QUESTIONS FOR PRACTICE SOLUTIONS - VK Global ...

1Questions for Practice Solutions

QUESTIONS FOR PRACTICE SOLUTIONS

CHAPTER-1: ELECTROMAGNETIC WAVES 1. (i) A plane electromagnetic wave travels in vacuum along X-direction. What can you say about

the direction of electric and magnetic field vectors? [CBSE Delhi 2011]

(ii) Write the relation for the speed of electromagnetic waves in terms of the amplitudes of electric and magnetic fields. [CBSE (AI) 2017]

(iii) In which directions do the electric and magnetic field vectors oscillate in an electromagnetic wave propagating along the X-axis? [CBSE (AI) 2017]

Ans. (i) Electric field vector along Y-axis Magnetic field vector along Z-axis. (ii) Speed of electromagnetic wave in terms of amplitudes of electric and magnetic field,

c B

E

0

0=

(iii) Electric field vector ( )E oscillates along Y-axis and magnetic field vector ( )B along Z-axis.

2. (i) Name the electromagnetic radiation to which waves of wavelength in the range of 10–2 m belong. Give one use of this part of EM spectrum. [CBSE Delhi 2009]

(ii) Name the electromagnetic radiation which can be produced by klystron or a magnetron valve. [CBSE (F) 2009]

Ans. (i) Range of microwave vary from 0.3 m to 10–3 m. They are used in aircraft navigation. (ii) Electromagnetic radiation product by a klystron or a magnetron valve is microwave.

3. Write the speed of electromagnetic waves in terms of µ0 and e0.

Ans. Speed of electromagnetic wave, ,c c1

0 0n f= = speed of EM wave in vaccum.

4. Write two uses of microwaves. [CBSE (F) 2011]

Ans. Uses of microwaves: (i) In long distance communications (ii) In radar

5. What is the phase difference between electric and magnetic field vectors in electromagnetic wave?

Ans. 2zr=

6. (i) How are radio waves produced? [CBSE (AI) 2011]

(ii) Which part of the electromagnetic spectrum corresponds to wavelengths (a) 10–10 m (b) 10–12 m?

(iii) Name the electromagnetic spectrum involving wavelength

(a) 1 Å (b) 4000 Å

2 Physics–XII: Term–2

Ans. (i) Radio waves are produced by the accelerated motion of charges in conducting wires. (ii) (a) X – rays (b) X – rays (iii) (a) X – rays (b) Gamma rays

7. (i) To which part of the electromagnetic spectrum does a wave of frequency 5×1011 Hz belong? [CBSE (AI) 2014]

(ii) Which part of the electromagnetic spectrum is used in satellite communication? [CBSE (F) 2010]

Ans. (i) Microwaves or short radiowaves (ii) Short radiowaves m or10 30< >m y MHz are used in satellite communication.

8. Name the electromagnetic radiation used to destroy cancer cells and write its frequency range. [CBSE (F) 2010]

Ans. c -rays. Frequency range: 1019 Hz—1022 Hz.

9. To which part of the electromagnetic spectrum do the waves emitted by radioactive nuclei belong? What is its frequency range? [CBSE (F) 2010]

Ans. c -rays.

Frequency range: 1019 Hz—1022 Hz.

10. (i) What evidence is there that sound wave is not an electromagnetic wave?

(ii) What is the relationship between amplitude of electrical and magnetic fields in free space?

Ans. (i) Sound wave require material medium for their propagation. It shows that the sound waves are mechanical waves in nature and not electromagnetic waves.

(ii) Relation between E and B in free space.

c BE=

11. Identify the part of the electromagnetic spectrum which is

(i) suitable for RADAR systems used in aircraft navigation.

(ii) adjacent to low frequency end of the electromagnetic spectrum.

(iii) produced in nuclear reactions.

(iv) produced by bombarding a metal target by high speed electrons.

Ans. (i) Microwave (ii) Radiowaves (iii) Gamma rays

(iv) X–rays

12. What is the range of frequencies used in satellite communication? What is common between these waves and light waves? [CBSE Delhi 2010]

Ans. Short radio waves m or10 30< >m y MHz are used in satellite communication.

Speed of waves is same for satellite communication and light waves.

13. How are electromagnetic waves produced? What is the source of the energy carried by a propagating electromagnetic wave?

Identify the electromagnetic radiations used (i) in remote switches of household electronic devices; and (ii) as diagnostic tool in medicine. [CBSE Ajmer 2015]


Ans. Electromagnetic waves are produced by oscillating charges which produce oscillating electric field and magnetic field. Source of the energy carried by a propagating electromagnetic wave is the energy of the accelerated charge.

The electromagnetic radiations used are (i) Infra-red radiations (ii) X-rays

14. Write any four characteristics of electromagnetic waves. Give any two uses each of

(i) Radio waves (ii) Microwaves. [CBSE Delhi 2010, 2011]

Ans. Characteristics of electromagnetic waves: (a) The do not require any material medium for their propagation. (b) The oscillations of vector E and vector B are perpendicular to each other and are in same

phase. (c) They are transverse in nature. (d) They travel through vacuum with same speed, c = 3 × 108 ms–1

(i) Radio waves are used in (a) Radio transmission (b) Radio astronomy (ii) Microwaves are used in (a) Microwave -oven (b) Radar systems

15. Name the constituent radiation of electromagnetic spectrum which [CBSE (F) 2010]

(a) is used in satellite communication.

(b) is used for studying crystal structure.

(c) is similar to the radiations emitted during the decay of radioactive nuclei.

(d) has its wavelength range between 390 nm and 700 nm.

(e) is absorbed from sunlight by ozone layer.

(f) produces intense heating effect.

Ans. (a) Radiowaves (b) X-rays (c) c -rays (d) Visible rays (e) UV rays

(f) Infrared

16. Name the electromagnetic waves with their frequency range, produced in

(a) some radioactive decay

(b) sparks during electric welding

(c) TV remote [CBSE 2020 (55/4/1)]

Ans. (a) 1019 – 1023 Hz, Gamma rays

(b) 1015 to 1017 Hz, UV rays

(c) 107 – 108 Hz, Radio waves


Numerical Questions 1. Electromagnetic waves travelling in a medium having relative permeability .1 3rn = and relative

permittivity .2 14rf = . Find the speed of electromagnetic waves in medium.

Ans. Speed of EM ware, . .

. × /m svc

2 14 1 33 10

1 8 10×

×

0 0

88

n f= = =

2. The rms value of the electric field of light coming from the sun is 720 N/C. Find the average total energy density of the electromagnetic wave.

Ans. Total energy density, u = ue + um

= 2ue ( ue = um)

E E2 21

× 02

02f f= =

= 8.85 × 10–12 × 720 = 6.37 × 10–9 Jm–3

3. The electric field of an electromagnetic wave in free space is given by E = 10 cos (107t + kx) jt V/m where x is in meters and t in seconds. Find

(a) the wavelength, λ (b) the propagation constant, k (c) the wave amplitude, E0 .

Ans. As given equation, ( )cosE t kx jE 0 ~= + t

( )cos t kx jE 10 107= + t

From comparing both, we get, E0= wave amplitude = 10 V/m. Angular velocity, ~= 107 rad/s

Time period, T = × s2

10

22 10

77–

~r r

r= =

Wavelength, × × × × . mv t 3 10 2 10 188 48 7–m r= = =

Hence, (a) . m188 4m= (b) k = . . ( ) /m V mc E2

188 42

0 033 1010

–

m

r r= = =

4. Electromagnetic waves travelling in a medium has speed 2 × 108 m/s. If the relative permeability is 1, then calculate the relative permittivity of medium.

Ans. Speed of EM wave, vc

r rn f=

⇒ .vc 1

2 103 10

1 2 25××

×r r

2

8

8 2

f n= = =b fl p

z z z


CHAPTER-2: RAY OPTICS AND OPTICAL INSTRUMENTS

1. (a) For the same angle of incidence, the angles of refraction in three different media A, B and C are 15°, 25° and 35° respectively. In which medium the velocity of light is minimum? [CBSE (AI) 2012]

(b) A ray of monochromatic light passes from medium (1) to medium (2). If the angle of incidence in medium (1) is i and the corresponding angle of refraction in medium (2) is i /2, which of the two media is optically denser? Give reason. [CBSE (F) 2013]

Ans. (a) From Snell's law, sinsin

nri

vc= =

For given i,v \ sin r ; r is minimum in medium A, so velocity of light is minimum in medium A.

(b) Medium (2) is optically denser.

Reason - angle of refraction is less than the angle of incidence.

2. (a) When monochromatic light travels from one medium to another, its wavelength changes but frequency remains the same. Explain. [CBSE Delhi 2011]

(b) An object is held at the principal focus of a concave lens of focal length f. Where is the image formed? [CBSE (AI) 2008]

Ans. (i) When monochromatic light travels from one medium to another its wavelength and speed both change such that

v v

1

1

2

2

m m= frequency

so frequency remains unchanged.

(ii) From lens formula, f v u1 1 1

–=

Here, u = – f and f = – ve (concave lens)

So, v f fv

f1 1 12–

– –&= =

That is image will be found between optical centre and focus of lens, towards the side of the object.

3. (a) The line AB in the ray diagram represents a lens. State whether the lens is convex or concave. [CBSE Patna 2015]

A

B

(b) A diverging lens of focal length ‘f ’ is cut into two identical parts, each forming a plano concave lens. What is the focal length of each part?


Ans. (a) Concave lens (b) For a complete diverging lens

( )f

n R R1

11 1

––

–g= c m

⇒ ( )f nR

2 1––

g=

For each plano concave lens,

f

nR

1 1 1 1– – –g 3=l` cj m

⇒ f nR

f1 2––

g= =l f p

i.e., focal length of each half part will be twice the focal length of initial diverging lens. 4. A biconvex lens made of a transparent material of refractive index 1.5 is immersed in water of

refractive index 1.33. Will the lens behave as a converging or a diverging lens? Give reason. [CBSE (AI) 2014]

Ans. As a diverging lens

– – – – – – – – – – – – – – – – – –– – – – – – – – – – – – – – – – – –– – – – – – – – – – – – – – – – – –– – – – – – – – – – – – – – – – – –– – – – – – – – – – – – – – – – – –– – – – – – – – – – – – – – – – – –– – – – – – – – – – – – – – – – – –– – – – – – – – – – – – – – – – – –– – – – – – – – – – – – – – – – – –– – – – – – – – – – – – – – – – – –– – – – – – – – – – – – – – – – – –– – – – – – – – – – – – – – – – – –– – – – – – – – – – – – – – – – – –– – – – – – – – – – – – – – – – – –– – – – – – – – – – – – – – – – – –– – – – – – – – – – – – – – – – – –– – – – – – – – – – – – – – – – – –– – – – – – – – – – – – – – – – – –– – – – – – – – – – – – – – – – – –– – – – – – – – – – – – – – – – – –– – – – – – – – – – – – – – – – – –– – – – – – – – – – – – – – – – – –

ng =1.25

medium nw =1.33

As the light travels from rarer to denser medium, it diverges from its path. Alternate method On using thin lens maker formula

f n

nR R

11

1 1– –

w w

g

1 2= f ep o

On using sign convention R1 = + ve, R2 = –ve, and ng = 1.25 and nw = 1.33

.

f R R1

1331 25

11 1

–w 1 2

= +d en o

R R1 1

1 2+ =e o +ve value and

.– ve133

1 251 –=d n value

Hence fw is –ve. So, it behaves as a diverging lens. 5. (a) A glass lens of refractive index 1.5 is placed in a liquid. What must be the refractive index of

the liquid in order to make the lens disappear ? [CBSE Delhi 2008] (b) A glass lens of refractive index 1.45 disappears when immersed in a liquid. What is the value

of refractive index of the liquid? [CBSE Delhi 2010] Ans. (a) For disappearance of glass lens in liquid, refractive index of liquid should be equal to

refractive index of lens = 1.5. (b) For disappearance of glass lens in liquid, refractive index of liquid = refractive index of lens = 1.45


6. When light travels from a rarer to a denser medium, the speed decreases. Does this decrease in speed imply a decrease in the energy carried by the light wave? Justify your answer. [CBSE (AI) 2010]

Ans. No; when light travels from a rarer to denser medium, its frequency remains unchanged. According to quantum theory, the energy of a light beam depends on frequency and not on speed.

