PRE SEM SOLUTION 2013-B.TECH IST YEAR

Solved Pre-SEM Exam Paper-Engineering Physics-I Paper Code-AS102

2012

Dr. Jitendra Pal Singh, KEC Ghaziabad

SECTION- A

Short type question answer (50-75 words)

1. Attempt all parts. All parts carry equal

marks: (2x10=20)

A. Write the postulates of Special theory

of relativity.

Ans.

There are two postulates of special theory of

relativity.

(i)Laws of physics are same in all inertial

frame of references.

(ii) The speed of light in free space is same in

all inertial frame of reference.

B. What are massless particles?

Ans.

According to the theory of relativity, mass (m)

of any particle varies with velocity (v) of this

object by the following relation

2

2

1c

v

mm o

Where, mo is rest mass of the particle and c is

velocity of light.

2

2

1c

vmmo

If v=c, then

0

1112

2

o

o

m

mc

cmm

If any particle is moving with velocity of light,

then rest mass of the particle is zero. These

particles are called photons.

C. What is Displacement current?

Ans.

Current produced by changing electric field is

known as displacement current (Id). Idea of

displacement current is originated from the

parallel plate capacitor. Consider, a parallel

plate capacitor of plate area A

Electric field inside capacitor

DAQ

EDA

QD

A

QE

A

QE

o

o

o

Current due to this charge,

dsJdsdt

dDI

waygeneralIn

Adt

dD

dt

DAd

dt

dQI

dd

d

..

.)(

Jd is known as displacement current density.

D. Define Proper Length and Proper

Time.

Ans.

Proper Length- The length of an object in its

rest frame is called proper length and

represented by Lo. The length of any object in

motion with respect to an observed always

appears to the observer to be shorter than its

proper length.

Proper Time- The time which is determined

by events that occur at the same place in

observer‟s frame of reference is called proper

time.

E. In diffraction grating spectrum

,which orders of interference will not

be observed if opaque space is double

the slit widths.

Ans.-

Position of principal maxima in a grating

spectrum is given by

)1(sin nde

Where, n is the order of the maximum

Position of minima in the pattern is given


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......3,2,1,

)2(sin

mwhere

me

If both conditions are simultaneously satisfied,

a particular order of maximum will be absent

in grating spectrum, these are known as absent

spectra.

Dividing equation (1) by (2)

)3(

sin

sin

m

n

e

de

m

n

e

de

As per question, opaque space is double of slit

widths i.e. d=2e

....9,6,3

3

3

2

mn

m

n

e

e

m

n

e

ee

Hence, 3rd

, 6th

and 9th

.. order spectra will be

absent.

F. What is specific rotation?

Ans.

Specific rotation for given wavelength of light

at a given temperature is defined as the rotation

produced by one decimeter of the solution

containing one gram of optically active

substances per cc solution.

If θ is the rotation produced by l decimeter

length of solution and C is concentration in

gram per cc, then specific rotation (S) is given

by

lCS

G. What is quarter wave plate?

Ans.

It is crystal plate of double referacting material

of a uniaxial crystal cut with its optical axis

parallel to the refracting faces. If thickness of

the plate is such that it may produce a path

difference of λ/4 between the O-ray and E-ray

when monochromatic plane polarized light of

wavelength λ is incident normal to the plate,

then plate is known as quarter wave plate.

If t is thickness of the plate and μE,μo are the

refractive index corresponding to E-ray and O-

ray, then for calcite crystal,

Eo

t

4

Similarly for quartz crystal

oE

t

4

H. Define population inversion.

Ans.

The term population inversion describes an

assembly of atoms in which the majority are in

energy levels above the ground state.

If N1 and N2 are the number of atoms in

ground state and higher state, then for

population inversion

N2>N1

This is a non equilibrium state and generally

forbidden in the systems as it is the tendency of

atoms to occupy the state of minimum energy

i.e. ground state.

I. How numerical aperture is related

with acceptance angle?

Ans.

Numerical aperture (NA) is a measure of the

amount of light that can be accepted by a fiber.

It is defined as the sine of acceptance angle.

This angle is measure of the light gathering

power of the fiber. If θm is the acceptance

angle, then 2

2

2

1sin nnNA m

Where, n1 and n2 are the refractive indices of

core and cladding

J. Give the main conditions of

interference.

Ans.


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Main conditions of interference are as follows

(i)The interfering waves should be coherent i.e.

phase difference between them must remain

constant with time.

