object itself. If value of x is ( )N N f.- Then find value of N.

4. A concave mirror forms the real image of a point source lying on the optical axis at a distance of 50 cmfrom the mirror. The focal length of the mirror is 25 cm. The mirror is cut into two halves and its halvesare drawn a distance of 1 cm apart (from each other) in a direction perpendicular to the optical axis.How far from each other the images formed by the two halves of the mirror be (in cm) ?

5. The average size of an Indian face is 24 × 16 cm2. Find the minimum size of a plane mirror required tosee the face completely by:(i) one eyed man (ii) two eyed man (Distance between eyes is = 4 cm)

6. Two plane mirrors form an angle of 120°. The distance between the two images of a point source formedin them is 20 cm. If the distance from the light source to the point where the mirrors touch if it lies onthe bisector of angle formed by the mirrors is

x

3. Find x.

7. In the figure shown find the total magnification after two successive reflections first on M1 and thenon M2 .

8. A cylindrical vessel, whose diameter and height both are equal to 30 cm, is placed on a horizontalsurface and a small particle P is placed in it at a distance of 5.0 cm from the corner. An eye is placedat a position such that the edge of the bottom is just visible (see figure). The particle P is in the plane

of drawing. If water is poured in the vessel upto a height 5 x cmx 3

æ ö÷ç ÷ç ÷ç ÷÷ç -è øto make the particle P visible. Then

x is :

9. A container contains water upto a height of 20 cm and there is a point source at the centre of the bottomof the container. A rubber ring of radius r floats centrally on the water. The ceiling of the room is 2.0 m

above the water surface. (a) If the radius of the shadow of the ring formed on the ceiling is 169

xm if

r = 15 cm. (b) If the maximum value of r for which the shadow of the ring is formed on the ceiling is y

5 7m. Then x + y is : (Refractive index of water = 4/3).

10. See the figure.

Find the distance (from the mirror) of final image formed by the system.

JEE (Adv.)-Physics Geometrical Optics11. Light is incident from glass to air. The variation of the angle of deviation with the angle of incidence i

for 0 < i < 90° is shown. Find the values of x,y and z, (if critical angle C = 60º), shown in the figure.

12. A pole of length 2.00 m stands half dipped in a swimming pool with water level 1 m higher than the bed(bottom). The refractive index of water is 4/3 and sunlight is coming at an angle of 37° with the vertical.If the length of the shadow of the pole on the bed is x/4. Find x. Use sin–1 (0.45) = 26.8°, tan(26.8°) = 0.5

13. The x-y plane is the boundary between two transparent media. Medium-1 with z > 0 has refractive index

2 and medium -2 with z < 0 has a refractive index 3 A ray of light in medium-1 given by the vector

A = 6 3 i + 8 3 j 10 k is incident on the plane of separation. If the unit vector in the direction of

refracted ray in medium 2 is a b b cˆ ˆ ˆr i j k

55 2 2= + - . Then value of a + b + c is :

14. A small piece of wood is floating on the surface of a m deep lake. When the sun is vertically above thepiece its shadow is formed at A. When the sun is just setting the shadow of the piece is formed at B. If therefractive index of water is 4/3 then find the distance between A and B.

15. An insect at point ‘P’ sees its two images in the water-mirror system as shown in the figure. One image isformed due to direct reflection from water surface and the other image is formed due to refraction, reflection& again refraction by water mirror system in order. Find the separation between the two images. M has focallength 60 cm. (nw = 4/3) :

16. An object lies in front of a thick parallel glass slab, the bottom of which is polished. If the distancebetween first two images formed by bottom surface is 4cm then find the thickness of the slab.[Assume nglass = 3/2 and paraxial rays]

17. In the given figure, a triangular prism of glass is located inside water. A ray, incident normally, on one of the

faces, is totally reflected from face BC. If refractive index of glass is greater than 4 x

3. Find x.

18. Light travelling in air falls at an incidence angle of 2° on one refracting surface of a prism of refractiveindex 1.5 and angle of refraction 4º. The medium on the other side is water (n = 4/3). Find the deviationproduced by the prism.

JEE (Adv.)-Physics Geometrical Optics19. A hemispherical portion of the surface of a solid glass sphere (µ = 1.5)

of radius r (surrounding is air) is silvered to make the inner sidereflecting. An object is placed on the axis of the hemisphere at a distance3r from the centre of the sphere. The light from the object is refracted atthe unsilvered part, then reflected from the silvered part and againrefracted at the unsilvered part. If final image of formed at distance Nrfrom D. Then N is :

20 The internal surface of the walls of a sphere is specular (i.e. reflecting). The radius of the sphere is R = 36 cm.A point source S is placed at a distance R/2 from the centre of the sphere and sends light to the remote partof the sphere. Then where will the image of the source be after two successive reflections from the remotepart and then the nearest wall of the sphere will be located from nearest wall (in cm) ?

21. A quarter cylinder of radius R and refractive index 1.5 is placed on a table. A pointobject P is kept at a distance of mR/3 from it.Find the value of m for which a ray fromP will emerge parallel to the table as shown in the figure.

22. A converging lens of focal length 15 cm and a converging mirror of focal length 10 cm are placed 50 cmapart with common principal axis. A point source is placed in between the lens and the mirror at adistance of 40 cm from the lens. Find the distance between the final two images formed.

PART - III : ONE OR MORE THAN ONE CORRECT OPTIONS1. The image (of a real object) formed by a concave mirror is twice the size of the object. The focal length of the

mirror is 20 cm.The distance of the object from the mirror is (are) :(A) 10 cm (B) 30 cm (C) 25 cm (D) 15 cm

2. Which of the following statements are incorrect for spherical mirrors.(A) a concave mirror forms only virtual images for any position of real object(B) a convex mirror forms only virtual images for any position of a real object(C) a concave mirror forms only a virtual diminished image of an object placed between its pole and the focus

(D) a convex mirror forms a virtual enlarged image of an object if it lies between its pole and the focus.3. A flat mirror M is arranged parallel to a wall W at a distance L from it as shown in the figure. The light

produced by a point source S kept on the wall is reflected by the mirror and produces a light patch on the wall.The mirror moves with velocity v towards the wall.

LS

wall w

VM

(A) The patch of light will move with the speed v on the wall.(B) The patch of light will not move on the wall.(C) As the mirror comes closer the patch of light will become larger and shift away from the wall with

speed larger than v.(D) The width of the light patch on the wall remains the same.

4. Two plane mirrors are inclined to each other with their reflecting faces making acute angle. A light ray isincident on one plane mirror. The total deviation after two successive reflections is:(A) independent of the initial angle of incidence(B) independent of the angle between the mirrors(C) dependent on the initial angle of incidence(D) dependent on the angle between the mirrors.

JEE (Adv.)-Physics Geometrical Optics5. An object is kept on the principal axis of a convex mirror of focal length 10 cm at a distance of 10 cm

from the pole. The object starts moving at a velocity 20 mm/sec towards the mirror at angle 30º with theprincipal axis. What will be the speed of its image and direction with the principal axis at that instant.

(A) speed = 57

4 mm/sec (B) speed =

275

mm/sec

(C) tan 1 (2

3) with the principal axis (D) none of these

6. A ray of monochromatic light is incident on the plane surface of separation between two media x and y withangle of incidence ‘i’ in the medium x and angle of refraction ‘r’ in the medium y. The graph shows the relationbetween sin r and sin i.

(A) the speed of light in the medium y is (3)1/2 times than in medium x.

(B) the speed of light in the medium y is (1/3)1/2 times than in medium x.

(C) the total internal reflection can take place when the incidence is in x.

(D) the total internal reflection can take place when the incidence is in y.

7. In the figure shown a point object O is placed in air on the principal axis. The radius of curvature of thespherical surface is 60 cm. If is the final image formed after all the refractions and reflections.

(A) If d1 = 120 cm, then the ‘ If ‘ is formed on ‘ O ‘ for any value of d2.

(B) If d1 = 240 cm, then the ‘ If ‘ is formed on ‘ O ‘ only if d2 = 360 cm.

(C) If d1 = 240 cm, then the ‘ If ‘ is formed on ‘ O ‘ for all values of d2.

(D) If d1 = 240 cm, then the ‘ If ‘ cannot be formed on ‘ O ‘.

8. For the refraction of light through a prism kept in air(A) For every angle of deviation there are two angles of incidence.(B) The light travelling inside an isosceles prism is necessarily parallel to the base when prism is set for

minimum deviation.(C) There are two angles of incidence for maximum deviation.(D) Angle of minimum deviation will increase if refractive index of prism is increased keeping the outside

medium unchanged.

9. For refraction through a small angled prism, the angle of deviation (nsurrounding < nprism)

(A) increases with the increase in refractive index of the prism.

(B) will be doubled if refractive index of the prism is doubled.

(C) is directly proportional to the angle of the prism .

(D) will decrease with the increase in refractive index of the prism.

JEE (Adv.)-Physics Geometrical Optics10. Two refracting media are separated by a spherical interface as shown in the

figure. P P is the principal axis, 1 and 2 are the refractive indices of medium

of incidence and medium of refraction respectively. Then

(A) if 2 > 1, then there cannot be a real image of real object

(B) if 2 > 1, then there cannot be a real image of virtual object

(C) if 1 > 2, then there cannot be a virtual image of virtual object

(D) if 1 > 2, then there cannot be a real image of real object

11. An object O is kept infront of a converging lens of focal length 30 cm behind whichthere is a plane mirror at 15 cm from the lens as shown in the figure.(A) the final image is formed at 60 cm from the lens towards right of it(B) the final image is at 60 cm from lens towards left of it(C) the final image is real(D) the final image is virtual.