7. A converging lens is kept co-axially in contact with a diverging lens – both the lenses being of equal focal lengths. What is the focal length of the combination? [CBSE (AI) 2010]

Ans. Let focal length of converging and diverging lenses be +f and –f respectively.

Power of converging lens, ,Pf1

1 = Power of diverging lens, Pf1

–2 =

∴ Power of combination P P Pf f1 1

0–1 2= + = =

∴ Focal length of combination, F P1

01

3= = =

F 3=

8. To increase the magnifying power of a telescope, the objective and eyepiece of higher power can be taken. But in practice, the magnifying power cannot be increased beyond a certain limit. Explain.

Ans. The visual angle increases with increase of magnifying power. If we increase the magnifying power beyond a certain limit then the distinctness of the object will get lost and we will not be able to see detailed image of the object. The object will get blurred. So, magnifying power of the telescope cannot be increased indefinitely.

9. Two monochromatic rays of light are incident normally on the face AB of an isosceles right-angled prism ABC. The refractive indices of the glass prism for the two rays ‘1’ and ‘2’ are respectively 1.3 and 1.5. Trace the path of these rays after entering through the prism. Explain briefly. [CBSE (AI) 2014]

'1'

A

B C

45°

45°

'2'

Ans. As we know that the critical angle depends on refractive index n as

sin i n1

c =

If , thenc n45 2°+ = =

If . , . °sinn c1 3 1 31

45>1–+= = c m So the ray 1 is refracted out.

If . , . °sinn c1 5 1 51

45<1–+= = c m So the ray 2 is totally reflected back.


10. In the fig. given alongside, three light rays, red (R), green (G) and blue (B) are incident on an isosceles right-angled prism ABC at face AB. Explain with reason, which ray of light will be transmitted through the face AC. The refractive index of the prism for red, green, blue light are 1.39, 1.44, 1.47 respectively.

A

BGR

B C45°90°

Trace the path of rays after passing through face AB. [CBSE Delhi 2009, (F) 2011, 2013] Ans. If angle of incidence ‘i’ is less than the critical angle of glass-air interface AC then it will emerge

out. Critical angle sin iC = n

1 ...(i)

∴ .sin sin

ni

145

12 1 414

°C

Since nR = 1.39, nG = 1.44 and nB = 1.47, so from equation (i) angle of incidence for red colour iC > 45° while angle of incidence for blue and green colours iC < 45°, hence blue and green colour rays will emerge out.

11. Trace the path of ray (P) of light passing through the glass prism as shown in the figure. The prism is made of glass with critical angle ic= 40°. [CBSE (F) 2012]

P

A

B C45°

Ans. From the figure, the incident ray is normal to the surface of prism. So, incident angle = 0° Then, angle of refraction will be zero. It means that the ray of light will pass through the prism

undeviated, reaches the other end of prism. The second angle of incidence = 45° (greater than critical angle of prism). The ray of light

undergoes the phenomenon of total internal reflection and continues in the same manner. 12. Calculate the angle of emergence (e) of the ray of light incident normally on the face AC of a glass

prism ABC of refractive index 3 . How will the angle of emergence change qualitatively, if the ray of light emerges from the prism into a liquid of refractive index 1.3 instead of air?

[CBSE 2020 (55/5/1)] Ans. Given n 3g =

i = 0°

At the interface AC,

By Snell’s law

sinsin

ri

n

n

a

g=

A

B C

45°

B

G

R

A

B

e

C60°

30°


But sin i = sin 0° = 0, hence r = 0 At the interface AB, i = 30° Applying Snell’s law

° °sin

sinsin sin

e nn

e e30

31

3 30 60°

g

a & &= = = =

If light ray emerges from prism into a liquid of refractive index 1.3 instead of air, then it will bend away from normal by smaller extent as compared to air so angle of emergence will be less than 60°.

13. How is the working of a telescope different from that of a microscope? Ans. Difference in working of telescope and microscope: (i) The objective of a telescope forms the image of a very far off object at or within, the focus of

its eyepiece. The microscope does the same for a small object kept just beyond the focus of its objective.

(ii) The objective of a telescope has large focal length and large aperture, while the corresponding for a microscope have very small value.

(iii) The final image formed by a telescope is magnified relative to its size as seen by the normal eye, while the final image formed by a microscope is more magnified relative to its absolute size.

14. (i) State the principle on which the working of an optical fiber is based. (ii) What are the necessary conditions for this phenomenon to occur? [CBSE (AI) 2009] Ans. (i) Working of an optical fibre is based on total internal reflection. (ii) (a) Rays of light have to travel from optically denser medium to optically rarer medium and (b) Angle of incidence in the denser medium should be greater than critical angle. 15. (a) Draw a labelled ray diagram of a compound microscope. (b) Derive an expression for its magnifying power. Ans. (a) Labelled diagram of compound microscope. The objective lens form image A' B' near the first focal point of eyepiece.

AB′′

Eyepiece

B′E

A′

A′′

Objective lens

B Fo

h

u fo fe

D

h′β Fe

(b) Angular magnification of objective lens m0 = linear magnification = hhl

Since tan Lh

fh

0b= =

l

mhh

fL

00

= =l

...(i)


where L is the distance between second focal point of the objective and first focal point of eyepiece.

If the final image A" B" is formed at the near point.

Angular magnification, m fD1

e e= +e o

If the final image is formed at infinity, then angular magnification me = fD

e ...(ii)

Thus, total magnification of the compound microscope M = m0 × me

= fL

fD×e0

16. (a) Draw a labelled ray diagram of a refraction type telescope in normal adjustment. (b) Give its two shortcomings over reflection type telescope. (c) Why is eyepiece of a telescope of short focal length, while objective is of large focal length?

Explain. [CBSE (F) 2013] Ans. (a) Refraction type telescope in normal adjustment.

Eyepiece

OE

Objective lens

fo fe

h′α βB′

A′

(b) Shortcomings — (i) Very difficult and expensive to make large sized lenses, which can form images that are

free from any kind of chromatic aberration and distortions. (ii) Big lenses of diameter 1.02 m and above tend to be very heavy and therefore, are difficult

to make and need to be supported by their edges. (c) If eyepiece of short focal length and objective lens of large focal length are used in constructing

the telescope, magnifying power increases, as per relation M = f

f

e

0.

Thus telescope of high magnifying power can see path of stars of actual separation of one minute arc and even less values of arc.

17. Define the magnifying power of refractive telescope and write the expression for it. Write two important limitations of a refracting telescope over a reflecting type telescope. [CBSE (AI) 2013]

Ans. It is defined as the ratio of the angle subtended by the final image on the eye (b) to the angle subtended by the object on eye (a).

tantan

M ab

ab= = c m


Magnifying power Mf

f–

e

0= (for comfortable view)

f

f

D

f1

–

e

e0= +e o (for strained eye)

Limitations: (i) Image is not free from chromatic aberration and spherical aberration. (ii) Aperture of the objective lens should be large for high resolving power. 18. Draw a ray diagram to show the image formation in a refracting type astronomical telescope

in the near point adjustment. Write the expression for its magnifying power. Why should the diameter of the objective of telescope be larger?

Ans.

Magnifying Power The magnifying power of a telescope is measured by the ratio of angle (b) subtended by final

image on the eye to the angle (a) subtended by object on the eye, i.e.,

Magnifying power M ab

=

As a and b are very small angles, therefore, from figure. The angle subtended by final image A′′ B′′ on eye

b = angle subtended by image A′ B′ on eye =tan'

' 'C AAB

2b =

As the object is very far (at infinity) from the telescope, the angle subtended by object at eye is same as the angle subtended by object on objective lens.

'

' 'tan

C AAB

1a a= =

' '/ '

' '/ '

'

'M

AB C A

AB C A

C A

C A

1

2

2

1ab

= = =

If the focal lengths of objective and eye-piece be fo, and fe , distance of image A′ B′ from eye-piece be ue, then by sign convention

C1 A′ = + f0 , C2 A′ = – ue

–M uf

e

0= ...(i)

If ve is the distance of A''B'' from eye-piece, then by sign convention, fe is positive, ue and ve both

are negative. Hence by lens formula – ,f v u1 1 1= we have

– (– ) orf v u u f v1 1 1 1 1 1

–e e e e e e

= = +


Substituting this value in (i), we get

–M ff v1 1

e e0= +d n ...(ii)

This is the general formula for magnifying power. In this formula only numerical values of f0 , fe and ve are to be used because signs have already been used.

The aperture of the objective lens is preferred to be large so that it may collect sufficient light to form a brighter image of a distant object.

19. Draw ray diagram when incident ray falls normally on one of the two equal sides of a right angled isosceles prism having refractive index µ = 3. [CBSE South 2016]

Ans. On face AC, i = 45°

For, TIR to take place at face AC,

Ray 1

A

B C

i

45°

45°

Ray 2

Ray 3

Here, sin iC = 1

31

n =

⇒ ic = sin–1 .3

135 26°=e o

Hence, i > ic, so TIR takes place at face AC.

20. (a) Draw a ray diagram showing the formation of image by a reflecting telescope. (b) Write two advantages of a reflecting telescope over a refracting telescope. [CBSE (AI) 2017] Ans. (a)

Secondaryconvex mirror

Objectiveconcave mirror

Eyepiece

(b) Advantages: (i) Parabolic mirror is used to remove the spherical aberration. (ii) No chromatic aberration in mirror. (iii) Light mechanical support is required, because mirror weighs much less than a lens of

equivalent optical quality.


21. (a) Draw a ray diagram for the formation of image by a compound microscope. (b) You are given the following three lenses. Which to lenses will you use as an eyepiece and as

an objective to construct a compound microscope?

Lenses Power (D) Aperture (cm)

L1 3 8

L2 6 1

L3 10 1

[CBSE (AI) 2017] Ans. (a)

uo vo ue

B

A

B" ve

EO

B'

A'Fe'Fo

D

Fe

Eyepiece

A"

Objective L1

(b) The condition under which a large magnification can be achieved in an astronomical telescope is fo >> fe , focal length of objective must be greater than focal length of eyepiece.

Objective : Lens L1 Eyepiece : Lens L3 Reason: The objective lens should have large aperture (here, 8 cm) and large focal length

Powerf1

=d n while the eyepiece should have small aperture and small focal length.

22. A magician during a show makes a glass lens with n = 1.47 disappear in a trough of liquid. What is the refractive index of the liquid? Could the liquid be water?

Ans. The refractive index of the liquid must be equal to 1.47 in order to make the lens disappear. No, the liquid is not water because refractive index of water = 1.33. It could be glycerine. 23. (a) A point object ‘O’ is kept in a medium of refractive index n1 in front of a convex spherical

surface of radius of curvature R which separates the second medium of refractive index n2 from the first one, as shown in the figure.


Draw the ray diagram showing the image formation and deduce the relationship between the object distance and the image distance in terms of n1, n2 and R.