(ii)The two interfering sources should emit

light of same frequency

(iii)In case of polarized light, the two waves

should be in same state of polarization.

SECTION- B

2. Attempt any five parts. All parts carry

equal marks: (5x5=25)

(a)What must be the minimum number of

lines per cm in a half width grating to

resolve the wavelength 5890A0 and 5896 A

0.

Ans.

Given

A

Ad

ndN

cmperlinesofnumberN

spectrumofordern

nNd

gratingofpowersolving

AA

58932

58905896

2

658905896~

,1

,Re

5890,5890

21

21

21

So, number of lines per cm

9826

5893N

a) Light of wavelength 6000A0 falls

normally on a thin wedge shaped film of

refractive index 1.4 forming fringes that

are 2.0 mm apart. Find the angle of the

wedge

Ans.

For normal incidence, fringe-width is given

by

β = λ/(2μ. θ)

2

Given, μ=1.4,λ=6000Å=6000×10-10

m,

λ =6×10-7

m

β=2.0 mm=2×10-3

m

4

44

4

4

3

7

1063

1014.3

198

14.3

180101.1

101.1

6.5

106

1024.12

106

rad

b) Assuming that all the energy from a

1000 watt lamp is radiated

uniformly,calculate the average values of

the intensities of electric and magnetic

fields of radiation at a distance of 2m from

the lamp.

Ans.

P=1000 Watt=1000J/s

For uniform distribution of area, the surface

area will be given by

A= 4πr2=4π.2

2=16π m

2

Since, energy flow per unit area per second is

known pointing vector (S). So, magnitude of

this vector will be given by

16

1000

sec

S

Area

perEnergyS

According to the definition of poynting vector,

HES

In electromagnetic waves electric field vector

and magnetic field vector are perpendicular to

each other. So, magnitude of S will be given by


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)1(16

1000

16

100090sin

EH

EHEHHES o

Intrinsic impedance of free space is given by

)2(6.376

6.376

EH

H

E

Putting this value in eq (1)

mvoltE

E

E

EE

/51.86

02.7496

24.50

0.376600

16

0.376600

6.37616

1000

6.37616

1000

16

1000

6.376

2

From eq. (2)

mAH

EH

/23.0

6.376

51.86

6.376

c) A circular lamina moves with its

plane parallel to the X-Y plane of a

reference frame S at rest. Assuming its

motion to be along the axis of X. Calculate

the velocity at which its surface area would

appear to be reduced to half to an observer

in frame S at rest.

Ans.

If r and D are the radius and diameter of

circular lamina, its surface area (S) is given by

42

22

2 DDrS

Since, lamina is moving in X-direction, so its

diameter in X-direction will be contracted by

2

22

2

2

14

2.

2.

.min.

,

1

c

vD

DyD

radiusmajorradiusorellipseofArea

DD

directiony

insamebewilldiameteritsHowever

c

vDD

x

y

x

According to problem

Area of ellipse=(1/2).area of circle

smv

cv

ccv

cv

c

v

c

v

c

v

D

c

vD

/106.2

866.0

2

732.1

2

3

4

3

4

3

4

11

4

11

2

11

4.

2

11

4

8

22

2

2

2

2

2

2

2

2

22

d) Determine the specific rotation of the

given sample of the sugar solution if the

plane of polarization is turned through

13.20.The length of the tube containing

10% sugar solution is 20cm.

Ans.


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dcmcml

CGiven

lCS

220

1.0100

10%10,2.13,

11 )/()(662

132

1.02

2.13

ccgmdcmlc

S

e) A parallel beam of sodium light

(λ=5890A0)strikes a film of oil floating on

water .When viewed at an angle of 300 from

the normal 8th

dark band is seen

.Determine the thickness of the film

(refractive index of oil = 1.5)

Ans.

Condition for dark fringe of light reflected

from a film of thickness t is

2

35.1

1058905890,8,

)1(cos2

10

mAnGiven

nrt

From Snell‟s law

3

8.2cos

9

8

9

11

sin1cos

3

1

5.1

5.0sin

5.1

30sinsinsin

sin

sin

2

r

rr

r

n

ir

r

i

From eq (1)

8.2

1058908

10589083

8.2

2

32

10

10

t

t

mt

mmt

mt

6659.1

106659.11016659

1058908.2

610

10

f) The value of µe and µo for quartz is

1.5508 and 1.5418 respectively .Calculate

the phase retardation for λ = 5000A0 when

the plate thickness is 0.032 mm.