12. Which of the following cannot form real image of a real object ?(A) concave mirror (B) convex mirror (C) plane mirror (D) diverging lens

13. The values of d1 & d2 for final rays to be parallel to the principal axis are : (focal lengths of the lenses arewritten above the respective lenses in the given figure)

(A) d1 = 10 cm, d2 = 15 cm (B) d1 = 20 cm, d2 = 15 cm(C) d1 = 30 cm, d2 = 15 cm (D) None of these

14. An equiconvex lens of refractive index n2 is placed such that the refractive index of the surroundingmedia is as shown. Then the lens :

(A) must be diverging if n2 is less than the arithmetic mean of n1 and n3

(B) must be converging if n2 is greater than the arithmetic mean of n1 and n3

(C) may be diverging if n2 is less than the arithmetic mean of n1 and n3

(D) will neither be diverging nor converging if n2 is equal to arithmetic mean of n1 and n3

15. A man wants to photograph a white donkey as a Zebra after fitting a glass with black streaks onto the lensof his camera.(A) The image will look like a white donkey on the photograph.(B) The image will look like a Zebra on the photograph(C) The image will be more intense compared to the case in which no such glass is used.(D) The image will be less intense compared to the case in which no such glass is used.

JEE (Adv.)-Physics Geometrical Optics16. Which of the following quantities related to a lens depend on the wavelength of the incident light?

(A) power (B) focal length

(C) chromatic aberration (D) radii of curvature

17. A narrow beam of white light goes through a slab having parallel faces

(A) The light never splits in different colours

(B) The emergent beam is white

(C) The light inside the slab is split into different colours

(D) The light inside the slab is white

18. By properly combining two prisms made of different materials, it is possible to

(A) have dispersion without average deviation (B) have deviation without dispersion

(C) have both dispersion and average deviation (D) have neither dispersion nor average deviation

PART - IV : COMPREHENSIONSComprehension # 1

Chromatic Aberration

The image of a white object in white light formed by a lens is usually coloured and blurred. This defectof image is called chromatic aberration and arises due to the fact that focal length of a lens is differentfor different colours. As R.I. of lens is maximum for violet while minimum for red, violet is focusednearest to the lens while red farthest from it as shown in figure.

As a result of this, in case of convergent lens if a screen is placed at FV centre of the image willbe violet and focused while sides are red and blurred. While at FR, reverse is the case, i.e., centre willbe red and focused while sides violet and blurred. The difference between fV and fR is a measure of thelongitudinal chromatic aberration (L.C.A), i.e.,L.C.A. = fR – fV = – df with df = fV – fR .....(1)

However, as for a single lens,

21 R1

R1)1(

f1

.....(2)

212 R

1R1d

fdf

.....(3)

Dividing Eqn. (3) by (2) ;

)1(d

fdf

)1(d

= dispersive power ........(4)

And hence, from Eqns. (1) and (4),L.C.A. = –df = fNow, as for a single lens neither f nor can be zero, we cannot have a single lens free from chromaticaberration.

JEE (Adv.)-Physics Geometrical OpticsCondition of Achromatism :In case of two thin lenses in contact

21 f1

f1

F1

i.e., 2FdF

= 22

221

1

fdf

fdf

The combination will be free from chromatic aberration if dF = 0

i.e., 22

221

1

fdf

fdf

= 0

which with the help of Eqn. (4) reduces to

22

2221

11

ff

ff

= 0 i.e.,2

2

1

1

ff = 0 ...........(5)

This condition is called condition of achromatism (for two thin lenses in contact) and the lens combinationwhich satisfies this condition is called achromatic lens, from this condition, i.e., from Eqn. (5) it is clearthat in case of achromatic doublet :(1) The two lenses must be of different materials.

Since, if 1 = 2 , 0f1

f1

21i.e.,

F1

= 0 or F =

i.e., combination will not behave as a lens, but as a plane glass plate.(2) As 1 and 2 are positive quantities, for equation (5) to hold, f1 and f2 must be of opposite nature, i.e.if one of the lenses is converging the other must be diverging.(3) If the achromatic combination is convergent,

fC < fD and asD

C

ff

= D

C , C < D

i.e., in a convergent achromatic doublet, convex lens has lesser focal length and dispersive power thanthe divergent one.

1. Chromatic aberration in the formation of images by a lens arises because :(A) of non-paraxial rays.(B) the radii of curvature of the two sides are not same.(C) of the defect in grinding.(D) the focal length varies with wavelength.

2. Chromatic aberration of a lens can be corrected by :(A) providing different suitable curvatures of its two surfaces.(B) proper polishing of its two surfaces.(C) suitably combining it with another lens.(D) reducing its aperture.

3. A combination is made of two lenses of focal lengths and in contact ; the dispersive powers of thematerials of the lenses are and . The combination is achromatic when :(A) = 0, = 2 0, = 2 (B) = 0, = 2 0, = /2(C) = 0, = 2 0, = – /2 (D) = 0, = 2 0, = – 2

4. The dispersive power of crown and flint glasses are 0.02 and 0.04 respectively. An achromatic converginglens of focal length 40 cm is made by keeping two lenses, one of crown glass and the other of flintglass, in contact with each other. The focal lengths of the two lenses are :(A) 20 cm and 40 cm (B) 20 cm and –40 cm(C) –20cm and 40 cm (D) 10 cm and –20cm

5. Chromatic aberration in a spherical concave mirror is proportional to :(A) (B) 2 (C) 1/ (D) None of these

JEE (Adv.)-Physics Geometrical OpticsComprehension # 2

The ciliary muscles of eye control the curvature of the lens in the eye and hence can alter the effectivefocal length of the system. When the muscles are fully relaxed, the focal length is maximum. When themuscles are strained the curvature of lens increases (that means radius of curvature decreases) andfocal length decreases. For a clear vision the image must be on retina. The image distance is thereforefixed for clear vision and it equals the distance of retina from eye-lens. It is about 2.5 cm for a grown-upperson (Refer the figure below).

A person can theoretically have clear vision of objects situated at any large distance from the eye. Thesmallest distance at which a person can clearly see is related to minimum possible focal length. Theciliary muscles are most strained in this position. For an average grown-up person minimum distance ofobject should be around 25 cm.

A person suffering for eye defects uses spectacles (eye glass). The function of lens of spectacles is toform the image of the objects within the range in which person can see clearly. The image of thespectacle-lens becomes object for eye-lens and whose image is formed on retina.

The number of spectacle-lens used for the remedy of eye defect is decided by the power of the lensrequired and the number of spectacle-lens is equal to the numerical value of the power of lens with sign.

For example power of lens required is +3D (converging lens of focal length 3100

cm) then number of lens

will be + 3.

For all the calculations required you can use the lens formula and lens maker's formula. Assume thatthe eye lens is equiconvex lens. Neglect the distance between eye lens and the spectacle lens.

6. Minimum focal length of eye lens of a normal person is

(A) 25 cm (B) 2.5 cm (C) 925

cm (D) 1125

cm

7. Maximum focal length of eye lens of normal person is

(A) 25 cm (B) 2.5 cm (C) 925

cm (D) 1125

cm

8. A nearsighted man can clearly see object only upto a distance of 100 cm and not beyond this. Thenumber of the spectacles lens necessary for the remedy of this defect will be.

9. A farsighted man cannot see object clearly unless they are at least 100 cm from his eyes. The numberof the spectacles lens that will make his range of clear vision equal to an average grown up person

(A) + 1 (B) – 1 (C) + 3 (D) – 3


PART - I : JEE (ADVANCED) / IIT-JEE PROBLEMS (PREVIOUS YEARS)* Marked Questions may have more than one correct option.1. Two beams of red and violet colours are made to pass separately through a prism (angle of the prism is 60º).

In the position of minimum deviation, the angle of refraction will be [JEE' 2008; 3/163](A) 30º for both the colours (B) greater for the violet colour(C) greater for the red colour (D) equal but not 30º for both the colours

2. A light beam is traveling from Region I to Region IV (Refer Figure). The refractive index in Regions I, II, III and

IV are n0, 2n0 , 6

n0 and 8n0 , respectively. The angle of incidence for which the beam just misses entering

Region IV is [JEE' 2008; 3/163]

(A) sin–1 43

(B) sin–1 81

(C) sin–1 41

(D) sin–1 31

3. An optical component and an object S placed along its optic axis are given in Column I. The distancebetween the object and the component can be varied. The properties of images are given in Column II.Match all the properties of images from Column II with the appropriate components given in Column I.