(b) When the image formed above acts as a virtual object for a concave spherical surface separating the medium n2 from n1 (n2 > n1), draw this ray diagram and write the similar (similar to (a)) relation. Hence obtain the expression for the lens maker’s formula. [CBSE Delhi 2015]

Ans. (a) n1 < n2

M

ir

γPN C

α βO

u Rv

n2

I

n1

In the above figure, (i) all angles are small hence, sin , tani i b b= = (ii) curvature of the surface is small hence PN is negligible. Consider ∆OMC and ∆MCI i a c= + ...(i) rc b= + ...(ii) Using Snell’s law, n1 sin i = n2 sin r n1 i = n2 r n1 ( )a c+ = n2 –c b^ h n1 tan tana c+^ h = n2 ( – )tan tanc b [a for small angle sin i - i and sin r - r]

n NOMN

NCMN

n NCMN

NIMN

–1 2+ =c cm m

Now, NO . PO = – u NC . PC = + R NI . PI = + v

⇒ n u R n R v1 1 1 1– –1 2

+ =; ;E E ⇒ vn

un

R

n n–

–2 1 2 1= ...(iii)

(b) Ray diagram:

C

R′2

vI

v′

I′

n2 n1

vn

v

n

R

n n–

–2 1 1 2+ =l l

...(iv)

Adding (iii) and (iv), we get

un

vn

vn

v

nn n

R R1 1

– – – –1 2 2 12 1+ + + =

l l_ di n

⇒ u v nn n

R R1 1 1 1–

––

1

2 1+ =l

f dp n ⇒ f n

n nR R

1 1 1––

1

2 1=l

f dp n


24. (i) Draw a labelled schematic ray diagram of astronomical telescope in normal adjustment. (ii) Which two aberrations do objectives of refracting telescope suffer from? How are these

overcome in reflecting telescope? Ans. (i)

Final image

at infinity

(ii) (a) It suffers from chromatic aberration (b) It suffers from spherical aberration Reflecting telescope is preferred over refracting telescope because (a) No chromatic aberration, because mirror is used. (b) Spherical aberration can be removed by using a parabolic mirror. (c) Image is bright because no loss of energy due to reflection. (d) Large mirror can provide easier mechanical support. 25. (a) How does the refractive index of a transparent medium depend on the wavelength of

incident light used? (b) An equiconvex lens of focal length ‘f’ is cut into two identical plane convex lenses. How will

the power of each part be related to the focal length of the original lens? Ans. (a) Refractive Index: The refractive index of a medium is defined as the ratio of speed of light in vacuum to the

speed of light in a medium.

i.e., Speed of light in mediumSpeed of light in vacuum

n vc= =

medium

air

medium

air

om

om

m

m= = …(i)

λair and λmedium being wavelengths of light in air and medium respectively.

∴ /

/

sinsin

ri

nn

c v

c vvv

1

2

1

2

2

1

2

1

m

m= = = =f p ...(ii)

(b) For a biconcave lens

– –( – )f

n R R fn R P

1 11 1 1

12–

1 2= + = = =] eg o

When a lens is split,

–( – )f

n R1

11

1= ⇒ –

fn

R1 1 1–2

= ] g

f f f1 1

21

1 2= = ⇒ P P

P21 2= =


26. Write two characteristics of image formed when an object is placed between the optical centre and focus of a thin convex lens. Draw the graph showing variation of image distance v with object distance u in this case. [CBSE Sample Paper 2021]

Ans. Characteristics of the image formed: (i) Virtual and enlarged image (ii) Same side of the object We know that,

v u f1 1 1

– =

As both image and object lie on the same side, both v and u are negative,

.v u f v f u u f1 1 1 1 1 1 1 1–

or–

–+ = = + =

Comparing with y = mx + c, the graph is shown. 27. A ray of light PQ enters an isosceles right angled prism

ABC of refractive index 1.5 as shown in figure.

A

Q

CB

P

90°

45°

(i) Trace the path of the ray through the prism. (ii) What will be the effect on the path of the ray if the refractive index of the prism is 1.4?

[CBSE 2020 (55/4/1)] Ans. (i) n = 1.5

sini n1

c1–= b l

.sin sin32

0 661 1– –= =c ]m g = 41° i = 45°C Since i > ic Hence the ray will suffer total interval

reflection

(ii) If n = 1.4

sini n1

c1–= b l

= 45.58

i = 45°

Hence i < ic The ray will get refracted.

O

1v

y

x1u−1

f

45°45°

45° 45°45°

B

A

C

A

CB

Q

45°


Numerical Questions 1. A screen is placed 90 cm away from an object. The image of the object on the screen is formed

by a convex lens at two different locations separated by 20 cm. Determine the focal length of the lens. [CBSE Guwahati 2015, 2019 (55/5/1)]

Ans. Given separation between object and screen, D = 90 cm Separation between two positions of lens, x = 20 cm

∴ Focal length of lens, ( ) ( )

fD

D x4 4 90

90 204 90

8100 400– – –2 2 2 2

# #= = =

. cm4 907700

21 4×= =

2. A converging lens has a focal length of 20 cm in air. It is made of a material of refractive index 1.5. If it is immersed in water of refractive index 4/3, what will be the new focal length? [CBSE (F) 2010]

Ans. (a) Wavelength of light has no effect on focal length of a spherical mirror.

(b) Given, fa = 20 cm, ng = 1.5, nl = 34

×

...

× cm cmf

n

n

nf

1

1

1 331 5

1

1 5 120 80

–

–

–

–l

l

g

g

a c= =c m

3. A convex lens of focal length 20 cm is placed coaxially in contact with a concave lens of focal length 25 cm. Determine the power of the combination. Will the system be converging or diverging in nature? [CBSE Delhi 2013]

Ans. Convex lens and concave lens are in contact as shown in fig.

Power of convex lens, P1 = ( ) ( )in m in cmf f1 100

1 1+=

= 25

100+

= 4D

Power of convex lens, P2 = f

120

100–2+

=

= – 5D Power of combination, P = P1+ P2 = 4D + (–5D) = –1D 4. A diverging lens of refractive index 1.5 and of focal length 20 cm in air has the same radii of

curvature for both sides. If it is immersed in a liquid of refractive index 1.7, calculate the focal length of the lens in the liquid. [CBSE (AI) 2008]

Ans. As we know, fn

n

n

1

1

–

–

l

l

g

g

=

J

L

KKKKKKKKK

N

P

OOOOOOOOO ⇒ fl =

.

.

.( ) .

.( )

1 71 5

1

1 5 120 0 88 1

0 520–

–– – × –=

J

L

KKKKKK

N

P

OOOOOO

= +83.4 cm Here +ve sign shows that lens is converging in nature.

5. A biconvex lens has a focal length 32

times the radius of curvature of either surface. Calculate the

refractive index of lens material. [CBSE Delhi 2010]

Ans. ( )f

n R R1

11 1

– –1 2

= e o

L1 L2

f1= 25cm f2= –20cm


For biconvex lens, , –R R R R1 2=+ =

Given f R32

=

∴ ( ) ( )R n R n n23

12

1 43

1 43

47

– –& &= = = + =c m 6. Find the radius of curvature of the convex surface of a plano-convex lens, whose focal length is

0.3 m and the refractive index of the material of the lens is 1.5. [CBSE Delhi 2010] Ans. For a plano-convex lens ,R R R–1 23= =

∴ ( )f

n R R1

11 1

– –1 2

= e o gives

( – )f

n R1

11 13= +c m

or ( – )f R

nR n f

1 11

– &= =

Given f = 0.3 m, n = 1.5 ∴ R = (1.5 – 1) × 0.3 m = 0.15 m = 15 cm 7. The focal lengths of the objective and eyepiece of a microscope are 1.25 cm and 5 cm respectively.

Find the position of the object relative to the objective in order to obtain an angular magnification of 30 in normal adjustment. [CBSE Delhi 2012]

Ans. Given, fo = 1.25 cm, fe = 5 cm Angular magnification, m = 30 Now, m = me × mo

In normal adjustment, angular magnification of eyepiece

mfd

525

5ee

= =+ =

Hence m0 = 6

But m uv

0 0

0=

⇒ uv

6–0

0= ⇒ v u6–0 0=

Applying lens equation to the objective lens

.f v u u u

1 1 11 25

161 1

– – –0 0 0 0 0

&= =

⇒ u0 = –1.46 cm = – 1.5 cm

8. A ray of light, incident on an equilateral glass prism ( )n 3g = moves parallel to the base line of

the prism inside it. Find the angle of incidence for this ray. [CBSE Delhi 2012] Ans.

i

60°

r = 30°

60°

From the figure, we see r = 30°


We know that

sinsin

nri

21 = ⇒ sin

sin i3

30°=

sin i = sin3 30 3 21

° ×=

⇒ i = 60° 9. A small telescope has an objective lens of focal length 150 cm and an eye piece of focal length

5 cm. If this telescope is used to view a 100 m high tower 3 km away, find the height of the final image when it is formed 25 cm away from the eye piece. [CBSE Delhi 2012]

Ans. Using, the lens equation for objective lens.

f v u1 1 1

–0 0 0

=

⇒ v1501 1

3 101

–– ×0 5

= ⇒ v1501

3 101 1

–× 5 0

=

⇒ cm cmv 19993 10

150×

0

5b=

Hence, magnification due to the objective lens

mm

m uv

3000150 10

2010×

0 0

02 2– –-= =

m0 = 0.05 × 10–2

Using lens formula for eye piece

f v u1 1 1

–e e e

=

⇒ u51

251 1

– –e

= ⇒ cmu 625–

e=

∴ Magnification due to eyepiece, m

62525

6–

–e= =

Hence, total magnification, m = me × m0

m = 6 × 5 × 10–4 = 30 × 10–4

Hence, size of final image = 30 × 10–4 × 100 m = 30 cm 10. (a) Monochromatic light of wavelength 589 nm is incident from air on a water surface. If n for

water is 1.33, find the wavelength, frequency and speed of the refracted light. (b) A double convex lens is made of a glass of refractive index 1.55, with both faces of the same

radius of curvature. Find the radius of curvature required, if the focal length is 20 cm. [CBSE (AI) 2017] Ans. (a) Given, λair = 589 nm = 589 × 10–9 m Speed of light, c = 3 × 108 m/s anw = 1.33

Frequency of the refracted light, c

airo

m=

⇒ . Hz589 10

3 105 09 109

814

–#

##o = =

Wavelength of refracted light, nwaterair

aw

mm

=

⇒ . . m1 33589 10

4 42 10water

97

––#

#m = =


Speed of refracted light, v nc

water aw

=

⇒ . . msv 1 333 10

2 26 10water

88 1–#

#= =

(b) Given, f = 20 cm and n = 1.55 Let the radius of the curvature of each of the two surfaces of the lens be R. If R1 = R, then R2 = –R

( )f

n R R1

11 1

– –1 2

= < F

⇒ 120

1 55 11 1 1

200 55 2= − +

⇒ = ×( . )

.R R R

⇒ 120

1 1020 1 10= ⇒ = ×.

.R

R

∴ R = 22 cm 11. (a) Velocity of light in glass is 2 × 108 m/s and in air is 3 × 108 m/s. If the ray of light passes from

glass to air, calculate the critical angle. (b) A double convex lens of + 5 D is made of glass of refractive index 1.55 with both faces of

equal radii of curvature. Find the value of its radius of curvature. [CBSE (F) 2015] Ans. (a) Refractive index of a transparent medium decreases with increase in wavelength of the

incident light used. Refractive index of glass with respect to air is given

n = Speed of light in glasspeed of light in airS

= × /

× /.

m s

m s

2 10

3 101 58

8

=

Now, n = sin i

1

c ⇒ sin ic = n

1

.sin sin sini n1

1 51

32

c1 1 1– – –= = =b c cl m m

(b) Power of a lens, P = ( )in mf

1

After cutting, the power of each part will be half of the power of original lens. Therefore, focal length = 2f

∴ Power of each part, Pf2

1=l

P = f1 ⇒

f5

1=

m cmf 51

20= =

Now, f

n R R1 1

1 1– –

1 2=] eg o

Since R1 = + R and R2 = – R

∴ ( – )f

n R R1

11 1

= +c m

.R20

1 1 5 1 2– ×=] g

R = 20 cm


12. Light from a point source in air falls on a spherical glass surface (n = 1.5 and radius of curvature = 20 cm). The distance of the light source from the glass surface is 100 cm. At what position the image is formed?