Ans.

For a quartz crystal, path difference between O

and E-ray will be given by

tOE .

Phase difference between these two rays or

phase retardation

tOE .2

.2

A

Ammt

Given OE

5000

10032.0032.0

,5418.1,5508.1,

7

61728.3nretardatio Phase

5

1080864.1

5

10032.0914.32

5000

10032.010914.32

5000

10032.0009.014.32

10032.0)5418.15508.1(5000

14.32

.2

nretardatio Phase

73

7

7

tOE


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SECTION-C

Note: Attempt all questions. All questions

carry equal marks. (5x7=35)

1. Attempt any one part of the following:

a) Show that the apparent length of a

rigid body in the direction of its motion with

uniform velocity v is reduced by the factor

√1-v2/c2. Discuss the result.

Ans.

Consider, a rod is moving with velocity (v) in

x-direction. The length of this rod in the frame

of references (S), which is at rest is L. If x1

and x2 are position of ends of the rods in this

frame of reference is L, then

L = x2-x1

(1)

Consider, another frame of reference (S‟),

which is moving with the velocity of rod i.e. v.

Let the position of ends of this rod in this

frame be x’2 and x’1 . If L‟ is length of rod in

this frame of reference, then

L‟ = x’2-x’1

(2)

Since, relative velocity of the rod in S‟ frame is

zero, hence length of the rod in this frame of

reference will be proper length (Lo) of the rod.

i.e.

L‟ =Lo= x’2-x’1

(3)

Lorentz transformation is given by

22

1'

1

1 cv

vtxx

22

2'

2

1 cv

vtxx

From equation (3)

)4(1

1

)1(

1

11

22

22

22

12

22

1

22

2

cvLL

cv

LL

equationFrom

cv

xxL

cv

vtx

cv

vtxL

o

o

o

o

Since, v<c

So, L<Lo

Thus the length of the rod seems to be

contracted by a factor of 221 cv .

Discussion:

There may be three cases

(i) if v<<c

From equation (4)

o

o

LL

c

v

cvLL

1

12/122

If the velocity of rod is very much small

compared to velocity of light, then length of

rod will be same.

(ii) if v is comparable to c, then

L<Lo

(iii) if v=c

0

11

1

2/1

2/122

L

LL

ccLL

o

o

This is impossible. Hence, no object can never

attain the velocity of light.

b) Deduce the relativistic velocity

addition theorem .Show that it is consistent

with Einstein’s second postulate Ans.

Velocity addition theorm


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Consider, a particle is moving with If any

particle in moving in X-direction with velocity

V, then Lorentz transformation for this particle

may be written as

22

2

22

'

1'

'

'

1

cv

c

vxt

t

zz

yy

cv

vtxx

(1)

(x, y, z, t) and (x‟, y‟, z‟, t‟) are the coordinates

of points in S and S‟ frame. v is the velocity of

S‟ frame in X-direction.

Components of velocity of particle in X, Y and

Z-direction Vx, Vy, and Vz in S-frame.

Corresponding components are Vx‟, Vy

‟ and

Vz‟ in S‟ frame.

By the definition of velocity,

'

''

'

''

'

'' .

,

dt

dzVand

dt

dyV

dt

dxV

dt

dzVand

dt

dyV

dt

dxV

xyx

zyx

(2)

Differentiating 1 with respect to t

22

2

'

22

'

1

'

'

1

cv

c

vxt

t

zz

yy

cv

vtxx

22

2

''

''

22

'

'

22

'

'

1

'

'

1

1

cv

c

vxt

t

dt

dz

dt

dz

dt

dy

dt

dy

cv

vV

dt

dx

cv

vdt

dx

dt

dx

x

2.Attempt any one part of the following:

a) Explain the concept of Maxwell’s

displacement current and show how it lead

to the modification of Ampere’s law.

Ans.

Displacement Current

According to Maxwell, a changing electric

field is equivalent to a current which flows as

long as electric field is changing. This

equivalent current produces the same magnetic

effect as an ordinary current in a conductor.

This equivalent current is known as

displacement current.

The idea of displacement current was

originated from the study of discharge of a

condenser. During the charging of a parallel

plate capacitor charge starts accumulating

gradually on the plate of capacitor and

accordingly current start decreasing in the

circuit. When the capacitor is fully charged,

current stops. The is no actual flow of charge

between the space of plates during charging. If

a compass is placed between the plates, the

needle deflects. This indicates that there is a

magnetic field between the plates though there

is no source of magnetic field in the gap. The

other source is nothing but the changing


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electric field between the plates and is

equivalent to displacement current.