[JEE' 2008, 6/163, –1]Column I Column II

(A) (p) Real image

(B) (q) Virtual image

(C) (r) Magnified image

(D) (s) Image at infinity

JEE (Adv.)-Physics Geometrical Optics4. A ball is dropped from a height of 20 m above the surface of water in a lake. The refractive index of water is

4/3. A fish inside the lake, in the line of fall of the ball, is looking at the ball. At an instant, When the ball is12.8 m above the water surface, the fish sees the speed of ball as [Take g = 10 m/s2]

[JEE' 2009, 3/160, –1](A) 9 m/s (B) 12 m/s (C) 16 m/s (D) 21.33 m/s

5.* A student performed the experiment of determination of focal length of a concave mirror by u-v method usingan optical bench of length 1.5 meter. The focal length of the mirror used is 24 cm. The maximum error in thelocation of the image can be 0.2 cm. The 5 sets of (u, v) values recorded by the student (in cm) are : (42, 56),(48, 48), (60, 40), (66, 33), (78, 39). The data set(s) that cannot come from experiment and is (are) incorrectlyrecorded, is (are) [JEE' 2009, 4/160, –1](A) (42, 56) (B) (48, 48) (C) (66, 33) (D) (78, 39)

6.* A biconvex lens of focal length 15 cm is in front of a plane mirror. The distance between the lens and the mirroris 10 cm. A small object is kept at a distance of 30 cm from the lens. The final image is(A) virtual and at a distance of 16 cm from the mirror [IIT-JEE 2010](B) real and at a distance of 16 cm from the mirror(C) virtual and at a distance of 20 cm from the mirror(D) real and at a distance of 20 cm from the mirror

7. A ray OP of monochromatic light is incident on the face AB of prism ABCD nearvertex B at an incident angle of 60° (see figure). If the refractive index of the materialof the prism is 3 , which of the following is (are) correct? [IIT-JEE 2010](A) The ray gets totally internally reflected at face CD(B) The ray comes out through face AD(C) The angle between the incident ray and the emergent ray is 90°(D) The angle between the incident ray and the emergent ray is 120°

8. Two transparent media of refractive indices 1 and 3 have a solid lens shaped transparent material of refractiveindex 2 between them as shown in figures in Column II. A ray traversing these media is also shown in thefigures. In Column I different relationships between 1, 2 and 3 are given. Match them to the ray diagramsshown in Column II. [IIT-JEE 2010]

Column I Column II

(A) 1 < 2 (p)

(B) 1 > 2 (q)

(C) 2 = 3 (r)

(D) 2 > 3 (s)

(t)

JEE (Adv.)-Physics Geometrical Optics9. The focal length of a thin biconvex lens is 20 cm. When an object is moved from a distance of 25 cm in front of

it to 50 cm, the magnification of its image changes from m25 to m50. The ratio 25

50

mm

is [IIT-JEE 2010]

10. Image of an object approaching a convex mirror of radius of curvature 20 m along its optical axis is observed to

move from 253 m to

507 m in 30 seconds. What is the speed of the object in km per hour ? [IIT-JEE 2010]

11. A large glass slab 53

of thickness 8 cm is placed over a point source of light on a plane surface. It is seen

that light emerges out of the top surface of the slab from a circular area of radius R cm. What is the value of R?[IIT-JEE 2010]

12. A light ray traveling in glass medium is incident on glass-air interface at an angle of incidence . The reflected

(R) and transmitted (T) intensities, both as function of , are plotted. The correct sketch is : [IIT-JEE 2011]

(A)

T100%

Inte

nsity

0 90°

R(B)

T100%

Inte

nsity

0 90°

R

(C)

T100%

Inte

nsity

0 90°

R

(D)

T100%

Inte

nsity

0 90°

R

13. Water (with refractive index = 43

) in a tank is 18 cm deep. Oil of refractive index 74

lies on water making a

convex surface of radius of curvature ‘R = 6 cm’ as shown. Consider oil to act as a thin lens. An object ‘S’ isplaced 24 cm above water surface. The location of its image is at ‘x’ cm above the bottom of the tank.Then ‘x’ is : [IIT-JEE 2011]

=4/3

=7/4

=1.0

S

R=6cm

JEE (Adv.)-Physics Geometrical Optics14. A biconvex lens is formed with two thin plano-convex lenses as shown in the figure, Refractive index n of the

first lens is 1.5 and that of the second lens is 1.2. Both the curved surfaces are of the same radius of curvatureR = 14 cm. For this biconvex lens, for an object distance of 40 cm, the image distance will be :-

n=1.2n=1.5

R=14cm(A) –280.0 cm (B) 40.0 cm (C) 21.5 cm (D) 13.3 cm

Paragraph for Questions 15 and 16Most materials have the refractive index, n>1. So, when a light ray from air enters a naturally occurring

material, then by Snell's law, 1 2

2 1

sinsin

n

n , it is understood that the refracted ray bends towards the normal.

But it never emerges on the same side of the normal as the incident ray. According to electromagnetism, the

refractive index of the medium is given by the relation, r r

cn

v, where c is the speed of

electromagnetic waves in vacuum, v its speed in the medium, r and r are the relative permittivity andpermeability of the medium respectively.In normal materials, both r and r are positive, implying positive n for the medium. When both r and r arenegative, one must choose the negative root of n. Such negative refractive index materials can now be artificiallyprepared and are called meta-materials. They exhibit significantly different optical behaviour, without violatingany physical laws. Since n is negative, it results in a change in the direction of propagation of the refractedlight. However, similar to normal materials, the frequency of light remains unchanged upon refraction even inmeta-materials. [IIT-JEE-2012]

15. For light incident from air on a meta-material, the appropriate ray diagram is

(A)

1

2

Meta-material

Air

(B)

1

Meta-material

Air

2

(C)

1

2

Meta-material

Air

(D)

1

2

Meta-material

Air

JEE (Adv.)-Physics Geometrical Optics16. Choose the correct statement.

(A) The speed of light in the meta-material is v = c|n|

(B) The speed of light in the meta-material is v = cn

(C) The speed of light in the meta-material is v = c.(D) The wavelength of the light in the meta-material ( m) is given by m = air|n|, where air is the wavelength of

the light in air. [IIT-JEE 2012]

17. A ray of light travelling in the direction 1 ˆ ˆi 3j2

is incident on a plane mirror. After reflection, it travels along

the direction 1 ˆ ˆi 3 j2

. The angle of incidence is :- [JEE-Advance-2013](A) 30° (B) 45° (C) 60° (D) 75°

18. The image of an object, formed by a plano-convex lens at a distance of 8 m behind the lens, is real and is one-

third the size of the object. The wavelength of light inside the lens is 23

times the wavelength in free space.The radius of the curved surface of the lens is :- [JEE-Advance-2013](A) 1 m (B) 2 m (C) 3 m (D) 6 m

19. A right angled prism of refractive index µ1 is placed in a rectangular block of refractive index µ2, which issurrounded by a medium of refractive index µ3, as shown in the figure. A ray of light 'e' enters the rectangularblock at normal incidence. Depending upon the relationships between µ1, µ2, and µ3, it takes one of the fourpossible paths 'ef', 'eg', 'eh' or 'ei'. [JEE-Advance-2013]Match the paths in List I with conditions of refractive indices in List II andselect the correct answer using the codes given below the lists :

45°

µ1

µ2 µ3

e

i

f

g

h

List I List IIP. e f 1. 1 22

Q. e g 2. µ2 > µ1 and µ2 > µ3

R. e h 3. µ1 = µ2

S. e i 4. 2 1 22 and µ2 > µ3

Codes :P Q R S P Q R S

(A) 2 3 1 4 (B) 1 2 4 3(C) 4 1 2 3 (D) 2 3 4 1

20. A transparent thin film of uniform thickness and refractive index n1 = 1.4 iscoated on the convex spherical surface of radius R at one end of a long solidglass cylinder of refractive index n2 = 1.5, as shown in the figure. Rays oflight parallel to the axis of the cylinder traversing through the film from air toglass get focused at distance f1 from the film, while rays of light traversingfrom glass to air get focused at distance f2 from the film. Then Air n2

n1

[JEE-Advance-2014]

(A) |f1| = 3R (B) |f1| = 2.8 R (C) |f2| = 2R (D) |f2| = 1.4 R21. A point source S is placed at the bottom of a transparent block of height 10 mm and refractive index 2.72. It

is immersed in a lower refractive index liquid as shown in the figure. It is found that the light emerging from theblock to the liquid forms a circular bright spot of diameter 11.54 mm on the top of the block. The refractiveindex of the liquid is :- [JEE-Advance-2014]

Liquid

Block

S(A) 1.21 (B) 1.30 (C) 1.36 (D) 1.42

JEE (Adv.)-Physics Geometrical Optics22. Four combinations of two thin lenses are given in List-I. The radius of curvature of all curved surfaces is r and

the refractive index of all the lenses is 1.5. Match lens combinations in List-I with their focal length in List-IIand select the correct answer using the code given below the lists. [JEE-Advance-2014]

List–I List–II

(P) (1) 2r

(Q) (2)r2

(R) (3) –r

(S) (4) r

Code :(A) P-1, Q-2, R-3, S-4 (B) P-2, Q-4, R-3, S-1 (C) P-4, Q-1, R-2, S-3 (D) P-2, Q-1, R-3, S-4

23. Consider a concave mirror and a convex lens (refractive index = 1.5) of focal length 10 cm each, separated bya distance of 50 cm in air (refractive index = 1) as shown in the figure. An object is placed at a distance of15 cm from the mirror. Its erect image formed by this combination has magnification M1. When the set-up is

kept in a medium of refractive index 7/6 the magnification becomes M2. The magnitude 2

1

MM is :

[JEE-Advance-2015]

\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

15cm

50cm

24. Two identical glass rods S1 and S2 (refractive index = 1.5) have one convex end of radius of curvature 10 cm.They are placed with the curved surfaces at a distance d as shown in the figure, with their axes (shown by thedashded line) aligned. When a point source of light P is placed inside rod S1 on its axis at a distance of 50 cmfrom the curved face, the light rays emanating from it are found to be parallel to the axis inside S2. Thedistance d is : [JEE-Advance-2015]

d(A) 60 cm (B) 70 cm (C) 80 cm (D) 90 cm

25. A monochromatic beam of light is incident at 60° on one face of an equilateral prism of refractive index n and

emerges from the opposite face making an angle (n) with the normal (see the figure). For n = 3 the value

of is 60° and d

mdn

. The value of m is : [JEE-Advance-2015]

60°

JEE (Adv.)-Physics Geometrical OpticsParagraph for Question No. 26 and 27

Light guidance in an optical fiber can be understood by considering a structure comprising of thin solid glasscylinder of refractive index n1 surrounded by a medium of lower refractive index n2. The light guidance in thestructure takes place due to successive total internal reflections at the interface of the media n1 and n2 asshown in the figure. All rays with the angle of incidence i less than a particular value of im are confined in themedium of refractive index n1. The numerical aperture (NA) of the structure is defined as sin im.