Ans. As we know, vn

un

R

n n–

–2 1 2 1= (by lens maker's formula)

⇒ .

( ).

v1 5

1001

201 5 1

– ––=

⇒ . .v

1 5200 5

1001

–= ⇒ .v

1 52003=

∴ v = × . .cm3200

1 5 100=

13. A convex lens of focal length 20 cm and a concave lens of focal length 15 cm are kept 30 cm apart with their principal axes coincident. When an object is placed 30 cm in front of the convex lens, calculate the position of the final image formed by the combination. [CBSE 2019 (55/5/1)]

Ans. For image formed by convex lens:

f v u1 1 1

–=

Here, u = – 30 cm, f1 = + 20 cm

⇒ v201 1

301= + ⇒ cmv 30 20

20 3060–

×= =

Here, u for concave lens = + 30 cm and, f = – 15 cm

f v u1 1 1

–=

⇒ –v151 1

301

–= ⇒ – cmv 15 30

15 3030–

×= =

No, the result will not change from principle of reversibility. 14. An astronomical telescope has an objective lens of focal length 20 m and eyepiece of focal length

1 cm. (i) Find the angular magnification of the telescope. (ii) If this telescope is used to view the moon, find the diameter of the image formed by the

objective lens. Given the diameter of the moon is 3.5 × 106 m and radius of lunar orbit is 3.8 × 108 m. [CBSE 2020 (55/1/1)]

Ans. (i) m

mm

f

f

1020

2000e

02–

= = =

(ii)

d = ?

f = 20 md = 3.5 x 106m

u = 3.8 x 108

α

α

m

0 0

i

tan ud

f

di0

0a= = ⇒

.

.× . md

3 8 103 5 10

20 0 18××

i 8

6= =

K 30cm 30cm

f1 = 20 cm

IO

f2 = 15 cm

30cm


15. An object is placed 30 cm in front of a plano-convex lens with its spherical surface of radius of curvature 20 cm. If the refractive index of the material of the lens is 1.5, find the position and nature of the image formed. [CBSE 2020 (55/1/1)]

Ans. According to lens maker's formula

( )f

n R R1

11 1

– –1 2

= e o

. .f1 1 5 1 20

1 1200 5

2005

401– – 3= = = =] cg m [For plano convex lens, R1 → R and R2 → ∞]

∴ f = 40 cm

Now –f v u1 1 1= ⇒

( )v

f ufu

40 3040 30

–× –

=+

=

⇒ – cmv 1040 30

120– ×= =

Image is virtual, erect and enlarged in front of lens 120 cm away. 16. A biconcave lens of power P vertically splits into two identical plano concave parts. Find the

power of each part. [CBSE 2020 (55/5/1)] Ans. For a biconcave lens

– –( – )f

n R R fn R P

1 11 1 1

12–

1 2&= + = =] eg o

When a lens is split,

–( – )f

n R1

11

1= ⇒ –

fn

R1 1 1–2

= ] g

f f f1 1

21

1 2= =

P PP21 2= =

17. An optical instrument uses a lens of 100 D for the objective lens and 50 D for its eye piece. When the tube length is kept at 20 cm, the final image is formed at infinity.

(a) Identify the optical instrument. (b) Calculate the magnification produced by the instrument. [CBSE 2020 (55/5/1)] Ans. (a) P0 = 100 D Pe = 50 D L = 20 cm The focal length of objective,

cm

cmf 100100

10 = =

The focal length of eyepieces,

cm cmf 50100

2e= =

The focal length of objective is less than the focal length of eyepiece also both the lenses have small focal lengths so it is a compound microscope.

(b) Magnification produced by instrument is,

MfL

fD×

o e= = ×1

202

25

M = 250


18. A screen is placed 80 cm from an object. The image of the object on the screen is formed by a convex lens placed between them at two different locations separated by a distance 20 cm. Determine the focal length of the lens. [CBSE 2020 (55/5/1)]

Ans. For the first position of the lens we have

–f y x y x1 1 1 1 1–= = +b l

where x is the object distance and y is the image distance. for the second position of the lens

( ) ( )f y x1

201

201

– – –= +

y x201

201

–= + +

y x y x1 1

201

201

–+ = + +

( ) ( )( )

xyx y

y xx y

20 20–

+= +

+

xy = (y – 20) (x + 20) xy = xy – 20x + 20y – 400 20x – 20y = – 400 x –y = – 20 x + y = 80 2x = 60, x = 30, y = 50

f x y1 1 1

301

501

50 3080×

= + = + =

. cmf 850 3

18 75×= =

19. For a glass prism ( )µ = 3 the angle of minimum deviation is equal to the angle of the prism. Calculate the angle of the prism.

Ans. For minimum derivation,

( )

sin

sin

A

A

2

2m

n

d

=

+

Given, , A3 mn d= =

Now, .

sin

sin

sin

sin

sin

sin cos

A

A A

AA

A

A A

3

2

2

2 2

2 2 2=

+

= =d n

⇒ cosA

3 2 2=

⇒ cosA2 2

3= ⇒

A2 30°=

∴ A = 60° 20. In normal adjustment position, find the magnitude of (i) The length of the telescope (ii) the magnification of the telescope If the focal length of the objective lens is 15 m and the focal length of an eye lens is 5 cm.


Ans. In normal adjustment, image front at infinity then, magnifying power or magnification,

Mf

f

e

0=

and length of telescope, L = f0 + fe

(i) L = f0 + fe = (15 m + 0.05 m) 15.05 m

(ii) M = .f

f

0 0515

300e

0 = =

21. If a convex lens of focal length 50 cm is placed in contact coaxially with a concave lens of focal length 20 cm, what is the power of the combination?

Ans. For convex lens, cmf 501 =+ For concave lens, – cmf 202 =

To combination of lenses, f f f1 1 1

1 2= +

⇒ f1

501

201

1002 5

1003

–– –= = =

∴ – cmf 3100=

Now, Pcombination = ( ) / –in cm

Df

100100 3100

3–= =

22. A converging lens of refractive index 1.5 has a power of 10 D. When it is completely immersed in a liquid, it behaves as a diverging lens of focal length 50 cm. Find the refractive index of the liquid. [CBSE 2020 (55/4/1)]

Ans. Focal length of lens in water,

f

n

n

nf

1

1

–

–l

l

g

g

a=

m cmf 101

10a= =

fl = –50 cm

–.

( . )

n

50 101 5

1

1 5 1

–

–

l

=d n

–.

.

n

51 5

1

0 5

–l

=d n

.

–n101 1 5

1–

l=

.n10

9 1 5l

=

n 35

l=

z z z


CHAPTER-3: WAVE OPTICS 1. (i) Differentiate between a ray and a wavefront. [CBSE Delhi 2009] (ii) Sketch the shape of wavefront emerging from a point source of light and also make the

rays. [CBSE (F) 2009] (iii) What is the geometrical shape of the wavefront when a plane wave passes through a convex

lens? [CBSE (AI) 2008] Ans. (i)

Ray Wavefront

The perpendicular line drawn at any point on the wavefront that represents the direction of propagation of the wave is called ray.

The locus of a particle of a medium vibrating in phase is called a wavefront.

(ii) The wavefront will be spherical of increasing radius.

Rays

Spherical wavefront

S

(iii) A plane wavefront emerging after refraction through convex lens is spherical and converges to focus F.

Convex lens

F

Spherical wavefrontat radius R

2. (i) State the reason, why two independent sources of light cannot be considered as coherent sources. [CBSE Delhi 2008]

(ii) How does the fringe width of interference fringes change, when the whole apparatus of Young’s experiment is kept in a liquid of refractive index 1.3? [CBSE Delhi 2008]

Ans. (i) Coherent sources are defined as the sources in the which initial phase difference remains constant.

In the case of two independent sources, the initial phase difference cannot remain constant because light is emitted due to millions of atoms and their number goes on changing in a quite random manner.

(ii) Fringe width, dD& \b

mb m= for same D and d. When the whole apparatus is immersed

in a transparent liquid of refractive index n, the wavelength decreases to .n 1 3mm m= =l . So,

fringe width decreases to .1 31 times.


3. How does the fringe width, in Young’s double-slit experiment, change when the distance of separation between the slits and screen is doubled? [CBSE (AI) 2012]

Ans. Angular separation is /

D DD d

di

b m m= = =

Since i is independent of D, angular separation would remain same. 4. Define a wavefront. Using Huygens’ principles, draw the shape of a refracted wavefront, when

a plane wave is incident on a convex lens. [CBSE Ajmer 2015] Ans. Wavefront: A wavefront is a locus of particles of medium all vibrating in the same phase. A plane wavefront emerging after refrection through convex lens is spherical and convergs to

focus F.

R

5. Define the term ‘wavefront of light’. A plane wavefront AB propagating from denser medium (1) into a rarer medium (2) is incident on the surface P1P2 separating the two media as shown in fig.

Using Huygens’ principle, draw the secondary wavelets and obtain the refracted wavefront in the diagram. [CBSE 2020 (55/5/1)]

B

P2P1 A

1

2

Ans. A wavefront is a locus of particles of medium all vibrating in the same phase.

1

221

6. (a) A monochromatic source of light of wavelength l illuminates a narrow slit of width d to produce a diffraction pattern on the screen. Obtain the conditions when secondary wavelets originating from the slit interfere to produce maxima and minima on the screen.

(b) How is the diffraction pattern be affected when (i) the width of the slit is decreased? (ii) the monochromatic source of light is replaced by a source of white light?

[CBSE (F) 2013]


Ans. (a) Let AB be a slit of width ‘d’ and a parallel beam of monochromatic light is incident on it. According to Fresnel the diffraction pattern is the result of superposition of a large number of waves, starting from different points of illuminated slit.

Let i be the angle of diffraction for waves reaching at point P of screen and AN the perpendicular dropped from A on wave diffracted from B.

The path difference between rays diffracted at points A and B, BP AP BN–T = =

In , ° andANB ANB BAN90T + + i= =

∴ sin sinorABBN

BN ABi i= =

As AB = width of slit = d ∴ Path difference, sindT i= ...(i) To find the effect of all coherent waves at P, we have to sum up their contribution, each with

a different phase. This was done by Fresnel by rigorous calculations, but the main features may be explained by simple arguments given below:

At the central point C of the screen, the angle θ is zero. Hence the waves starting from all points of slit arrive in the same phase. This gives maximum intensity at the central point C.

Minima: Now we divide the slit into two equal halves AO and OB, each of width d2 . Now

for every point, M1 in AO, there is a corresponding point M2 in OB, such that ;M Md21 2 =

then path difference between waves arriving at P and starting from M1 and M2 will be

.sind2 2i

m= (ii) This means that the contributions from the two halves of slit AO and OB are

opposite in phase and so cancel each other. Thus equation (ii) gives the angle of diffraction

at which intensity falls to zero. Similarly it may be shown that the intensity is zero for sin

bn

im= , with n as integer. Thus the general condition of minima is

d sin θ = nl ...(iii) Secondary Maxima: Let us now consider angle θ such that

sind2

3i i

m= =

which is midway between two dark bands given by

sin sinandd d

2i i

mi i

m= = = =


I0

0–4λd

–3λd

–2λd

–λd

λd

2λd

3λd

4λd

The intensity of secondary maxima decreases with increase of order n because with increasing n, the contribution of slit decreases.