Consider, the circuit element shown in Figure

which consists of two wires connected to the

plates of a parallel plate capacitor. If a current

Ic flows in the wires, then the charge on the

plates of the capacitor

dt

dQIc

From ampere‟s circuital law

L

cILdB .

i.e. line integral of B around a closed path

equals to the conduction currents passes

through a surface S bounded by L.

However, in the above circuit, there are two

choices of surface S.

(i) S1 passing through one of the wires and

hence gives a contribution Ic to the circuital

law. i.e.

L

cILdB . for S1

(ii) S2 passes through the space between

the capacitor plates. It does not pass through

any conductor and hence does not give a

contribution to circuital law.

L

LdB 0. for S2

b) Show that the velocity of plane

electromagnetic waves in free space is given

by c = 1/√µ0 ε0

Ans.

Maxwell‟s equations are given by

t

EJB

t

BE

B

E

oo

o

0

For free space

000 J

)4(

)3(

)2(0

)1(0

t

EB

t

BE

B

E

oo

Taking curl of eq 3

)5(

,0

)4(&)1(

2

22

2

22

2

2

t

EE

t

EE

t

E

tE

eqeqFrom

t

BEE

t

BE

oo

oo

oo

Similarly by taking curl of eq 4

)6(2

22

t

BB oo

Equation (5) and (6) are the general wave

equation.

General wave equation is given by


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)7(1

2

2

2

2

t

u

vu

Comparing eq(5), eq(6) and eq(7)

oo

oo

oo

v

v

v

1

1

1

2

2

lightofvelocitysmv

v

v

mFNA

oo

oo

/103

109

1

1

1094

1104

11

.1094

1,104

8

16

9

7

1

9

27

Hence, the velocity of electromagnetic waves

in free space is equal to the velocity of light.


a) . Discuss the phenomenon of

interference of light due to thin films and

find the condition of maxima and minima.

Show that the interference patterns of

reflected and transmitted monochromatic

source of light are complimentary

Ans.

Consider, a parallel plate transparent film of

thickness t and refractive index μ. A light

beam SA is entering into this film from air.

One part of this beam is reflected in direction

AR1 and other part is refracted in AB direction.

Beam AB is again reflected in BC direction

which is further refracted in direction CR2.

Two beams AR1 and CR2 reflected and

interfere.

Figure1: Interference in reflected light.

CD is perpendicular drawn from C to line

AR1. Now path difference (Δ) will be given by

Δ = Path ABC in film – Path AN in air

= μ (AB+BC)- AN (1)

From triangle ABM and BCM,

ir

rtAN

irtAN

iACAN

iAC

AN

sin.cos

sin.2

sin.tan2

sin

sin

In triangle ANC

rtCMAMAC

rBMCMAM

r

tBCAB

r

tBC

r

BMAB

rBC

BMr

AB

BM

tan2

tan

cos

2

coscos

cos,cos

From Snell‟s law

ri

r

i

sinsin

sin

sin

i.e.


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r

rtAN

rr

rtAN

cos

sin2

sin.cos

sin.2

2

Putting values of AB, BC and AN in equation

1,

rt

rr

t

rr

t

r

rt

r

t

cos2

.cos.cos

2.

)sin1(cos

2.

cos

sin.2

cos

2.

2

2

2

Reflection at the surface of denser medium

gives additional path difference of λ/2 in ray

AR1, Now total path difference between two

rays

2cos2

rt

(i) Condition for maxima

For constructive interference,

Path difference =2 n.λ/2

i.e.

..2,1,0,2

).12(cos2

2cos2

2cos2

nnrt

nrt

nrt

This is condition for maxima.

When this condition satisfied, then the film

will appear in bright in the reflected light.

(ii) Condition for maxima

For Destructive interference,

Path difference should be odd multiple of λ/2

i.e.

....2,1,0,.cos2

2).112(cos2

22).12(cos2

2).12(

2cos2

nnrt

nrt

nrt

nrt

This is condition for minima.

When this condition is satisfied, then the film

will appear dark in the reflected light.

Interference in Transmitted light-

A light beam SA is entering into this film from

air. One part of this beam is refracted in AB

direction. Beam AB reflected in BC direction

which is refracted in direction CD. Refracted

part of AB and CD beam interfere and give

interefernce pattern

Similar to the phenomena of reflection, path

difference in transmitted may be calculated as

Δ = Path BCD in film – Path BL in air

=μ(BC+CD)-BL (1)

rt cos2

Condition of bright and dark fringes may be

given as

....2,1,0,).12(cos2 nnrt

These conditions are complementary to each

other.