[JEE-Advance-2015]

n > n1 2

CladdingAir

i

n2

Core

n1

26. For two structures namely S1 with n1 = 45 / 4 and n2 = 3/2, and S2 with n1 = 8/5 and n2 = 7/5 and taking the

refractive index of water to be 4/3 and that of air to be 1, the correct option(s) is (are)

(A) NA of S1 immersed in water is the same as that of S2 immersed in liquid of refractive index 16

3 15.

(B) NA of S1 immersed in liquid of refractive index 6

15 is the same as that of S2 immersed in water..

(C) NA of S1 placed in air is the same as that of S2 immersed in liquid of refractive index 4

15.

(D) NA of S1 placed in air is the same as that of S2 placed in water.27. If two structures of same cross-sectional area, but dif ferent numerical apertures NA1 and

NA2 (NA2 < NA1) are joined longitudinally, the numerical aperture of the combined structure is

(A) 1 2

1 2

NA NANA NA (B) NA1 + NA2 (C) NA1 (D) NA2

28. A parallel beam of light is incident from air at an angle on the side PQ of a right angled

triangular prism of refractive index n 2 . Light undergoes total internal reflection in the

prism at the face PR when has a minimum value of 45°. The angle of the prism is:

[JEE-Advance-2016]n= 2–

Q R

P

(A) 15° (B) 22.5° (C) 30° (D) 45°29. A transparent slab of thickness d has a refractive index n(z) that increases

with z. Here z is the vertical distance inside the slab, measured from thetop. The slab is placed between two media with uniform refractive indicesn1 and n2 (> n1), as shown in the figure. A ray of light is incident with angle

i from medium 1 and emerges in medium 2 with refraction angle f witha lateral displacement . Which of the following statement(s) is(are) true?

[JEE-Advance-2016]

n = constant1 i 1

z

n = constant2

n(z)d

f2

(A) is independent of n2 (B) is dependent on n(z)

(C) n1sin i = (n2 – n1)sin f (D) n1sin i = n2sin f

JEE (Adv.)-Physics Geometrical Optics30. A plano-convex lens is made of a material of refractive index n. When a small object is placed 30 cm away in

front of the curved surface of the lens, an image of double the size of the object is produced. Due to reflectionfrom the convex surface of the lens, another faint image is observed at a distance of 10 cm away from the lens.Which of the following statement(s) is(are) true ? [JEE-Advance-2016](A) The refractive index of the lens is 2.5(B) The radius of curvature of the convex surface is 45 cm(C) The faint image is erect and real(D) The focal length of the lens is 20 cm.

31. For an isosceles prism of angle A and refractive index µ, it is found that the angle of minimum deviation

m = A. Which of the following options is/are correct ? [JEE-Advance-2017](A) At minimum deviation, the incident angle i1 and the refracting angle r1 at the first refracting surface are

related by r1 = (i1/2)

(B) For this prism, the refractive index µ and the angle of prism A are related as 11

A cos2 2

(C) For this prism, the emergent ray at the second surface will be tangential to the surface when the angle

of incidence at the first surface is 1 21

Ai sin sin A 4cos 1 cos A

2(D) For the angle of incidence i1 = A, the ray inside the prism is parallel to the base of the prism.

32. A monochromatic light is travelling in a medium of refractive indexn = 1.6. It enters a stack of glass layers from the bottom side at an angle

= 30°. The interfaces of the glass layers are parallel to each other. Therefractive indices of different glass layers are monotonically decreasingas nm = n – m n, where nm is the refractive index of the mth slab and

n = 0.1 (see the figure). The ray is refracted out parallel to the interfacebetween the (m – 1)th and mth slabs from the right side of the stack. Whatis the value of m ? [JEE-Advance-2017]

m n – m nm – 1 n – (m – 1) n

321

n – 3 nn – 2 nn – nn

33. Sunlight of intensity 1.3 kW m–2 is incident normally on a thin convex lens of focal length 20 cm. Ignore theenergy loss of light due to the lens and assume that the lens aperture size is much smaller than its focallength. The average intensity of light, in kW m–2, at a distance 22 cm from the lens on the other side is_________. [JEE-Advance-2018]

34. A wire is bent in the shape of a right angled triangle and isplaced in front of a concave mirror of focal length f, as shown inthe figure. Which of the figures shown in the four optionsqualitatively represent(s) the shape of the image of the bentwire ? (These figures are not to scale.)? [JEE-Advance-2018]

ff2

45°

(A) > 45° (B)

(C) 0< <45° (D)

JEE (Adv.)-Physics Geometrical Optics35. A thin convex lens is made of two materials with refractive indices n1 and n2, as shown in figure. The radius

of curvature of the left and right spherical surfaces are equal. f is the focal length of the lens whenn1 = n2 = n. The focal length is f + f when n1 = n and n2 = n + n. Assuming n << (n–1) and1 < n < 2, the correct statement(s) is/are : [JEE-Advance-2019, P-1]

n1 n2

(A) The relation between f

f and

nn

remains unchanged if both the convex surfaces are

replaced by concave surfaces of the same radius of curvature.

(B) f n

f n

(C) For n = 1.5, n = 10–3 and f = 20 cm, the value of | f| will be 0.02 cm (round off to 2nd decimal place)

(D) If n

0n

then f

0f

36. A planar structure of length L and width W is made of two different optical media of refractive indicesn1 = 1.5 and n2 = 1.44 as shown in figure. If L >> W, a ray entering from end AB will emerge from endCD only if the total internal reflection condition is met inside the structure. For L = 9.6 m, if the incident

angle is varied, the maximum time taken by a ray to exit the plane CD is t × 10–9 s, where t is ____.

[Speed of light c = 3 × 108 m/s] [JEE-Advance-2019, P-1]

n2

n2

n1

C

DB

AAir

L

W

37. A monochromatic light is incident from air on a refracting surface of a prism of angle 75° and refractive

index 0n 3 . The other refracting surface of a prism is coated by a thin film of material of refractive index

n as shown in figure. The light suffers total internal reflection at the coated prism surface for an incidenceangle of 60 . The value of n2 is_______. [JEE-Advance-2019, P-2]

75°

Air

0n 3=


PART - II : JEE(MAIN) / AIEEE (PREVIOUS YEARS)1. A student measures the focal length of a convex lens by putting an object pin at a distance |u| from the lens

and measuring the distance ‘v’ of the image pin. The graph between ‘u’ and ‘v’ plotted by the student shouldlook like - [AIEEE-2008, 3/105]

(1)

v(cm)

O u(cm)

(2)

v(cm)

O u(cm)

(3)

v(cm)

O u(cm)

(4)

v(cm)

O u(cm)

2. A transparent solid cylindrical rod has a refractive index of 32

. It is surrounded by air. A light ray is incident

at the mid-point of one end of the rod as shown in the figure. [AIEEE-2009, 4/144]

The incident angle for which the light ray grazes along the wall of the rod is:

(1) sin-1 23

(2) sin-1 32

(3) sin-1 31

(4) sin-1 21

3. In an optics experiment, with the position of the object fixed, a student varies the position of a convex lensand for each position, the screen is adjusted to get a clear image of the object. A graph between the objectdistance u and the image distance v, from the lens, is plotted using the same scale for the two axes. Astraight line passing through the origin and making an angle of 45° with the x-axis meets the experimentalcurve at P. The coordinates of P will be: [AIEEE-2009, 4/144]

(1) 2f,

2f

(2) (f, f) (3) (4f, 4f) (4) (2f, 2f)

4. A car is fitted with a convex side–view mirror of focal length 20 cm. A second car 2.8 m behind the first car isovertaking the first car at a relative speed of 15 m/s. The speed of the image of the second car as seen in themirror of the first one is : [AIEEE 2011, 1 MAY; 4/120, –1]

(1) 101

m/s (2) 151

m/s (3) 10 m/s (4) 15 m/s

5. Let the x - y plane be the boundary between two transparent media. Medium 1 in z 0 has refractive index of

2 and medium 2 with z < 0 has a refractive index of 3 . A ray of light in medium 1 given by the vector

k10–j38i36A in incident on the plane of separation. The angle of refraction in medium 2 is:[AIEEE 2011, 1 MAY; 4/120, –1]

(1) 30º (2) 45º (3) 60º (4) 75º6. A beaker contains water up to a height h1 and kerosene of height h2 above water so that the total height of

(water + kerosene) is (h1 + h2). Refractive index of water is 1 and that of kerosene is 2. The apparent shift inthe position of the bottom of the beaker when viewed from above is : [AIEEE 2011, 11 MAY; 4/120, –1]

(1) 22

11

h11–h11 (2) 22

11

h1–1h1–1

(3) 12

21

h11–h11 (4) 12

21

h1–1h1–1

JEE (Adv.)-Physics Geometrical Optics7. When monochromatic red light is used instead of blue light in a convex lens, its focal length will :

[AIEEE 2011, 11 MAY; 4/120, –1](1) increase (2) decrease(3) remain same (4) does not depend on colour of light

8. An object 2.4 m in front of a lens forms a sharp image on a film 12 cm behind the lens. A glass plate 1 cm thick,of refractive index 1.50 is interposed between lens and film with its plane faces parallel to film. At what distance(from lens) should object be shifted to be in sharp focus onfilm ? [AIEEE- 2012](1) 5.6 m (2) 7.2 m (3) 2.4 m (4) 3.2 m