For n = 2, it is one-fifth, for n = 3, it is one-seventh and so on. (b) (i) Effect of the width of the slit — For given monochromatic waves, if slit width is decreased, the fringe pattern becomes

broader.

. [ ]d D

yn

D

y21n n

am i= + =c m ⇒ d.yn = constant

⇒ yd1

n \

(ii) If monochromatic source of light is replaced by white light, instead of white fringes we have few coloured fringes on either side of central white fringe, and then uniform illumination on the screen.

( )for VIBGYORyn \ m m

7. In a single slit diffraction experiment, the width of the slit is made double the original width. How does this affect the size and intensity of the central diffraction band? Draw a plot of the intensity distribution. [CBSE (AI) 2008]

Ans. The angular size of central diffraction band, a a22 1\i

m= . When width of slit ‘a’ is doubled, the

size of central band becomes half and the intensity is doubled.

IntensityI0

–3λ/a 0–2λ/a –λ/a 2λ/a 3λ/aλ/a


8. Draw the intensity pattern for single slit diffraction and double slit interference. Hence, state two differences between interference and diffraction patterns. [CBSE (AI) 2017]

Ans. I

(i) Interference (ii) Dif fraction

– 2 20 –2 2– 0

I

–

Intensity Patterns Differences between interference and diffraction

Interference Diffraction

(a) It is due to the superposition of two waves coming from two coherent sources.

(a) It is due to the superposition of secondary wavelets originating from different parts of the same wavefront.

(b) The width of the interference bands is equal. (b) The width of the diffraction bands is not the same.

(c) The intensity of all maxima (fringes) is same. (c) The intensity of central maximum is maximum and goes on decreasing rapidly with increase in order of maxima.

9. Answer the following: (a) In what way is diffraction from each slit related to the interference pattern in a double slit

experiment? (b) When a tiny circular obstacle is placed in the path of light from a distant source, a bright spot

is seen at the centre of the shadow of the obstacle. Explain, why. [CBSE Bhubaneshwar 2015] Ans. (a) In YDSE, the interference pattern is modulated by diffraction from each slit. The pattern is

the result of the interference of the diffracted wave from each slit. If slit width in interference pattern is reduced to size of wavelength of light used; the

differaction will also take place along with interference. (b) When a tiny circular obstacle is placed in the path of light from a distant source, a bright spot

is seen at the centre of the shadow of the obstacle. This is because light waves are diffracted from the edge of the circular obstacle, which interfere constructing at the centre of the shadow. This constructive interference produces a bright spot.

10. Some of factors, which could possibly influence the speed of wave propagation are: (i) nature of source (ii) direction of propagation (iii) motion of source and/or observer (iv) wavelength (v) intensity of wave On which of these factors, if any, does (a) the speed of light in vacuum (b) the speed of light in medium (say glass or water); depend?


Ans. (a) The speed of light in a vacuum i.e., 3 × 108 m/s (approximately) is a universal constant. It is not affected by the motion of the source, the observer, or both. Hence the given factors does not affect the speed of light in a vacuum.

(b) Out of the listed factors, the speed of light in a medium depends on the wavelength of light in that medium.

11. State three characteristic features which distinguish between interference pattern due to two coherently illuminated sources as compared to that observed in a diffraction pattern due to a single slit. [CBSE (F) 2014]

Ans. Difference between interference and diffraction

Interference Diffraction

(i) It is due to the superposition of two waves coming from two coherent sources.

(i) It is due to the superposition of secondary wavelets originating from different parts of the same wavefront.

(ii) The width of the interference bands is equal.

(ii) The width of the diffraction bands is not the same.

(iii) The intensity of all maxima (fringes) is same.

(iii) The intensity of central maximum is maximum and goes on decreasing rapidly with increase of order of maxima.

12. A monochromatic light of wavelength l is incident normally on a narrow slit of width ‘a’ to produce a diffraction pattern on the screen placed at a distance D from the slit. With the help of a relevant diagram, deduce the conditions for obtaining maxima and minima on the screen. Use these conditions to show that angular width of central maximum is twice the angular width of secondary maximum. [CBSE (F) 2017]

Ans. Let AB be a slit of width ‘a’ and a parallel beam of monochromatic light is incident on it. According to Fresnel the diffraction pattern is the result of superposition of a large number of waves, starting from different points of illuminated slit.

Let θ be the angle of diffraction for waves reaching at point P of screen and AN the perpendicular dropped from A on wave diffracted from B.

The path difference between rays diffracted at points A and B,

∆ = BP – AP = BN In ∆ ANB , ∠ANB = 90° and ∠BAN = θ

∴ sin sinorABBN

BN ABi i= =

As AB = width of slit = a ∴ Path difference, ∆ = a sin θ ....(i) To find the effect of all coherent waves at P, we have to sum up their contribution, each with a

different phase. This was done by Fresnel by rigorous calculations, but the main features may be explained by simple arguments given below:

At the central point C of the screen, the angle θ is zero. Hence the waves starting from all points of slit arrive in the same phase. This gives maximum intensity at the central point C.

Minima: Now we divide the slit into two equal halves AO and OB, each of width a2 . Now for

every point, M1 in AO, there is a corresponding point M2 in OB, such that ;M Ma21 2 = then

path difference between waves arriving at P and starting from M1 and M2 will be .sina2 2i

m=

(ii) This means that the contributions from the two halves of slit AO and OB are opposite in phase and so cancel each other. Thus equation (ii) gives the angle of diffraction at which intensity falls


to zero. Similarly it may be shown that the intensity is zero for sin ,an

im= with n as integer.

Thus the general condition of minima is a sin θ = nl ...(iii)

Secondary Maxima: Let us now consider angle θ such that sin a23

i im= =

Which is midway between two dark bands given by sin sinanda a2

i im

i im

= = = =

Angular width of central maxima = a2m

We have a sin θ = nl for minima ⇒ a.θ = nl (Here, for small angle θ - sin θ) For first minima a 1i m= ⇒ a1i

m=

For 2nd minima a 22i m= ⇒ a

22i

m=

Hence angular width of secondary maxima

= –2 1i i = am

Hence, angular width of central maximum is twice the angular width of secondary maximum. 13. (a) Using Huygens’ construction of secondary wavelets explain how a diffraction pattern is obtained

on a screen due to a narrow slit on which a monochromatic beam of light is incident normally. (b) Show that the angular width of the first diffraction fringe is half that of the central fringe.

(c) Explain why the maxima at n a21

im= +c m become weaker and weaker with increasing n.

[CBSE Delhi 2015] Ans. (a) When a plane wavefront from a distance source

illuminates a slit, each point within the slit becomes a source of secondary spherical wavelets. These spherical wavelets go on increasing in size, and superpose on each other, in the region between slit and screen.

(i) If all path differences are zero, then all the parts of the slit contribute in phase. This gives maximum intensity at point P.

(ii) If all path differences are non-zero, these secondary wavelets may either superpose in phase or out of phase, resulting in either occurrence of intensity of light or in zero occurrence.

These variation in the light intensity on the screen is termed as diffraction of light.

θ

θ

θB

A

To P1

PFrom S


(b) The angular width of central maxima is the angular separation between the first minima on the two sides of the central maximum.

Position of first minima d sin θ = nl

d1 !im=

Similarly d

22 !i

m=

Angular width of first diffraction fringe

d

bm=

i

Angular width of central maxima

to tod d

–m m+

= d

2b

m=il

Hence angular width of the first diffraction fringe is half of the central fringe. (c) The path difference between the extreme rays can be given as BC = BP1 – AP1 = a sin θ where ‘a’ is the size of the slit. If the slit is divided into three equal parts, first two-third of the slit which have a path

difference of 2m

will contribute no intensity at the point. However, the remaining one-third

of the slit contributes to the intensity at the point between two minima. Hence, the intensity of light will be much weaker than the central maxima.

So, path difference, sina n 21

i m= +c m where n = 1, 2, 3, 4, etc.

⇒ .a n 21

i m= +c m

⇒ n a21

im= +c m

14. (a) Consider two coherent sources S1 and S2 producing monochromatic waves to produce interference pattern. Let the displacement of the wave produced by S1 be given by

y1 = a cos wt and the displacement by S2 be y2 = a cos (wt + f) Find out the expression for the amplitude of the resultant displacement at a point and show

that the intensity at that point will be cosI a4 22 2 z=

Hence establish the conditions for constructive and destructive interference. (b) What is the effect on the interference fringes in Young’s double slit experiment when (i) the

width of the source slit is increased; (ii) the monochromatic source is replaced by a source of white light? [CBSE Allahabad 2015]

Ans. (a) The resultant displacement will be given by y = y1 + y2

= a cos wt + a cos(wt + f) = a[cos wt + cos(wt + f)] = 2a cos(f/2) cos(wt + f/2) The amplitude of the resultant displacement is 2a cos(f/2)


The intensity of light is directly proportional to the square of amplitude of the wave. The resultant intensity will be given by

cosI a4 22 2 z=

∴ Intensity ,cosI4 202 z= d n where I0 = a2 is the intensity of each harmonic wave

At the maxima, f = ±2nπ

∴ cos 2 12 z =

At the maxima, I = 4I0 = 4 × intensity due to one slit

cosI I4 202 z= d n

For constructive interference, I is maximum.

It is possible when ; ;cos n n2 1 2 22 z zr z r= = =d n

For destructive interference, I is minimum, i.e., I = 0

It is possible when ;–

;cosn

n2 0 2 22 1

2 1 22 !z z r

zr= = =d ^ ^n h h

(b) (i) When the width of the slit is increased: From the relation sin ,aim= we find that if the

width of the slit (a) is increased, then for a given wavelength, sin θ is small and hence θ is small. Hence diffraction maxima and minima are quite close on either side of θ.

(ii) With monochromatic light, the diffraction pattern consists of alternate bright and dark bands. If white light is used central maximum is white and on either side, the diffraction bands are coloured.

15. (i) State the essential conditions for diffraction of light. (ii) Explain diffraction of light due to a narrow single slit and the formation of pattern of fringes

on the screen. (iii) Find the relation for width of central maximum in terms of wavelength ‘l’, width of slit ‘a’,

and separation between slit and screen ‘D’. (iv) If the width of the slit is made double the original width, how does it affect the size and

intensity of the central band? [CBSE (F) 2016] Ans. (i) Essential conditions for diffraction of light: (a) Source of light should be monochromatic. (b) Wavelength of the light used should be

comparable to the size of the obstacle. (ii) Single slit diffraction is explained by treating

different parts of the wavefront at the silt as sources of secondary wavelets.

At the central point C on the screen, θ is zero. All path differences are zero and give maximum intensity at C.

At any other point P, the path difference between two edges of the slit is NP – LP = NQ

= sina a.i i

Any point P, in direction Q, is a location of minima if a ni m= This can be explained by dividing the slit into even number of parts. The path difference

between waves from successive parts is 180° out of phase and hence cancel each other leading to a minima.

M1

MM2

L

N

Q

θ

θ

θ

To C

To P

From S


Any point P, in direction Q, is a location of maxima if a n 21

i m= +c m This can be explained by dividing the slit into odd number of parts. The contributions

from successive parts cancel in pairs because of 180° phase difference. The unpaired part produces intensity at P, leading to a maxima.

(iii) If Q is the direction of first minima, then a a&i m im= =

Angular width of central maxima = a22

im

=

Linear width of central maxima, . D aD

22

b im

= =

Intensity, sin

I I I1

0 2

2

2& \b

b

b= (roughly)

(iv) If ‘a’ is doubled, β becomes half and intensity becomes 4 times.