..2,1,0,cos2 nnrt


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b) Describe and explain the formation

of Newton’s rings in reflected

monochromatic light. Prove that the

diameter of bright rings are proportional to

the square roots of odd natural numbers.

A plano-convex lens of large radius of

curvature R is placed on a plane glass plate

with its curved surface downwards and is

illuminated from above with a parallel beam of

monochromatic light. Some of the light is

reflected from the upper surface of the glass

plate and some from the lower surface of the

lens; interference thus occurs by division of

amplitude, the fringes being localised in the air

gap between the lens and plate.

Condition for interference- Wedge shaped

film of air forms between the plano-convex

lens and glass plate. In this case, path

difference between two reflected rays will be

given by

2cos2

rt

For normal incidence, r=0

For air film, μ=1

Path difference will be given by the following

equation

22

t

Since, t is variable in this case, so there is need

to estimate this in terms of any fix value.

Since, radius of curvature R is fixed, so let us

calculate „t‟ in terms of „R‟.

...2

11

,,

)()(

,

,

2

22

22

R

rRR

rRRt

rQMRCQtMO

QMCQMO

CMQIn

RCO

MOCOt

2

22

1.2

2

1

.2

1

,2

11

2

2

2

2

2

2

R

r

R

rdifferencePath

R

rt

R

rRRRt

RrastermsorderhighergNegelectin

R

rRRt


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(i) For bright fringes

Path difference should be an even multiple of

λ/2.

i. e.

2

)12(

2)12(

2)12(

22)2(

2)2(

2

2

2

2

Rnr

Rnr

Rnr

nR

r

nR

r

Diameter of bright fringes

.........3,2,1,)12(

tan2

2.)12(2).12(

2

)12(22

nnD

tconsR

RnRnD

RnrD

n

n

Thus the diameter of bright fringes is

proportional to the square-root of the odd

natural numbers.

Ratio of diameters corresponding to bright

fringes

646.2:236.2:732.1:1:::

7:5:3:1:::

4321

4321

DDDD

DDDD

Separation between successive rings =

0.732:0.504:0.410

Thus seperation between consecutive bright

rings decreases as the order increases.

(ii) For dark fringes

Path difference should be an odd multiple of

λ/2.

i. e.

....3,2,1,0,

.2

.2

.2

.22

)112(

22)12(

2)12(

2

2

2

2

nRnr

Rnr

RnRnr

nR

r

nR

r

Diameter of dark fringes

.........3,2,1,0,

tan4

2..4

22

nnD

tconsR

RnRnD

RnrD

n

n

Thus the diameter of bright fringes are

proportional to the square-root of natural

numbers.

Ratio of diameter of few dark fringes

2:732.1:414.1:1:0:::

4:3:2:1:0:::

4321

4321

DDDD

DDDD

Separation between successive rings =

1:0.414:0.318:0.268

Thus separation between consecutive dark

rings also decreases as the order increases.

Thus, central ringes are broader and peripheral

rings are closer.


a) Write short notes on

(i) Interference filters

Ans.

An intereference filter is a optical system that

transmits a very narrow range of wavelength

and provides a monochromatic beam of light.

They are composed of transparent glass or

quartz substrate on which multiple thin layers


2012


of dielectric material, sometimes separated by

spacer layers

Permit great selectivity

Constructive and destructive interference

occurs between reflections from various

layers

Transmission determined by :

(i)thickness of the dielectric layers

(ii)number of these layers

(iii)angle of incident light on the filters

Advantage

(i)They can be used as reflectors in two and

three color analysis.

(ii)They usually do not themselves produce

fluorescence.

(iii)They are available in short pass versions.

(iv)They are excellent as primary barrier

filters.

Disadvantage

(i)Have lower blocking properties

(ii)Reduced passing properties

(iii)Their reflecting and passing properties are

not absolute, this should be considered while

dealing with multiple wavelengths

(ii) Antireflection coatings

An antireflective or antireflection (AR) coatin

g is a type of optical coating applied to the

surface of lenses and other optical devices to

reduce reflection.

This improves the efficiency of the system

since less light is lost. In complex systems

such as a telescope, the reduction in

reflections also improves the contrast of the

image by elimination of stray light.