9. A beam of light consisting of red, green and blue colours is incident on a right-angledprism on face AB. The refractive indices of the material for the above red, green andblue wavelength are 1.39, 1.44 and 1.47 respectively. A person looking on surfaceAC of the prism will see : [AIEEE - 2012 (Online)]

A

BC45°

(1) red and green colours (2) No light (3) green and blue colours (4) red colour only10. Which of the following processes play a part in the formation of a rainbow ? [AIEEE - 2012 (Online)]

(a) refraction (b) total internal reflection(c) dispersion (d) interference(1) a, b and c (2) a and b (3) c and d (4) a, b and d

11. A glass prism of refractive index 1.5 is immersed in water (refractive index 4/3) asshown in the figure. A light beam incident normally on the face AB is totally reflectedto reach the face BC, if:- [AIEEE - 2012 (Online)]

C

B A

(1) sin > 59

(2) sin > 13

(3) sin > 23

(4) sin > 89

12. Diameter of a plano-convex lens is 6cm and thickness at the centre is 3 mm. If speed of light in material of lensis 2 × 108 m/s, the focal length of the lens is : [JEE(Main)-2013](1) 15 cm (2) 20 cm (3) 30 cm (4) 10 cm

13. The graph between angle of deviation ( ) and angle of incidence (i) for a triangular prism is represented by :- [JEE(Main)-2013]

(1) (2) (3) (4)

14. A thin convex lens made from crown glass 3

µ2

has focal length ƒ. When it is measured in two different

liquids having refractive indices 43

and 53

, it has the focal length ƒ1 and ƒ2 respectively. The correct relation

between the focal lengths is : [JEE(Main)-2014](1) ƒ2 > ƒ and ƒ1 becomes negative (2) ƒ1 and ƒ2 both become negative(3) ƒ1 = ƒ2 < ƒ (4) ƒ1 > ƒ and ƒ2 become negative

15. A green light is incident from the water to the air - water interface at the critical angle ( ). Select the correctstatement :- [JEE(Main)-2014](1) The spectrum of visible light whose frequency is more than that of green light will come out to the air

medium.(2) The entire spectrum of visible light will come out of the water at various angles to the normal.(3) The entire spectrum of visible light will come out of the water at an angle of 90° to the normal.(4) The spectrum of visible light whose frequency is less than that of green light will come out to the air medium.

JEE (Adv.)-Physics Geometrical Optics16. Assuming human pupil to have a radius of 0.25cm and a comfortable viewing distance of 25cm, the minimum

separation between two objects that human eye can resolve at 500 nm wavelength is :- [JEE(Main)-2015](1) 100 µm (2) 300 µm (3) 1 µm (4) 30 µm

17. On a hot summer night, the refractive index of air is smallest near the ground and increases with height from theground. When a light beam is directed horizontally, the Huygens' principle leads us to conclude that as ittravels, the light beam : [JEE(Main)-2015](1) bends downwards (2) bends upwards(3) becomes narrower (4) goes horizontally without any deflection

18. Monochromatic light is incident on a glass prism of angle A. If the refractive index of the material of the prismis µ, a ray, incident at an angle , on the face AB would get transmitted through the face AC of the prismprovided : [JEE(Main)-2015]

C

A

B

(1) 1 1 1

cos µ sin A sinµ (2)

1 1 1cos µ sin A sin

µ

(3) 1 1 1

sin µ sin A sinµ (4)

1 1 1sin µ sin A sin

µ

19. An observer looks at a distant tree of height 10 m with a telescope of magnifying power of 20. To the observerthe tree appears : [JEE(Main)-2016](1) 20 times nearer (2) 10 times taller (3) 10 times nearer (4) 20 times taller

20. In an experiment for determination of refractive index of glass of a prism by i – , plot, it was found that a rayincident at angle 35°, suffers a deviation of 40° and that it emerges at angle 79°. In that case which of thefollowing is closest to the maximum possible value of the refractive index ? [JEE(Main)-2016](1) 1.8 (2) 1.5 (3) 1.6 (4) 1.7

21. To find the focal length of a convex mirror, a student records the following data :

Object pin Convex Lens Convex Mirror Image Pin

22.2cm 32.2cm 45.8cm 71.2cm

The focal length of the convex lens is f1 and that of mirror is f2. Then taking index correction to be negligiblysmall, f1 and f2 are close to : [JEE(Main)-2016 (online)](1) f1 = 7.8 cm f2 = 12.7 cm(2) f1 = 15.6 cm f2 = 25.4 cm(3) f1 = 12.7 cm f2 = 7.8 cm(4) f1 = 7.8 cm f2 = 25.4 cm

22. A hemispherical glass body of radius 10 cm and refractive index 1.5 is silveredon its curved surface. A small air bubble is 6 cm below the flat surface insideit along the axis. The position of the image of the air bubble made by the mirroris seen : [JEE(Main)-2016 (online)]

\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

\\\\\\

\\\ 6cmo

Silvered

10cm

(1) 16 cm below flat surface (2) 20 cm below flat surface(3) 30 cm below flat surface (4) 14 cm below flat surface

JEE (Adv.)-Physics Geometrical Optics23. A convex lens, of focal length 30 cm, a concave lens of focal length 120 cm, and a plane mirror are arranged

as shown. For an object kept at a distance of 60 cm from the convex lens, the final image, formed by thecombination, is a real image, at a distance of : [JEE(Main)-2016 (online)]

|Focal length| |Focal length| = 30 cm = 120 cm

60 cm 20 cm

70 cm

(1) 70 cm from the convex lens (2) 60 cm from the convex lens(3) 60 cm from the concave lens (4) 70 cm from the concave lens

24. A diverging lens with magnitude of focal length 25 cm is placed at a distance of 15 cm from a converginglens of magnitude of focal length 20 cm. A beam of parallel light falls on the diverging lens. The final imageformed is : [JEE(Main)-2017](1) real and at a distance of 40 cm from the divergent lens(2) real and at a distance of 6 cm from the convergent lens(3) real and at a distance of 40 cm from convergent lens(4) virtual and at a distance of 40 cm from convergent lens.

25. A particle is oscillating on the X-axis with an amplitude 2 cm about the pointx0 = 10 cm, with a frequency . A concave mirror of focal length 5 cm is placedat the origin (see figure). [JEE(Main)-2018 (online)]Identify the correct statements.

x =10 cm0

x = 0

(a) The image executes periodic motion(b) The image executes non-periodic motion(c) The turning points of the image are asymmetric w.r.t. the image ofthe point at x = 10 cm.

(d) The distance between the turning points of the oscillation of the image is 10021

cm.

(1) b, d (2) b, c (3) a, c, d (4) a, d26. A planoconvex lens becomes an optical system of 28 cm focal length /////////////// / // ////////////////////

when its plane surface is silvered and illuminated from left to right asshown in Fig-A.If the same lens is instead silvered on the curved surface and illuminatedfrom other side as in Fig-B. it acts like an optical system of focal length10 cm. The refractive index of the material of lens is : [JEE(Main)-2018 (online)](1) 1.75 (2) 1.51 (3) 1.55 (4) 1.50

27. A convergent doubled to separated lenses, corrected for spherical aberration, has resultant focal length of10 cm. The separation between the two lenses is 2 cm. The focal lengths of the component lenses are :

[JEE(Main)-2018 (online)](1) 18 cm, 20 cm (2) 12 cm, 14 cm (3) 16 cm, 18 cm (4) 10 cm, 12 cm

28. A ray of light is incident at an angle of 60° on one face of a prism of angle 30°. The emergent ray of light makesan angle of 30° with incident ray. The angle made by the emergent ray with second face of prism will be :-

[JEE(Main)-2018 (online)](1) 0° (2) 45° (3) 90° (4) 30°

JEE (Adv.)-Physics Geometrical Optics29. A convex lens is put 10 cm from a light source and it makes a sharp image on a screen, kept 10 cm from

the lens. Now a glass block (refractive index 1.5) of 1.5 cm thickness is placed in contact with the light source.To get the sharp image again, the screen is shifted by a distance d. Then d is :

(1) 0.55 cm away from the lens [JEE(Main)-2019 (online)](2) 1.1 cm away from the lens

(3) 0.55 cm towards the lens

(4) 0

30. The eye can be regarded as a single refracting surface . The radius of curvature of this surface is equal to thatof cornea (7.8 mm). This surface separates two media of refractive indices 1 and 1.34. Calculate the distancefrom the refracting surface at which a parallel beam of light will come to focus. [JEE(Main)-2019 (online)](1) 2 cm (2) 1 cm (3) 3.1 cm (4) 4.0 cm

31. The variation of refractive index of a crown glass thin prism with wavelength of the incident light is shown. Whichof the following graphs is the correct one, if Dm is the angle of minimum deviation? [JEE(Main)-2019 (online)]

1.535

1.530

1.525

1.520

1.5151.510

400 500 600 700(nm)

(1)

400 500 600 700

Dm

(nm)

(2)

400 500 600 700

Dm

(nm)

(3)

400 500 600 700

Dm

(nm)

(4)

400 500 600 700

Dm

(nm)

32. Formation of real image using a biconvex lens is shown below : If the whole set up is immersed in water without disturbing the object

and the screen position, what will one observe on the screen ?