Numerical Questions 1. In Young’s double slit experiment, the two slits are separated by a distance of 1.5 mm and

the screen is placed 1 m away from the plane of the slits. A beam of light consisting of two wavelengths 650 nm and 520 nm is used to obtain interference fringes. Find

(a) the distance of the third bright fringe for l = 520 nm on the screen from the central maximum. (b) the least distance from the central maximum where the bright fringes due to both the

wavelengths coincide. [CBSE Ajmer 2015] Ans. (a) Fringe width from central maximum

.× ×

dD

1 5 10520 10 1

× 3

9

–

–

bm= =

For 3rd bright fringe, .

31 5 103 520 1

×× ×

–3b = = 1.04 × 10–3 m = 1.04 mm.

(b) For least distance of coincidence of fringes, there must be, (yn)l1 = (yn+1)l2 As, l1 > l2

∴ (yn)l1 = (yn+1)l2

⇒ ( )

d

nD

d

n D11 2m m=

+

⇒ nl1 = (n + 1)l2

∴ n 650 520520

130520

4– –1 2

2

m m

m= = = =

Hence, least distance, .

yd

nD

1 5 104 1 650 10

×× × ×

–

–

min1

3

9m= =

= 1.73 × 10–3 m = 1.7 mm


2. A parallel beam of light of 600 nm falls on a narrow slit and the resulting diffraction pattern is observed on a screen 1.2 m away. It is observed that the first minimum is at a distance of 3 mm from the centre of the screen. Calculate the width of the slit. [CBSE (AI) 2013]

Ans. For diffraction, condition of minima. .d ni m=

⇒ dDy

nm=d n For first minima n = 1

d yDm=

.

3 101 2 600 10

×× ×

–

–

3

9=

= 2.4 × 10–4 m 3. A parallel beam of light of wavelength 500 nm falls on a narrow slit and the resulting diffraction

pattern is observed on a screen 1 metre away. It is observed that the first minimum is at a distance of 2.5 mm from the centre of the screen, find the width of the slit. [CBSE (F) 2010]

Ans. Given l = 500 nm = 5 × 10–7 m, D = 1 m If a is width of slit, then for first minimum

sin a1im=

For small θ1, sin θ1 = θ1 = D

y1

∴ D

ya

1 m=

y1 = 2.5 mm = 2.5 × 10–3 m

∴ .

a yD

2 5 105 10 1

×× ×

–

–

1 3

7m= = = 2 × 10–4 m = 0.2 mm

4. The distance between two slits in Young’s interference experiment is 0.03 cm. The fourth bright fringe is obtained at a distance of 1 cm from central fringe on a screen placed at a distance of 1.5 m from slits. Calculate the wavelength of light used.

Ans. In YDSE, for nth maximum bright fringe, yd

n Dn

m= , n = 1, 2, 3, . . . .

For, n = 4, .

.y

0 03 104 1 5

×× ×

–4 2m= ,

..

where given cm

mcm

y

Dd

1

1 50 03

4 =

==

R

T

SSSSSSSSS

V

X

WWWWWWWWW ⇒ .

.1 10

0 03 104 1 5

××

× ×––

22

m=

⇒ ..

× m4 1 51 10 0 03 10

5 10×× × ×2 2

7– –

–m = =

= 500 nm 5. A slit of width ‘a’ is illuminated by light of wavelength, 700 nm. What will be the value of slit

width ‘a’ when (a) first minimum falls at an angle of diffraction 30°? (b) first maximum falls at an angle of diffraction 30°? Ans. (a) For nth minima, a sin θ = nl Given, l = 700 nm = 700 × 10–9 m, n = 1, θ = 30°

∴ . . ×sin sin

man

301 700 10

0 57 10

1 4 10°

× × ×– ––

9 76

i

m= = = =

d

ToP

ToC�


(b) For nth maxima, a sin θ = (2n + 1) 2m

Given, n = 1, θ = 30°, l = 700 × 10–9 m.

∴ ( ) ( )

sin sina

n2

2 12 30

2 1 1 7 101

3 7 10× °

× × × × ×7 7– –

i

m=

+=

+=

= 21 × 10–7 m = 2.1 × 10–6 m. 6. Find the intensity at a point on a screen in Young’s double slit experiment where the interfering

waves of have a path difference of (i) l/4, and (ii) l/3. [CBSE (F) 2017]

Ans. cosI I4 202 z=

(i) If path difference = 4m

⇒ x2

×T Tm

rz=

⇒ x2

4 2×Tm

r m r= =

Also, cos cosI I I I4 2 4 4 202

02

0

Tz r= = =

(ii) If x 3Tm=

⇒ 2

3 32

×Tzm

r m r= =

∴ cosI I4 202 Tz=

cosI I4 3 22×0

20

r= =d n

7. Two wavelengths of sodium light 590 nm and 596 nm are used, in turn, to study the diffraction taking place at a single slit of aperture 2 × 10–4 m. The distance between the slit and the screen is 1.5 m. Calculate the separation between the positions of the first maxima of the diffraction pattern obtained in the two cases. [CBSE Delhi 2012]

Ans. For maxima other than central maxima

a.θ = n 21m+c m

and θ = Dy

∴ .a Dy

= n 21m+c m

For light of wavelength l1 = 590 nm

.

y2 10 1 5× ×–14 1 = 1 2

1590×+c m

y1 = .

23

2 10590 10 1 5

××

× ×–

–

4

9

= 6.64 mm For light of wavelength l2 =596 mm

.

y2 10 1 5× ×–4 2

= 1 21+c m × 596 nm

P

Ca = 2×10 m–4

D = 1.5m

Screen

y

�


⇒ y2 = .

23

2 10596 10 1 5

××

× ×–

–

4

9

= 6.705 mm Separation between two positions of first maxima ∆y = y2 – y1 = 6.705 – 6.64 = 0.065 mm 8. In Young’s double slit experiment, using monochromatic light of wavelength l the intensity of

light at a point on the screen where path difference is l, is K units. Find out the intensity of light at a point where path difference is l/3. [CBSE Delhi 2014]

Ans. We know

cosI I4 202 z=

for path difference l, phase difference, f = 2π Intensity of light = K Hence, K = 4I0 cos2 π = 4I0

For path difference λ3

, Phase difference, φ π= 23

Intensity of light, ′= = =I I I I42

430

20

20cos cos

φ π

⇒ ′=I K4

9. The intensity at the central maxima in Young’s double slit experiment is I0. Find out the intensity

at a point where the path difference is λ λ λ6 4 3

, .and [CBSE North 2016]

Ans. As we know, intensity of interfering wave, cos cosI I I4 2 22

02z z

= =

(i) If, ,6Tm= ( )

2 26 3×Tz

m

r

m

r m r= = =

∴ cosI I II

6 43

4

3×0

20

0r= = =

(ii) If, ,4Tm= ( ) ×

2 24 2Tz

m

r

m

r m r= = =

∴ ×cosI I II

4 21

202

00r= = =

(iii) If, ,3Tm= ( ) ×

2 23 3

2Tz

m

r

m

r m r= = =

∴ ×cosI I II

62

41

402

00r

= = =

Note: Intensity at central maxima I0 = 4I, where I is the intensity of fringe due to one slit.


10. Two slits are made 1 mm apart and the screen is placed 1 m away. What should be the width of each slit to obtain 10 maxima of the double slit pattern within the central maximum of the single slit pattern when blue green light of wavelength 500 nm is used? [CBSE Guwahati 2015]

Ans. Fringe width at central maxima, aD

aD2

100bm

b bm= = =; E

⇒ a10 5 10

2 500 10 1× ×

× × ×–

–

4

9= = 2 × 10–4 m = 0.2 mm.

11. Monochromatic light of wavelength 588 nm is incident from air to water interface. Find the

wavelength and speed of the refracted light. The refractive index of water is 34 .

[CBSE 2020(55/3/1)]

Ans. n c

cw w w w

0 0 0

om

om

m

m= = =

∴ /n 4 3588

w w

0m

m= =

= 441 nm Now,

c nc

w w

0=

43 10 3× ×8

=

= 2.25 × 108 m/s

12. Two waves from two coherent sources S and S’ superimpose at X as shown in the figure. If X is a point on the second minima and SX – S’X is 4.5 cm. Calculate the wavelength of the waves. [CBSE Sample Paper 2021]

X

S’S

Ans. Path difference = .Dxd

4 5 29= =

For 2nd minima, x = dn

DdD2

2 123–

m m=d n

` Path difference = dDDd

23

23

m m=

Now, , cm23

29

39

3m

m= = =

z z z


CHAPTER-4: DUAL NATURE OF MATTER AND RADIATION

1. An electron and alpha particle have the same de Broglie wavelength associated with them. How are their kinetic energies related to each other? [CBSE Delhi 2008]

Ans. Given electronm m=a

de Broglie wavelength associated with a particle of mass m and energy E is

mEh

2m =

\ m Eh

m Eh

2 2e e

=a a

⇒ E

Emme

e=

a

a

That is kinetic energy of electron and -particle are in inverse ratio of these masses. 2. (i) Show graphically, the variation of the de-Broglie wavelength (l) with the potential (V) through

which an electron is accelerated from rest. [CBSE Delhi 2011] (ii) How does the stopping potential applied to a photocell change, if the distance between the

light source and the cathode of the cell is doubled? [CBSE (AI) 2008] Ans. (i) According to de Broglie wavelength for accelerated electrons,

V1

?m

V√

λ

(ii) Stopping potential remains unchanged. Reason: On doubling the distance between the light source and the cathode of the cell, the

intensity of light incident on the photocell becomes one-fourth. As stopping potential does not depend on intensity, the stopping potential remains unchanged.

3. (i) Plot a graph showing the variation of photoelectric current with anode potential for two light beams of same wavelength but different intensity.

(ii) What affects the maximum kinetic energy of photoelectrons: intensity or frequency of incident radiation?

Ans. (i) For two beams of same wavelength, the frequency will also be same or constant.

i.e.,

I1I2

ν1ν2

Photo current

Anode potential (V)ν =

(I1 > I2)

λc


(ii) According to Einstein's photo electric equation, K.Emax = hν – hν0 (a) Electrons are emitted from metal surface which having frequency above the threshold

frequency (ν0) of that metal surface. Hence, maximum kinetic energy of electrons depends on frequency of radiation (i.e., ν increases, K.Emax also increases).

(b) Maximum kinetic energy of emitted electrons is independent of the intensity of incident radiations.

4. (i) Plot a graph showing the variation of photoelectric current with collector plate potential at a given frequency but for two different intensities I1 and I2 where I2 > I1 [CBSE (F) 2014]

(ii) Show the variation of photocurrent with collector plate potential for different intensity but same frequency of incident radiation. [CBSE (F) 2011]

Ans. (i)

V0

I2I1

I2>I1

Collector plate potential

Pho

tocu

rren

t

Intercept of the graph with potential axis gives the stopping potential. (ii)

PhotoelectricCurrent

O-Vo

StoppingPotential

Collector platePotential

RetardingPotential

5. The graph below shows variation of photo-electric current with collector plate potential for different frequencies of incident radiations.

(i) Which physical parameter is kept constant for the three curves? (ii) Which frequency (ν1, ν2 or ν3) is the highest? [CBSE (F) 2009] Ans. (i) Intensity of incident radiations was kept constant. (ii) Frequency ν1 is highest. 6. (i) If the frequency and intensity of incident radiation on a metallic surface are doubled, do the

maximum kinetic energy and photocurrent get doubled? Explain with reason. (ii) Green light ejects photoelectrons from a given photosensitive surface whereas yellow light

does not. What will happen in the case of violet and red light? Give reason for your answer.

�1


Photo-electriccurrent

�2 �3


Ans. (i) The number of photoelectrons emitted is independent of incident light frequency but is dependent on intensity. If both the intensity and frequency of the incident light are doubled then saturation current or photo current is also be doubled.