Figure shows the phenomena of interference

that lead to antireflecting coating. In this the

refractive index of the coating material is

choosen that it has an intermediate value

between the glass and air.

For example that refractive index of MgF2 is

1.38.

The ray AB is incident on the system and at

point B it suffers reflection on the surface of

AR coating (denser medium) and proceeds

along BC. A part of it moves along BD in the

same medium and at D it suffers reflection

again at the surface of denser (glass) and

emerges out of along EF in air.

Hence, same phase change occurs in both the

rays at reflection.

Reflection at glass/AR coating and AR

coating/air bounaries will produce phase

change of π.

Path difference =2

.2

Total path difference between rays

rt

rt

rt

cos2

cos2

22cos2

Term „λ‟ may be omitted as addition and

subtraction of λ will not affect the path

difference.

For normal incidence,

Path difference = 2μ.t

http://en.wikipedia.org/wiki/Optical_coating

http://en.wikipedia.org/wiki/Lens_(optics)

http://en.wikipedia.org/wiki/Reflection_(physics)

http://en.wikipedia.org/wiki/Light

http://en.wikipedia.org/wiki/Telescope

http://en.wikipedia.org/wiki/Contrast_(vision)

http://en.wikipedia.org/wiki/Stray_light


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Where, μ and t are refractive index and

thickness of AR coating material respectively.

For condition of destructive interference,

2μ.t= (2n+1).(λ/2),where, n=0,1,2..

For n=0

2μ.tmin = λ/2

tmin= λ/4 μ

Minimum thickness of the coating required

for no reflection at the centre of visible

spectrum (λ = 5.5×10-5

cm)

cmt

cmt

5

min

5

min

10966.0

38.14

105.5

4

b) Describe the construction and

principle of Nicol Prism. Explain its

working as a polarizer and analyser.

Ans.

The basic principle behind Nicol Prism is

based on its unique behaviour on the event of

incidence of light rays on its surface.

When an ordinary ray of light is passed

through a calcite crystal, it is broken up into

two rays:

• An „Ordinary ray‟ which is polarized and

has its vibrations perpendicular to the

principle section of the crystal and

• An extra-ordinary ray which is polarized

and whose vibration is parallel to the

principle section of the prism.

• If by some optical means, one of the two

rays eliminates, the ray emerging through

the crystal will be Plane polarized.

• In Nicol Prism, ordinary ray is eliminated

and Extra-ordinary ray, which is plane

polarized, is transmitted through the prism.

A calcite crystal‟s length is three times

its breadth. Let ADFGBC be such a

crystal having ABCD as a principle

section of the crystal with BAD = 700.

The end faces of the crystal are cut in

such a way that they make angles of

680 and 112

0 in the principle section

instead of 710 and 109

0. The crystal is

then cut into two pieces from one blunt

corner to the other along two pieces

from one blunt corner to the other along

a plane perpendicular to the extra

ordinary rays.

Refractive index of Calcite for O ray,

Refractive index of Canada balsam,

Refractive index Calcite of E ray,

Thus we see that the Canada Balsam is

optically denser than calcite for E ray

and rarer for O ray. Finally the crystal

is enclosed in a tube blackened inside.

Working as polariser

Working as an analyser

When principal section of both the Nicols are

parallel, then emitted E-ray from polarizer P

has vibrations parallel to principal section of

analyser A, so get freely transmitted through it.

In this setting of Nicols the intensity of emitted

light is maximum. This position and the

position when the angle between the principle

sections of two prisms is 180° is known as

“Parallel nicols


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When Nicol A is rotated from its position,

intensity of light emitted from it decrease

and becomes zero when principal sections of

two planes at right angle to each other. In this

situation light emitted from polarising Nicol P

has vibration in a plane normal to principal

section of analysing Nicol A and is totally

internally reflected back from Canada Balsam

layer and no light is emitted. In this setting,

two Nicols are said to be “Crossed Nicols”.


a) Discuss the phenomenon of

Fraunhoffer diffraction at a single slit and

show that the relative intensities of the

successive maximum are nearly1: 4/9π2:4/25

π2:4/49 π2

Ans.

A slit is a rectangular aperture whose length is

large compared to its breadth. AB is a slit in

the figure shown below.

Let a parallel beam of monochromatic light of

wavelength λ be incident normally on upon a

narrow slit, AB=e. Let the diffracted light be

focused by a convex length L.