[JEE(Main)-2019 (online)(1) Image disappears (2) No change(3) Erect real image (4) Magnified image

JEE (Adv.)-Physics Geometrical Optics33. What is the position and nature of image formed by lens combination shown in figure? (f1, f2 are focal lengths)

[JEE(Main)-2019 (online)]

B

O

20cm f1 = +5cm f2 = –5cm

2 cm

A

(1) 70 cm from point B at left; virtual (2) 40 cm from point B at right; real

(3) 203

cm from point B at right , real (4) 70 cm from point B at right, real

34. In figure, the optical fiber is = 2m long and has a diameter of d = 20 µm. If a ray of light is incident on oneend of the fiber at angle 1 = 40°, the number of reflection it makes before emerging from the other end is closeto : (refractive index of fibre is 1.31 and sin 40° = 0.64) [JEE(Main)-2019 (online)]

40°2

d

(1) 55000 (2) 57000 (3) 66000 (4) 4500035. A thin convex lens L (refractive index = 1.5) is placed on a plane mirror M. When

a pin is placed at A, such that OA = 18 cm, its real inverted image is formed at Aitself, as shown in figure. When a liquid of refractive index µ1 is put between the lensand the mirror, The pin has to be moved to A', such that OA' = 27 cm, to get itsinverted real image at A' itself. The value of µ1 will be:-[JEE(Main)-2019 (online)]

A'

A

L

OM

(1) 2 (2) 43

(3) 3 (4) 32

36. A concave mirror has radius of curvature of 40 cm. It is at the bottom of a glass thathas water filled up to 5 cm (see figure). If a small particle is floating on the surfaceof water, its image as seen, from directly above the glass, is at a distance d fromthe surface of water. The value of d is close to : (Refractive index of water = 1.33)

[JEE(Main)-2019 (online)] 5cm

particle

(1) 8.8 cm (2) 11.7 cm(3) 6.7 cm (4) 13.4 cm

37. The magnifying power of a telescope with tube 60 cm is 5. What is the focal length of its eye piece ? [JEE(Main)-2020 (online)]

(1) 30 cm (2) 40 cm (3) 20 cm (4) 10 cm38. An object is gradually moving away from the focal point of a concave mirror along the axis of the mirror. The

graphical representation of the magnitude of linear magnification (m) versus distance of the object from themirror (x) is correctly given by : [JEE(Main)-2020 (online)](Graphs are drawn schematically and are not to scale)

(1)1

f 2fx

m

(2) 1

f 2fx

m

(3) 1

f 2fx

m

(4) 1

f 2fx

m


Exercise # 1

PART - ISection (A) :A-1. 2A-2. 2A-3. 120º anticlockwise and 240º clockwise.A-4. 30º clockwiseA-5. (a) Position of image = (1 cos 60º, – 1 sin 60º)

(b) Velocity of image = (1 cos 60º i , + 1 sin 60º j ) m/s.

A-6. (a) 1 ; (b) (4, 0) ; (c) No

Section (B) :

B-1. 16 mm3 B-2. d/2

B 3. 10.35 cm = 3803933

cm

B-4. 84 cm, 0.5 cmB-5. 0.2 m from the mirrorB-6. 60 cmB-7. (a) 40 cm/s opposite to the velocity of object.,

(b) 20 cm/s opposite to the velocity of object.B-8. 3 mSection (C) :

C-1. cm71

213 = 9.9 mm, 45°

C-2. 2/3 × 10–8 sec C-3. 30 cmC-4. 35 cm , Shift = 5 cm.C-5. 9 cm/s C-6. 25 cm.

C-7. 368

cm C-8. 3 cm

C-9.1

h2

2

C-10. 0.9 cm above P

C-11. n > 2 C-12. 45°

Section DD-1. = 60° D-2. 90º

D-3. (i) 1.5°, (ii) 83

D-4. 1 12 sin

Section (E)E-1. 50 cm right of BE-2. 40 cm from pole in the medium of refractive index

1, virtual, erect and 4 cm in size.E-3. 80 cm

E-4.2

27 = 13.5 cm below the surface of water

E-5. 8/3 mm, virtual at v = - 20, no inversionE-6. (a) 2, (b) not possible, it will focus close to the

centre if the refractive index is largeSection (F)F-1. Converging ; realF-2. ± 12 cm, ± 60 cmF-3. 360 cm; ; – 600 cmF-4. (i) 7/5

(ii) In this liquid the 1st lens will be diverging &the 2nd a converging one

F-5. (a) 312

3µµµ2

Rµ (b)

312

1µµµ2

Rµ

F-6. 20 cm, 1 m, –4, 4 cmF-7. 30 cm F-8. 1.5 cmF-9. 0.4 cmF-10. 60 cm from the lens further away from the mirror

F -11. 31 cm from the lens

F-12. 35

cm from the lensF-13. 1.0 cm if the light is incident from the side of

concave lens and 2.5 mm if it is incident fromthe side of the convex lens and the correspondingratio of intensities are 1/4 and 4.

Section (G)G-1. 10 cm for convex lens and 60 cm for concave

lensG-2. 10 D, Optical power of each lens = 5 D.

G-3. (a) 15 cm from the lens on the axis (b) 1.14 cmtowards the lens

Section (H)

H-1. (a) rv

rvµµ

)µµ(2 , (b) 1µ

)1µ(2

y

y

H-2. (a) 41

= 0.25 (b) 0.90°

H-3. 4° H-4. 490099

H-5. (a) 3° (b) 0.015º (c) 3º ( d) 0.225°

JEE (Adv.)-Physics Geometrical OpticsSection (I)I-1. u0 = –1.5 cm; v0 = 7.5 cm. |ue| = (25/6) cm = 4.17

cm. The separation between the objective andthe eye-piece should be (7.5 + 4.17) cm = 11.67cm. Further the object should be placed 1.5 cmfrom the objective to obtain the desiredmagnification.

I -2. 24; 150 cm

I-3. (a) ve = – 2.5 cm and fe = 6.25 cm giveue = – 5 cm ; v0 = (15 – 5) cm = 10 cm.

f0 = u0 = – 2.5 cm; Magnifying power = 525

5.210

= 20

(b) ue = – 6.25 cm, v0 = (15 – 6.25) cm = 8.75, f0 = 2.0 cm. Therefore u0 = – (70/27) = – 2.59 cm.

Magnifying power = |u|v

0

0× (25/6.25)

= 827

× 4 = 13.5

PART - IISection (A)

A-1. (A) A-2. (B) A-3. (C)

A-4. (A) A-5. (B) A-6. (C)

A-7. (A) A-8. (C) A-9. (B)A-10. (B)

Section (B)B-1. (B) B-2. (D) B-3. (C)B-4. (B) B-5. (C) B-6. (A)B-7. (C) B-8. (C) B-9. (A)B-10. (C) B-11. (A) B-12. (D)B-13. (B) B-14. (C)

Section (C)C-1. (D) C-2. (C) C-3. (C)C-4. (C) C-5. (B) C-6. (A)

C-7. (B) C-8. (D) C-9. (A)C-10. (A)Section (D)D-1. (B) D-2. (C) D-3. (B)D-4. (C) D-5. (a)-(A), (b)-(C), (c) -(C) 30º

Section (E)E-1. (A) E-2. (A) E-3. (B)E-4. (A)

Section (F)F-1. (C) F-2. (A) F-3. (A)F-4. (A) F-5. (C) F-6. (D)F-7. (C) F-8. (D) F-9. (D)F-10. (B) F-11. (A) F-12. (A)F-13. (B) F-14. (A)

Section (G)G-1. (A) G-2. (D) G-3. (B)G-4. (A) G-5. (C)

Section (H)

H-1. (D) H-2. (C) H-3. (A)

H-4. (C) H-5. (C) H-6. (D)H-7. (C) H-8. (D) H-9. (B)H-10. (D) H-11. (D)

Section (I)

I-1. (C) I-2. (D) I-3. (A)

I-4. (B) I-5. (B) I-6. (C)I-7. (D) I-8. (B)

PART - III1. (A)-(Q), (B)-(P), (C)-(S), (D)-(R)

2. (A)-(P,S,T);(B)-(Q,S);(C)-(Q, R) (D)-(Q, R)

3. (A)-Q; (B)-P; (C)-S; (D)-R

4. (A)-P; (B)-Q; (C)-RT; (D)-PT

5. (B)

Exercise # 02PART - I

1. (B) 2. (C) 3. (A)

4. (A) 5. (B) 6. (C)7. (C) 8. (D) 9. (A)10. (B) 11. (C) 12. (B)13. (C) 14. (C) 15. (B)16. (B) 17. (C) 18. (B)19. (D) 20. (A) 21. (D)22. (D) 23. (C) 24. (C)25. (A) 26. (A)


PART - II

1. 86 cm 2. 26

3. 3 4. 2 cm

5. (i) 96 cm2 (12 × 8 cm2)

(ii) 72 cm2(12 × 6 cm2)

6. 20 7. +2

8. 23 9. 62

10. 16 cm on the right side of the mirror.