(ii) Photo emission is possible only when energy of incident photon is more than work function of metal. The given surface will emit electrons with violet light but not with red light. The reason is energy of photon of violet light is more than work functions while red light photon has energy less than work function.

7. The graph below shows the variation of stopping potential Vs with the frequency (ν) of the incident radiation for two photosensitive metals X and Y.

yx

�0.5 1.0

Vs

(x 10 s )15 –1O

(i) Which of the metals has larger threshold wavelength? Give reason. (ii) Explain giving reason which metal gives out electrons having larger kinetic energy, for the

same wavelength of the incident radiation. [CBSE Guwahati 2015] (iii) If the distance between the light source and metal X is halved, what will be the kinetic energy

of electrons emitted due to this change? Give reason. Ans. (i) From the graph, the threshold frequency of metal Y is greater than metal X,

Here, c

00

m o= l0 = threshold wavelength

ν0 = threshold frequency. Here, then, (l0)Y < (l0)X (ν0)X < (ν0)Y So, metal X has larger threshold wavelength. (ii) If radiation of frequency (ν) greater than threshold frequency (ν0) irradiate the metal surface,

electrons are emitted out from the metal. Also energy of photon is directly proportional to the frequency (or inversely proportional to the wavelength).

Here metal X has lower threshold frequency than Y so electron ejected from X have larger kinetic energy than Y.

(iii) Emission of electron depends on frequency of photon. It does not depend on distance between light source and metal so kinetic energy of electrons ejected from metal X does not change.

8. (a) Describe briefly three experimentally observed features in the phenomenon of photoelectric effect.

(b) Discuss briefly how wave theory of light cannot explain these features. Ans. (a) The three experimentally observed features in the phenomenon of photoelectric effect are: (i) Threshold frequency: The minimum frequency required to eject an electron from

metalic surface is called threshold frequency. The photoelectric effect will occur when the incident frequency is greater or equal to the

threshold frequency for a given metal i.e., (ν ≥ ν0). (ii) The maximum kinetic energy of electron: The maximum kinetic energy of photo-

electrons is given by Emax = hν – W or hν – hν0. When the incident frequency is greater than the threshold frequency, the maximum kinetic energy is proportional to (ν – ν0).

(iii) Intensity of light or photon: The number of photon incident per unit time per unit area increases with increase of intensity of incident light. Thus, with increase in number of photoelectrons, the photocurrent also increases till it becomes saturated.


(b) The three characteristic features which cannot be explained by wave theory are: (i) Kinetic energy of emitted electrons is found to be independent of the intensity of incident

light. (ii) There is no emission of electrons if frequency of incident light is below a certain frequency

(threshold frequency). (iii) Photoelectric effect is an instantaneous process. 9. Define the terms (a) threshold frequency and (b) stopping potential. How were these terms

incorporated in Einstein’s photoelectric equation? [CBSE 2020 (55/3/1)] Ans. Threshold Frequency: The minimum frequency of incident light which is just capable of ejecting

electrons from a metal is called the threshold frequency. It is denoted by ν0. Stopping Potential: The minimum retarding potential applied to anode of a photoelectric tube

which is just capable of stopping photoelectric current is called the stopping potential. It is denoted by V0 (or VS)

The Einstein’s photoelectric equation is,

h m v21

max02o z= +

h h eVs0o o= +

10. Light of intensity ‘I’ and frequency ‘ν’ is incident on a photosensitive surface and causes photoelectric emission. What will be the effect on anode current when (i) the intensity of light is gradually increased, (ii) the frequency of incident radiation is increased, and (iii) the anode potential is increased? In each case, all other factors remain the same.

Explain, giving justification in each case. [CBSE Panchkula 2015] Ans. (i) Keeping the anode potential and the frequency of the incident radiation constant, if the

intensity of the incident light is increased, the photoelectric current or the anode current increases linearly. This is because photoelectric current is directly proportional to the number of photoelectrons emitted per second which is directly proportional to the intensity of the incident radiation.

(ii) For photoelectric emission to occur, there is a minimum cut off frequency of the incident radiation called the threshold frequency below which no photoelectric emission occurs. This frequency is independent of the intensity of the incident light. With an increase in the frequency of the incident radiation, the kinetic energy of the photoelectron ejected increases, whereas it is independent of the number of photoelectrons ejected. Hence, with increase in the frequency of incident radiation, there will not be any change in the anode current.

(iii) With an increase in the accelerating potential, the photoelectric current increases first, reaches maximum when all the electrons gets collected at the positive potential plate and then remains constant. The maximum value of the anode current is called the saturation current.

11. Draw a plot showing the variation of photoelectric current with collector plate potential for two different frequencies, ν1 > ν2 of incident radiation having the same intensity. In which case will the stopping potential be higher? Justify your answer. [CBSE (AI) 2011]

Ans. The plots are shown in fig. The stopping potential (Vs) is higher for higher radiations of frequency. Stopping potential is directly proportional to the frequency of incident radiation.

2

1

Saturationphotocurrent

(V )s 1 Collector potential(V )s 2

I�1 > �2


12. The following graph shows that variation of photocurrent for a photosensitive metal:

Photocurrent

A O X

(a) Identify the variable X on the horizontal axis. (b) What does the point A on the horizontal axis represent? (c) Draw this graph for three different values of frequencies of incident radiation ν1, ν2 and ν3

(ν1 > ν2 > ν3) for same intensity. (d) Draw this graph for three different values of intensities of incident radiation I1, I2, and I3

(I1 > I2 > I3) having same frequency. [CBSE (AI) 2017] Ans. (a) X : Collector plate potential (b) A : Stopping potential (c)

(d)

13. A proton and an α-particle move perpendicular to a magnetic field. Find the ratio of radii of circular paths described by them when both have (i) equal velocities, and (ii) equal kinetic energy.

Ans. (i) For v = same,

r = qBmv

r qm& ?

\ :r

r

m

m

qq

m

m

q

q

4

2

21

1 2× ×p p

p p

p

p

p= = = =

a a

a


Pho

tocu

rren

t

�1

V1 V2 V3

�2�3

�1 > �2 > �3

V0

I1I2I3

I1 >I2>I3


Pho

tocu

rren

t


(ii) For K.E = same,

.

r qBmK E

r qm2

& ?=

:r

r

m

m

qq

m

m

q

q

4

21 1× ×

p p

p p

p

p

p= = =

a a

a

14. Draw a plot showing the variation of photoelectric current with collector plate potential for two different frequencies, of incident radiation having the same intensity. In which case will the stopping potential be higher? Justify your answer. [CBSE (AI) 2011]

Ans. The plots are shown in fig. The stopping potential (Vs) is higher for radiations of frequency . Stopping potential is directly proportional to the frequency of incident radiation.

2

1

Saturationphotocurrent

(V )s 1 Collector potential(V )s 2

I�1 > �2

15. Obtain Einstein’s photoelectric equation. Explain how it enables us to understand the (i) linear dependence of the maximum kinetic energy of the emitted electrons on the frequency of incident radiation (ii) existence of threshold frequency for a given photoemitter.

Ans. In the photon picture, energy of the light is assumed to be in the form of photons each carrying energy.

When a photon of energy ‘hν’ falls on a metal surface, the energy of the photon is absorbed by the electrons and is used in the following two ways:

(i) A part of energy is used to overcome the surface barrier and come out of the metal surface. This part of energy is known as work function and is expressed as φ0 = hν0.

(ii) The remaining part of energy is used in giving a velocity ‘v’ to the emitted photoelectron

which is equal to the maximum kinetic energy of photoelectrons mv21

max2c m .

(iii) According to the law of conservation of energy,

h mv21

max02o z= +

⇒ h h mv h h KE21

max max02

0&o o o o= + = +

⇒ KEmax = hν – hν0

or KEmax = hν – φ0

This equation is called Einstein's photoelectric equation. According to above equation,

(i) Above threshold frequency (ν0), maximum K.E is directly proportional to ν, i.e., (K.E)max depends linearly on the frequency of the incident radiation.

(ii) When ν < ν0, (K.E)max becomes negative. The negative K.E has no physical meaning hence there is no photoelectric emission below the threshold frequency, ν0.


16. What is photoelectric effect? Explain experimentally the variation of photoelectric current with (i) intensity of light. (ii) the p.d. between the plates. (iii) frequency of incident light and hence state the laws of photoelectric emission. Ans. Photoelectric Effect The phenomenon of emission of electrons from a metallic surface by the use of light (or radiant)

energy is called photoelectric effect. The phenomenon was discovered by Lenard. For photoelectric emission, the metal used must have low work function, e.g., alkali metals. Caesium is the best metal for photoelectric effect.

(i) Characteristics of Photoelectric Effect Effect of Intensity: Intensity of light means the energy incident per unit area per second.

For a given frequency, if intensity of incident light is increased, the photoelectric current increases and with decrease of intensity, the photoelectric current decreases; but the stopping potential remains the same.

Intensity of radiations can be increased/decreased by varying the distance between source and metal plate (or emitter).

O

3I

2I

I

Cur

rent

(I)

–VsPotential difference (V)

Current (I)

OVs3 Vs2 Vs1

– (V) + (V)

ν3 > ν2 > ν1

ν3

( )a ( )b

ν2 ν1

This means that the intensity of incident light affects the photoelectric current but the maximum kinetic energy of photoelectrons remains unchanged as shown in fig (b).

(ii) Effect of Frequency: When the intensity of incident light is kept fixed and frequency is increased, the photoelectric current remains the same; but the stopping potential increases.

If the frequency is decreased, the stopping potential decreases and at a particular frequency of incident light, the stopping potential becomes zero. This value of frequency of incident light for which the stopping potential is zero is called threshold frequency ν0. If the frequency of incident light (ν) is less than the threshold frequency (ν0) no photoelectric emission takes place.

Thus, the increase of frequency increases the maximum kinetic energy of photoelectrons but the photoelectric current remain unchanged.

Law of photoelectric emission: (i) Kinetic energy of emitted electrons depends on the frequency of incident light on substance. (ii) There is a definite cut off value of frequency below which electron cannot be ejected by any

substances. (iii) Number of emitted electrons are directly proportional to the intensity of incident light. (iv) The logging time between the incident of light and emission of electrons is negligible.

17. (a) What is de Broglie hypothesis? Show that a moving particle of kinetic energy E has de

Broglie wavelength .mEh

2m =


(b) Show that the de Broglie wavelength associated with an electron accelerated through a p.d.

of V will be .

.AV

12 3m = c

Ans. (a) de Broglie Hypothesis Louis de Broglie postulated that the material particles (e.g., electrons, protons, α-particles,

atoms, etc.) may exhibit wave aspect. Accordingly, a moving material particle behaves as wave and the wavelength associated with material particle is

ph

mvh

m = = , where p is momentum.

If Ek is kinetic energy of moving material particle, then p mE2 k=

mEh

2 k

m =

i.e., ph

mvh

mEh

2 k

m = = =

The wave associated with material particle is called the de-Broglie wave or matter wave. The de-Broglie hypothesis has been confirmed by diffraction experiments.

(b) For charged particles associated through a potential of V volt, Ek = qV

mqVh

2m =

For electrons, q = e =1.6 × 10–19 C, m = 9 ×10–31 kg

. .

ÅmV V

12 2710

12 27–10#m = = (Only for electrons)

18. The work function of the following metals is given: Na = 2.75 eV, K = 2.3 eV, Mo = 4.17 eV and Ni = 5.15 eV. Which of these metals will not cause photoelectric emission for radiation of wavelength 3300 Å

from a laser source placed 1 m away from these metals? What happens if the laser source is brought nearer and placed 50 cm away? [CBSE Delhi 2017]

Ans. Here, l = 3300 Å = 3300 × 10–10 m = 3.3 × 10–7 m

Ehcm

= = . ..

eV3 3 10 1 6 106 63 10 3 10

× × ×× × ×

– –

–

7 19

34 8

= 3.77 eV The work function of Mo and Ni is more than the energy of the incident photons; so photoelectric

emission will not take place from these metals. Kinetic energy of photo electrons will not change, only photoelectric current will change if the laser is placed 50 cm away.