According to Huygen‟s theory, a plane wave

front is incident normally on the slit AB and

each point AB sends out secondary wavelets in

all directions. The rays preceding in the same

direction as the incident rays are focused O;

while those diffracted through an angle θ are

focused at point P.

Let the disturbance caused at P by the wavelet

from unit width of slit M be

tAy cos

Phase difference caused by the wavelet from

width dx at C when it reaches P,

sin.2

.2

x

encepathdiffer

Disturbance caused by this wavelet at P,

)sin.2

cos(.

)cos(.

xwtdxAdy

tdxAdy

For total disturbance

2

2

2

2

2

2

2

2

2

2

2

2

.sin.2

sin.sin

.sin.2

cos.cos

.

sin.2

sin.sin

sin.2

cos.cos

.)sin.2

cos(

)sin.2

cos(.

e

e

e

e

e

e

e

e

e

e

e

e

dxxwtA

dxxwtAy

dx

xwt

xwt

Ay

dxxwtAy

xwtdxAy

dyy


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)()(0.sin.

.cossin2

.cos

sin2

..sin.sin

sin2

..cos.cos

sin2

..sin.sin

sin2

..cos.cos

sin2.,

sin2,

.sin.2

sin.sin

.sin.2

cos.cos

2sin2

2sin2

2sin2

2sin2

2sin2

2sin2

2sin2

2sin2

2sin2

2sin2

2

2

2

2

ufufwtA

duuwtAy

dutwtA

duuwtAy

dutwtA

duuwtAy

dudx

ux

let

dxxwtA

dxxwtAy

e

e

e

e

e

e

e

e

e

e

e

e

e

e

sinsin.

sin.cos

2

sin2sin2.

sin2.cos

2

sin2sin.2

sin2.cos

.cos2.cos

)()(

2sin2

0

2sin2

2sin2

2sin2

0

2sin2

2sin2

ewtAy

ewtAy

e

uduu

duuduu

ufuf

ee

e

ee

e

wtAwtAeyFor

wte

e

Aey

e

e

ewtAy

ewtAy

cos.cos.,0

cossin

sinsin

.

sin

sinsin

..cos

sinsin.

sin.cos

0

wtAy

e

cos.sin

.

sin

0

2

22

0

2

0

sinsin.

sin.

oIAI

AI

AA

For principle maxima

Intensity to be maximum,α should be zero.

ateInderthen

if

minsin

,0sin

,0

So, for estimation of maximum value of I,

2

2

max

sin

OLim

oII

For minima

sinα = 0 and

α= n, n = 1, 2, 3,…

ne

ne

e

sin

sin)/(

sin)/(


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For secondary maxima

tansincos)(

0sin)(

0sincos

sin2

cos11

sinsin

2

sin

3

2

2

ii

or

i

I

Id

dI

II

O

O

O

The first condition (i) is just the condition for

minima that we have already seen (i.e. α = n,

n = 1, 2, 3, …)

The second condition (ii) results in a

transcendental equation whose graphical

solutions can be observed on the left. The

solutions can easily be found numerically: α

1.43, 2.46, 3.47…which approaches

(m+1/2) for large α.

c) Describe the method of production of

plane, circularly and elliptically polarized

light.

Production of plane polarized light-

When a beam of unpolarized light enters into a

nicol, it spilts up into ordinary ray and ordinary

ray

6.Attempt any one part of the

following:

a) Explain the action of He Ne Laser

.How is it superior to Ruby Laser.

A helium-neon laser, usually called a He-Ne

laser, is a type of small gas laser. HeNe lasers

have many industrial and scientific uses, and

are often used in laboratory demonstrations of

optics.

He-Ne laser is a four-level laser.

Its usual operation wavelength is 632.8 nm, in

the red portion of the visible spectrum.

It operates in Continuous Working (CW)

mode.

The setup consists of a discharge tube of length

80 cm and bore diameter of 1.5cm.

The gain medium of the laser, as suggested by

its name, is a mixture of helium and neon

gases, in a 5:1 to 20:1 ratio, contained at low

pressure (an average 50 Pa per cm of cavity

length ) in a glass envelope.


2012


The energy or pump source of the laser is

provided by an electrical discharge of around

1000 volts through an anode and cathode at

each end of the glass tube. A current of 5 to

100 mA is typical for CW operation.

The optical cavity of the laser typically

consists of a plane, high-reflecting mirror at

one end of the laser tube, and a concave output

coupler mirror of approximately 1%

transmission at the other end.