11. x = 60º, y = 30º, z = 60º

12. 5 13. 6

14. 3 m 15. 24 cm

16. t = 2 cm 17. 2

18. 1° 19. 2

20 30 21. m = 4/3

22. 9 cm

PART - III

1. (A), (B) 2. (A),(C), (D)

3. (B), (D) 4. (A), (D)

5. (B), (C) 6. (B), (D)

7. (A), (B) 8. (C), (D)

9. (A), (C) 10. (A), (C)

11. (B), (C) 12. (B),(C), (D)

13. (A), (B), (C) 14. (A), (B), (D)

15. (A), (D) 16. (A), (B), (C)

17. (B), (C) 18. (A), (B), (C)

PART - IV1. (D) 2. (C) 3. (D)

4. (B) 5. (D) 6. (D)

7. (B) 8. (B) 9. (C)

Exercise # 03PART - I

1. (A) 2. (B)3. (A) (p,q,r,s); (B) (q); (C) (p,q,r,s); (D) (p,q,r,s)4. (C) 5. (C) (D)6. (B) 7. (A), (B), (C)

8. (A)–pr, (B) –qst, (C) –prt, (D) –qs

9. 6 10. 3 11. 6

12. (C) 13. 2 14. (B)

15. (C) 16. (B) 17. (A)

18. (C) 19. (D) 20. (A), (C)

21. (C) 22. (B) 23. 7

24. (B) 25. 2 26. (A), (C)

27. (D) 28. (A) 29. (A), (B), (D)

30. (A), (D) 31. (A), (C), (D)

32. 8 33. 13.00 34. (D)

35. (A), (C), (D) 36. 49 to 51

37. 1.50

PART - II1. (2) 2. (3) 3. (4)

4. (2) 5. (2) 6. (2)

7. (1) 8. (1) 9. (4)

10. (1) 11. (4) 12. (3)

13. (3) 14. (4) 15. (4)

16. (4) 17. (2) 18. (3)

19. (1) 20. (2) 21. (1)

22. (2) 23. (2) 24. (3)

25. (3) 26. (3) 27. (1)

28. (3) 29. (1) 30. (3)

31. (2) 32. (1) 33. (4)

34. (2) 35. (2) 36. (1)

37. (4) 38. (2)


SUBJECTIVE QUESTIONS1. A point object is 10 cm away from a plane mirror while the eye of an observer (pupil diameter

5.0 mm) is 20 cm away. Assuming both the eye and point to be on the same line perpendicular to the mirror,then find the area of the mirror used in observing the reflection of the point :

2. A light ray, going through a prism with the angle of prism 60°, is found to deviate by 30°. What limit onthe refractive index can be put from these data?

3. In an experiment performed with a 60º prism where angle of minimum deviation for sodium light is 60° inair. The following experiment was done. When sodium light enters at one face at grazing incidencefrom a certain liquid, it emerges from the other face (in air) at 60° from the normal to edge of the prism.Are the observations correct ?

4. (i) A paper weight of refractive index n = 3/2 in the form of a hemisphere of radius 3.0 cm is usedto hold down a printed page. An observer looks at the page vertically through the paperweight.At what height above the page will the printed letters near the centre appear to the observer ?

(ii) Solve the previous problem if the paperweight is inverted at its place so that the sphericalsurface touches the paper.

5. In the given figure, a hollow sphere of glass of refractive index n hasa small mark M on its interior surface which is observed by anobserver O from a point outside the sphere. C is centre of the sphere.The inner cavity (air) is concentric with the external surface andthickness of the glass is everywhere equal to the radius of the innersurface. Find the distance by which the mark will appear nearerthan it really is, in terms of n and R assuming paraxial rays. 2R

4R

M Cair

glass

O

6. Two media each of refractive index 1.5 with plane parallel boundaries are separated by 100 cm. Aconvex lens of focal length 60 cm is placed midway between them with its principal axis normal to theboundaries. A luminous point object O is placed in one medium on the axis of the lens at a distance125 cm from it. Find the position of its image formed as a result of refraction through the system.

RANKER PROBLEMS

JEE (Adv.)-Physics Geometrical Optics7. Two converging lenses of the same focal length f are separated by distance 2 f as shown in figure.The

axis of the second lens is inclined at angle with respect to the axis of the first lens. A parallelparaxial beam of light is incident from left side on the lens. Find the coordinates of the final image withrespect to the origin of the first lens.

2f

8. A small object is placed 50 cm to the left of thin convex lens of focal length 30 cm. A convex spherical mirrorof radius of curvature 100 cm is placed to the right of the lens at a distance of 50 cm. The mirror is tilted suchthat the axis of the mirror is at an angle q = 30° to the axis of the lens, as shown in the figure. If the origin ofthe coordinate system is taken to be at the centre of the lens, the coordinates (in cm) of the point (x, y) atwhich the image is formed are :

||||||||||||||||||||||||||||||||||||||||||

50cm

(–50,0)

f=30cm

R=100 cm

(50+50 3, –50)–

x(0,0)

9. In the given figure, a stationary observer O looking at a fish F (in water of, = 4/3) through a converging lensof focal length 90 cm. The lens is allowed to fall freely from a height 62.0 cm with its axis vertical. The fish andthe observer are on the principal axis of the lens. The fish moves up with constant velocity 100 cm/s. Initiallyit was at a depth of 44.0 cm. Find the velocity with which the fish appears to move to the observer att = 0.2 sec. (g = 10 m/s2)

OLens

F

water 44 cm

62 cm

10. Two symmetric double-convex lenses L1 & L2 with their radii of curvature 0.2 m each are made fromglasses with refractive index 1.2 & 1.6 respectively. The lenses with a separation of 0.345 m aresubmerged in a transparent liquid medium with a refractive index of 1.4. Find the focal lengths of lensesL1 & L2. An object is placed at a distance of 1.3 m from L1, (away from L2) find the location of its imagewhile the whole system remains inside the liquid.

11. A kid of height 1.1 ft is sleeping straight between focus and centre of curvature along the principal axisof a concave mirror of small aperture. His head is towards the mirror and is 0.5 ft from the focus of themirror. How a plane mirror should be placed so that the image formed by it due to reflected light fromconcave mirror looks like a person of height 5.5 ft standing vertically. Draw the ray diagram. Find thefocal length of the concave mirror.


12. As shown in the figure, an object O is at the position ( 10, 2) with respect to the origin P. The concavemirror M1 has radius of curvature 30 cm. A plane mirror M2 is kept at a distance 40 cm infront of theconcave mirror. Considering first reflection on the concave mirror M1 and second on the plane mirror M2.Find the coordinates of the second image w.r.t. the origin P.

13. A point source S is moving with a speed of 10 m/s in x-y planeas shown in the figure. The radius of curvature of the concavemirror is 4m. Determine the velocity vector of the image formedby paraxial rays.

14. A beam of parallel rays of diameter ' b ' propagates in glass at an angle to its plane. Find the diameterof the beam when it goes to air through this face. (nglass = n)

15. A small ball is thrown from the edge of one bank of a river of width 100 m to just reach the other bank. The ballwas thrown in the vertical plane (which is also perpendicular to the banks) at an angle 37º to the horizontal.Taking the starting point as the origin O, vertically upward direction as positive y-axis and the horizontal linepassing through the point O and perpendicular to the bank as x-axis find :-

16. The figure shows the square front face (of side ‘a’) of a transparent cuboidalblock. The thickness or the third dimension of the block is negligible incomparison to ‘a’. The block has uniform refractive index µ equal to 2. Apoint source S which can emit light in all directions can move inside theblock. It is desired that no light of ‘S’ should pass through AB. Sketch theregion in which S should be present to satisfy this condition.

17. A ray of light is incident on a surface in a direction given by vector kj2i2A . The normal to that

surface passing through the point of incidence is along the vector k2jN . The unit vector in the

direction of reflected ray is given by R = kcjbia . Find three equations in terms of a, b, c usingwhich we can find the values of a, b and c.

18. In the given figure, the faces of prism ABCD made of glass with arefractive index n form dihedral angles A = 90°, B = 75°, C =135°and D = 60° ( The Abbe's prism ). A beam of light falls on face AB andafter total internal reflection from face BC escapes through face AD.Find the range of n and angle of incidence of the beam onto face AB,if a beam that has passed through the prism in this manner isperpendicular to the incident beam.

19. A point source of light is placed at a distance h below the surface of a large deep lake.(a) Show that the fraction f of the light energy that escapes directly from the water surface is indepen-

dent of h and is given by f = 1nn21

21 2 where n is the index of refraction of water..

(Note: Absorption within the water and reflection at the surface; except where it is total, have beenneglected)

(b) Evaluate this ratio for n = 4/3.

20. A glass hemisphere of refractive index 4/3 and of radius 4 cm is placedon a plane mirror. A point object is placed on axis of this sphere at adistance ' d ' from O as shown in the figure. If the final image is formedat infinity, then find the value of ' d ' in cm.

JEE (Adv.)-Physics Geometrical Optics21. In the given figure, O is a point object kept on the principal axis of a

concave mirror M of radius of curvature 20 cm. P is a prism of angle = 1.8 º. Light falling on the prism (at small angle of incidence) get

refracted through the prism and then fall on the mirror. Refractiveindex of prism is 3/2. Find the distance between the images formedby theconcave mirror due to this light.

22. A convex lens produces an image of a candle flame upon a screen whose distance from candle is D.When the lens is displaced through a distance x, (the distance between the candle and the screen iskept constant), it is found that a sharp image is again produced upon the screen. Find the focal lengthof the lens in terms of D and x.

23. A thin equiconvex lens of glass of refractive index = 3/2 & offocal length 0.3 m in air is sealed into an opening at oneend of a tank filled with water ( = 4/3). On the opposite sideof the lens, a mirror is placed inside the tank on the tank wallperpendicular to the lens axis, as shown in figure. The separationbetween the lens and the mirror is 0.8 m. A small object isplaced outside the tank infront of the lens at a distance of0.9 m from the lens along its axis. Find the position (relative tothe lens) of the image of the object formed by the system.

0.9m 0.8m

Mirr

or

24. A prism of refractive index n1 and another prism of refractive indexn2 are stuck together without a gap as shown in the figure. Theangles of the prisms are as shown. n1 and n2

d e p e n d o n , t h e wa v e l e n g t h o f l i g h t a c c o r d i n g t o

n1 = 1.20 + 2

4108.10 and n2 = 1.45 + 2

41080.1 where is in nm.

(i) Calculate the wavelength 0 for which rays incident at any angle on the interface BC passthrough without bending at that interface.

(ii) For light of wavelength 0, find the angle of incidence i on the face AC such that the deviationproduced by the combination of prisms is minimum.

25. A fly F is sitting on a glass slab S 45cm thick and of refractive index3/2. The slab covers the top of a container C containing water(R.I. 4/3) upto a height of 20 cm. Bottom of container is closed by aconcave mirror M of radius of curvature 40 cm. Locate the final imageformed by all refractions and reflection assuming paraxial rays.

26. A glass porthole is made at the bottom of a ship for observing sea life. The hole diameterD is much larger than the thickness of the glass. Determine the area S of the field of vision at the seabottom for the porthole if the refractive index of water is µw and the sea depth is h.

27. A light ray is incident on a plane mirror M. The mirror is rotated inthe direction as shown in the figure by an arrow at frequency 9/rps. The light reflected by the mirror is received on the wall W at adistance 10 m from the axis of rotation. When the angle of incidencebecomes 37º the speed of the spot (a point) on the wall is:


1.144

cm2 2. 32

µ 2 3. No

4. (i) No shift is observed (ii) 1 cm 5. (n 1)R/(3n 1)

6. 200 cm, right of the lens 7. 0,cos1

)cos21(f

8. 25, 25 3 9. 22.75 m/s (upwards)

10. f1 = 70 cm, f2 = 70 cm, 560 cm from L2 away from L1.

11. The plane mirror should be placed at an angle of 45º with negative x-axis; f = 2ft.

//////

//////

//////

//////

//////

//////

//////

×× C

//////////

//////////

//////////

/

FFA BB'

B''

A''

A'

1.1 0.5

x5.5

5.5

45°

12. Coordinates of I2 w.r.t. P = ( 46, 70) 13. iV = Vix i + jViy = – 2 i – 4 j

14. CD = sin

cosn1.b 22

15. (a) xx2

100 (b) v (t)image izfrfcEc = 205

3i + 20

53

23

t j

16.

17. a2 + b2 + c2 = 1; 3a + 4b + 2c = 0 ; b – 2c = 4/3 18. 2 < n 2, and 45º < 90º

19. (b) (4 – 7 ) / 8 20. 3 21. 20 cm.

22. f = (D2 x2)/4D 23. 90 cm from the lens towards right.

24. (i) 0 = 600 nm, n = 1.5 (ii) i = sin 1 (0.75) = 48.59º

25. 6135

cm = 22.5 cm below the upper surface of the glass slab

26.

2

2w

2D

1

h 27. 100 cm


SELF ASSESSMENT PAPERJEE (ADVANCED) PAPER

SECTION-1 : ONE OPTION CORRECT (Maximum Marks - 12)1. Given that, velocity of light in quartz = 1.5 108 m/s and velocity of light in

glycerine = (9/4) 108 m/s. Now a slab made of quartz is placed in glycerine

as shown. What is the shift produced by slab? (A) 6 cm

(B) 3.55 cm

(C) 9 cm

(D) 2 cm

2. A small rod ABC is put in water making an angle 6° with vertical. If it isviewed (paraxially) from above, it will look like bent shaped ABC'. The

angle of bending ( 'CBC ) will be in degree ..........34nw .

C

C'B

A6°

(A) 2° (B) 3°

(C) 4° (D) 4.5°

3. If a prism having refractive index 2 , has angle of minimum deviation equal to the angle of refraction of

the prism, then the angle of refraction of the prism is:

(A) 30º (B) 45º (C) 60º (D) 90º

4. An object approaches a fixed diverging lens with a constant velocity from infinity along the principalaxis. The relative velocity between object and its image will be :

(A) increasing (B) decreasing

(C) first increases then decreases (D) first decreases and then increases

SECTION-2 : ONE OR MORE THAN ONE CORRECT (Maximum Marks - 32)5. is the image of a point object O formed by spherical mirror, then which of the following statements is

correct:(A) If O and are on same side of the principal axis, then they have to be on opposite sides of the mirror.

(B) If O and are on opposite side of the principal axis, then they have to be on same side of the mirror.

(C) If O and are on opposite side of the principal axis, then they can be on opposite side of the mirror as well.

(D) If O is on principal axis then has to lie on principal axis only.6. It is found that all electromagnetic signals sent from A towards B reach

point C. The speed of electromagnetic signals in glass can be : (A) 1.0 × 108 m/s(B) 2.4 × 108 m/s(C) 2 × 107 m/s(D) 4 × 107 m/s

JEE (Adv.)-Physics Geometrical Optics7. A particle is moving towards a fixed convex mirror. The image also moves. If Vi = speed of image and

VO = speed of the object, then(A) Vi VO if |u| < |F| (B) Vi > VO if |u| > |F| (C) Vi < VO if |u| > |F| (D) Vi = VO if |u| = |F|

8. A small air bubble is trapped inside a transparent cube of size 12 cm. When viewed from one of thevertical faces, the bubble appears to be at 5 cm from it. When viewed from opposite face, it appears at3 cm from it.(A) The distance of the air bubble from the first face is 7.5 cm.(B) The distance of the air bubble from the first face is 9 cm.(C) Refractive index of the material of the prism is 2.0.(D) Refractive index of the material of the prism is 1.5.

9. A parallel beam of light is incident normally on the flat surface of a hemisphere of radius 6 cm andrefractive index 1.5, placed in air as shown in figure (i). Assume paraxial ray approximation.

(A) The rays are focused at 12 cm from the point P to the right, in the situation as shown in figure (i)(B) The rays are focused at 16 cm from the point P to the right , in the situation as shown in figure (i)(C) If the rays are incident at the curved surface (figure (ii)) then these are focused at distance 18 cm from point P to the right.(D) If the rays are incident at the curved surface (figure (ii)) then these are focused at distance 14 cm from point P to the right.

10. A glass prism is immersed in a hypothetical liquid. The curves showing the refractive index n as a function ofwavelength l for glass and liquid are as shown in the figure. When a ray of white light is incident on the prismparallel to the base :

Liquid

Glass

yellowO

n

Base

Vertex

(A) yellow ray travels without deviation (B) blue ray is deviated towards the vertex(C) red ray is deviated towards the base (D) there is no dispersion

11. A glass slab is of thickness d = a/2 is shown in figure. The refractive index of

the slab varies according to relation a

a xwhere a is positive constant.

A ray of light enters the slab at origin at an angle 0 /2, then choose theCORRECT statement (s):-

Y

Xa/2

(A) The path of the light ray inside medium is a circle given by (x – a)2 + y2 = a2

(B) The path of the light ray inside medium is a parabola given by (x – a)2 = y(C) The deviation of ray from its initial direction is = /3(D) Ray emerges grazing at the surface.

JEE (Adv.)-Physics Geometrical Optics12. Figure shows a convex lens cut symmetrically into two equal halves and separated laterally by a distance h. A point

object O placed symmetrically at a distance 30 cm, from the lens halves, forms two images separated by a distanced. A plot of d versus h is shown in figure. The focal lengthof the lens is :-

hO

30cm

d

6cm

3cm

1cm 2cmO h

(A) 22.5 cm (B) 40 cm (C) 45 cm (D) 20 cm

SECTION-3 : NUMERICAL VALUE TYPE (Maximum Marks - 18)13. In the figure shown L is a converging lens of focal length 10cm and M is a concave mirror of radius of curvature

20cm. A point object O is placed in front of the lens at a distance 15cm. AB and CD are optical axes ofthe lens and mirror respectively. Find the distance of the final image formed by this system from the opticalcentre of the lens. The distance between CD & AB is 1 cm.

14. In the given figure if observer sees the bottom of vessel at 8 cm, find the refractive index of the medium inwhich observer is present.

15. A man starting from point P crosses a 4 km wide lagoon and reaches point Q in the shortest possible timeby the path shown in the figure. If the person swims at a speed of 3 km/hr and walks at a speed of 4 km/hr,then find his time of journey (in minutes).

3km

Q

P

LAGOON(Salt water lake)

6kmLand

4kmLAGOON

(Salt water lake)

16. An observer observer a fish moving upwards in a cylindrical container of cross section area 1 m2 filledwith water up to a height of 5 m. A hole is present at the bottom of the container having cross section area1/1000 m2. Find out the speed of the image of fish observed by observer when the bottom hole is justopened. (Given: The fish is moving with the speed of 6 m/s towards the observer, of water = 4/3)

JEE (Adv.)-Physics Geometrical Optics17. A burning candle is placed in front of a concave spherical mirror on its principal optical axis at a distance of

(4/3)F from the pole of the mirror (here F is the focal length of the mirror). The candle is arranged at right angleto the axis. The image of the candle in the concave mirror impinges upon a convex mirror of focal length 2 F.The distance between the mirrors is 3F and their axes coincide. The image of the candle in the first mirrorplays the part of a virtual object with respect to the second mirror and gives a real image arranged betweenthe two mirrors. Plot this image and calculate the total linear magnification of the system.

18. A man is standing at the edge of a 1m deep swimming pool, completely filled with a liquid of refractive

index 3 2/ . The eyes of the man are 3 m above the ground. A coin located at the bottom of the poolappears to be at an angle of depression of 300 with reference to the eye of man. Then find horizontal distance(represented by x in the figure) of the coin from the eye of the man.

1. (A) 2. (A) 3. (D) 4. (B) 5. (A),(B),(D)6. (A),(C),(D) 7. (A), (C) 8. (A), (D) 9. (A), (D) 10. (A), (B), (C)

11. (A), (D) 12. (B), (D) 13. 6 26 cm 14. 1516

15. 250

16. 4.4975 m/s 17. –6 18. d = 4000 mm

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