HeNe lasers are normally small, with cavity

lengths of around 15 cm up to 0.5 m, and

optical output powers ranging from 1 mW to

100 mW.

The left side of the representation shows the

lower levels of the helium atoms.The energy

scale is interrupted and that there is a larger

difference in energy in the recombination

process than is evident in the diagram.

A characteristic of helium is that its first states

to be excited, 21S1 and 21S0 are metastable,

i.e. optical transitions to the ground state 11S0

are not allowed, because this would violate the

selection rules for optical transitions. As a

result of gas discharge, these states are

populated by electron collisions

A collision is called a collision of the second

type if one of the colliding bodies transfers

energy to the other so that a transition from the

previous energy state to the next higher or

lower takes place. Apart from the electron

collision of the second type there is also the

atomic collision of the second type. In the

latter, an excited helium atom reaches the

initial state because its energy has been used in

the excitation of a Ne atom. Both these

processes form the basis for the production of a

population inversion in the Ne system

Working

A description of the rather complex HeNe

excitation process can be given in terms of the

following four steps.

(a)When the power is switched on, An

energetic electron collisionally excites a He

atom to the state labeled 21So . A He atom in

this excited state is often written He*(21So),

where the asterisk means that the He atom is in

an excited state.

(b) The excited He*(21So) atom collides with

an unexcited Ne atom and the atoms exchange

internal energy, with an unexcited He atom and

excited Ne atom, written Ne*(3s2), resulting.

This energy exchange process occurs with high

probability only because of the accidental near

equality of the two excitation energies of the

two levels in these atoms. Thus, the purpose of

population inversion is fulfilled.

When the excited Ne atom passes from

metastable state(3s) to lower level(2p), it emits

photon of wavelength 632 nm.

This photon travels through the gas mixture

parallel to the axis of tube, it is reflected back

and forth by the mirror ends until it stimulates

an excited Ne atom and causes it to emit a

photon of 632nm with the stimulating photon.


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The stimulated transition from (3s) level to

(2p) level is laser transition.

This process is continued and when a beam of

coherent radiation becomes sufficiently strong,

a portion of it escape through partially silvered

end.

The Ne atom passes to lower level 1s emitting

spontaneous emission. and finally the Ne atom

comes to ground state through collision with

tube wall and undergoes radiationless

transition.

b) Explain the various types of losses in

optical fiber.

Ans.

Losses in optical fiber-


a) Explain the phenomenon of double

refraction calcite crystal. Define ordinary

and extraordinary rays.

Ans. Double refraction calcite crystal-

Certain crystals like quartz, mica, calcite

have the property of producing two refracted

rays for

every ray that is incident on them.

The phenomenon of refraction where in two

refracted rays are produced for a given

incident ray

is called double refraction or birefringence.

The two refracted rays are plane polarized in

mutual

perpendicular planes. By eliminating one of

the them, plane polarized light can be

obtained. The

crystals which exhibit this property are called

doubly refracting crystals.

b) Describe a half shade polarimeter

and explain how it is used to measure the

strength of sugar solution

Polarimeter is an instrument used to measure the

specific rotation of an optically active solution.

Construction: It consists of two Nicol prisms N1

and N2 which acts as polarizer and analyzer

respectively. A glass tube T is placed between N1

and N2. A half shade device „H‟ is placed

between N1

and T. Half shade device is a circular plate, half

of which is quartz and the other half is glass. A

telescope is used to observe the light emerging

out of the analyzer. A circular main scale S and

two

verniers V1 and V2 are fixed to the analyzer.

Working:-

Light from a monochromatic source is made to

fall on polarizer N1 using a condensing lens. The


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glass T is filled with distilled water. The light

emerging out of N1 is plane polarized. Half of

this beam

falls on quartz portion and the other half on the

glass portion of the half shade device. The plane

of

vibrations of light passing through glass portion

remains unaltered while that passing through

quartz

(Optically active) is rotated. Hence when viewed

through the telescope, we have two distinct

halves. By

rotating the analyzer both halves of field of view

are made equally bright. Reading R0 is noted.

Now,

the tube T is filled with optically active solution

of known concentration „c‟. When viewed

through the

telescope the brightness of the two halves are

different again. The analyzer is rotated to have

uniform

brightness again and the reading R is noted.

R~R0=θ gives the angle of rotation of the plane

of

polarization. The experiment is repeated for

different concentrations „c‟ and the results are

tabulated.

PRE SEM SOLUTION 2013-B.TECH IST YEAR

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