Numerical Questions 1. What is the stopping potential of a photocell, in which electrons with a maximum kinetic energy

of 6 eV are emitted? [CBSE (AI) 2008] Ans. Ek = eV0 ⇒ 6 eV = eV0 ⇒ V0 = 6 V The stopping potential V0= 6 volt (Negative).


2. What is the wavelength of a photon of energy 3.3 × 10–19 J? [CBSE 2020 (55/2/1)]

Ans. Energy of photon, Ehcm

=

⇒ .

.Ehc

3 3 106 63 10 3 10

×× × ×–

19

34 8

–m = = = 6.03 × 10–7 m.

3. Define the term “cut off frequency” in photoelectric emission. The threshold frequency of a metal is f. When the light of frequency 2f is incident on the metal plate, the maximum velocity of photo-electrons is v1. When the frequency of the incident radiation is increased to 5f, the maximum velocity of photo-electrons is v2. Find the ratio v1:v2. [CBSE (F) 2016]

Ans. Cut off frequency: It is that maximum frequency of incident radiation below which no photoemission takes place from a photoelectric material.

According to Einstein’s photoelectric equation

Khc

h– –max mz o z= =

Given that threshold frequency of the metal is f. If light of frequency, 2f is incident on metal plate, maximum velocity of photo-electron is v1 then,

( )mv h f f21

2 –12 = ⇒ mv hf2

112 = …(i)

If light of frequency, 5f is incident then maximum velocity of photoelectron is .

( )mv h f f21

5 –22 = ⇒ mv hf2

142

2 = …(ii)

Dividing (i) by (ii)

vv

41

2

12

=f p ⇒ vv

21

2

1 =

\ v1 : v2 = 1 : 2 4. A source of light is placed at a distance of 0.50 m from the photocell used and the cut-off potential

is found to be V0. If the distance between the light source and photocell is made 0.25 m, what will be the new cut-off potential?

Ans. By changing the locations of source of light from photo cell, there will be a change in the intensity of light falling on photocell. As cut-off potential does not depend upon the intensity of the incident light, hence cut-off potential remains same i.e., V0.

5. A mixture of three waves of wavelengths l1, l2 and l3 (such that l1 > l2 > l3) is made incident on a metal surface of threshold wavelength l2. If number of photons of each wavelength is 106 and efficiency of photoelectric effect is 1%, calculate the number of photoelectrons emitted.

Ans. As given, threshold wavelength = l2 and, l1 > l2 > l3 For photoelectric emission the wavelength of incident photon should be less than the threshold

wavelength. So, emission of electrons is not possible by wavelength l1 and only the photons of wavelength l3

will show photoelectric emission.

Now, ( )efficiencyNumber of photon incidentNumber of photon emitted

h =

⇒ Number of photon emitted = 0.01 × 106 = 104 photons. 6. A proton and an α-particle are accelerated, using the same potential difference. How are the de

Broglie wavelengths lP and lα related to each other?

Ans. de-Broglie wavelength mEh

mqVh

2 2m = =


For α-particle, m q Vh

2m =a

a a

For proton, m q Vh

2PP P

m =

\ m q

m q

P

P P

m

m=a

a a

But ,m

m

q

q4 2

p p= =a a

\ . .41

21

81

2 21

pm

m= = =a

7. The work function of caesium is 2.14 eV. Find (i) the threshold frequency for caesium and (ii) wavelength of incident light if the photocurrent is brought to zero by a stopping potential of 0.60 V.

Ans. (i) Work function, w = hν0

⇒ ν0 = .

. .hw

6 63 102 14 1 6 10

×× ×

–

–

34

19=

= 5.164 × 1014 Hz (ii) From Einstein's photoelectric equation, eV0 = hν – hν0

⇒ eV0 = –hc

Wm

⇒ hcm

= eV0 + W = 1.6 × 10–19 × 0.6 + 2.14 × 1.6 × 10–19

⇒ hcm

= 2.74 × 1.6 × 10–19

⇒ . . . .

.hc2 74 1 6 10 2 74 1 6 10

6 63 10 3 10× × × ×

× × ×– –

–

19 19

34 8m = =

= 4536 Å 8. The equivalent wavelength of a moving electron has the same value as that of a photon of energy

6 × 10–17 J. Calculate the momentum of the electron. Ans. Given, energy of photon, E = 6 × 10–17 J.

E = .hc

Ehc

6 106 63 10 3 10

×× × ×–

17

34 8

–&

mm = =

According to de Broglie wavelength, ph

m =

\ p = .

.h6 63 10 3 10

6 63 10 6 10× × ×

× × ×–

– –

34 8

34 17

m=

= 2 × 10–25 kg ms–1

9. Light of wavelength 2500 Å falls on a metal surface of work function 3.5 eV. What is the kinetic energy (in eV) of (i) the fastest and (ii) the slowest photoelectron emitted from the surface? [CBSE (F) 2011]

Ans. Wavelength of incident radiation (l) = 2500 Å Work function (W0) = 3.5 eV According to Einstein's photoelectric equation,


hc

W KEmax0m= +

⇒ KEmax = hc

W– 0m

= ( . ) ( )

..

2500 10

6 63 10 3 10

1 6 101

3 5×

× ××

×––

–

–10

34 8

19> H eV

= (4.97 – 3.5) eV = 1.47 eV (i) KE of fastest electron = 1.47 eV (ii) KE of slowest electron = 0 eV 10. Ultraviolet light of wavelength 2271 Å from a 100 W mercury source irradiates a photocell made

of molybdenum metal. If the stopping potential is – 1.3 V, estimate the work function of the metal. [CBSE (F) 2013]

Ans. From Einstein’s equation, hν = φ0 + K = φ0 + eVs

or φ0 = hν – eVs = –hc

eVsm (Equation is independent to the power of the source)

We have, .

Jhc

2271 106 6 10 3 10

×× × ×

–

–

10

34 8

m=

\ .

.. eV

2271 10 1 6 106 6 10 3 10

1 3× × ×× × ×

–0 10 19

34 8

– –

–

z = f p

= 5.5 eV – 1.3 eV = 4.2 eV 11. Monochromatic light of frequency 6.0 × 1014 Hz is produced by a laser. The power emitted is

2.0 × 10–3 W. Estimate the number of photons emitted per second on an average by the source.

Ans. Power of radiation, ,ν

νP tnh

Nh= = where N is number of photons per sec.

or .

.ν

NhP

6 63 10 6 102 0 10

34 14

3

–

–

# # #

#= =

= 5 × 1015 photons per second 12. An electron and a photon each have a wavelength of 1 nm. Find (i) their momenta (ii) the energy of the photon (iii) the kinetic energy of the electron. [CBSE (F) 2013] Ans. As given, le = lp = 1 nm = 1 × 10–9 m. (i) According de Broglie wavelength,

ph

m =

\ .

ph

1 106 63 10

××

e 9

34

–

–

m= = = 6.63 × 10–25 kg ms–1.

Also, pp = pe = 6.63 × 10–25 kg ms–1.

(ii) .

Ehc

106 63 10 3 10× × ×

p 9

34 8

–

–

m= = = 19.89 × 10–17 J

= ..

1 6 1019 89 10

××

–

–

19

17eV

= 1.24 × 103 eV = 1.24 keV


(iii) KE of electron = mh

m KEh

2 2e

2a

mm == G

= . ( )

( . )

2 9 1 10 10

6 63 10

× × ×

×– –

–

31 9 2

34 2

= 2.41 × 10–19 J

= ..

1 6 102 41 10

××

–

–

19

19

eV

= 1.51 eV 13. Light of wavelength 2000 Å falls on a metal surface of work function 4.2 eV. What is the kinetic

energy (in eV) of the fastest electrons emitted from the surface? [CBSE (F) 2011] Ans. Given, l = 2000 Å = 2000 × 10–10 m W0 = 4.2 eV According to Einstein's photoelectric equation:

hcm

= W0 + KEmax

or KEmax = hc

W– 0m

= ( )

( . ) ( )

.2000 10

6 63 10 3 10

1 6 101

×

× × ××

×–

–

–10

34 8

19 eV – 4.2 eV

= (6.2 – 4.2) eV = 2.0 eV

z z z


CHAPTER-5: ATOMS

1. (i) Name the famous experiment which led to the discovery of atomic nucleus. (ii) Why is neutron so effective as the bombarding particle? (iii) What is the ratio of radius of atom and radius of a nucleus? Ans. (i) Geiger-Marsden’s α-particle Scattering Experiment On the suggestion of Rutherford, in 1911, his two associates, H. Geiger and E. Marsden,

performed an experiment by bombarding α-particles (Helium nuclei Z = 2, A = 4) on a gold foil.

OR Rutherford's gold foil experiment. (ii) A neutron has no charge. It can easily penetrate even heavy nucleus without being repeled

or attracted by nucleus or electrons. So, it serve as ideal projectile for above experiment. (iii) The radius of an atom of any element is of the order of 10–10 m that of a nucleus is of the

order of 10–15 m (1 fermi). Hence, the ratio of radius of atom to that of nucleus;

rr

1010

10–

–

nucles

atom15

105= = .

2. Draw the graph of radius of orbit (rn) in hydrogen atom as a function of orbit number (n). [CBSE 2020 (55/2/3)]

Ans.

3. Define ionisation energy. What is its value for a hydrogen atom? [CBSE (AI) 2010] Ans. The minimum energy required to remove an electron from atom to infinitely is called the

ionisation energy. The ionisation energy for hydrogen atom is 13.6 eV. 4. State Bohr’s postulate of hydrogen atom which successfully explains the emission lines in the

spectrum of hydrogen atom.

Use Rydberg formula to determine the wavelength of Hα line.

[Given: Rydberg constant R = 1.03 × 107 m–1] [CBSE Panchkula 2015]

Ans. Bohr’s third postulate: It states that an electron might make a transition from one of its specified non-radiating orbits to another of lower energy. When it does so, a photon is emitted having energy equal to the energy difference between the initial and final states. The frequency of the emitted photon is given by

hν = Ei –Ef where Ei and Ef are the energies of the initial and final states and Ei > Ef . Hα line means first line of Balmer series (ni = 3)

R n n1 1 1

–H f i

2 2m

=a

f p

r1

rn

nn = 1


⇒ R RR1

21

31

41

91

365– –

H2 2m

= = =a

e o ; E

∴ .

. × mR5

365 1 03 10

366 990 10

× ×H 77–m = = =

a

= 6990 Å.

5. When an electron in hydrogen atom jumps from the third excited state to the ground state, how would the de Broglie wavelength associated with the electron change? Justify your answer. [CBSE Allahabad 2015]

Ans. de Broglie wavelength associated with a moving charge particle having a KE ‘K’ can be given as

ph

mKh

2m = = K mv m

p21

22

2

= => H …(i)

The kinetic energy of the electron in any orbit of hydrogen atom can be given as

K = – E = –. .eV eV

n n

13 6 13 6–2 2

=d n …(ii)

Let K1 and K4 be the KE of the electron in ground state and third excited state, where n1 = 1 shows ground state and n2 = 4 shows third excited state.

Using the concept of equation (i) and (ii), we have

K

K

n

n

4

1

1

4

2212

m

m= =

41

41

4

12

2

m

m= =

⇒ 414

mm

=

i.e., the wavelength in the ground state will decrease. 6. Using Bohr’s postulates, obtain the expression for (i) kinetic energy and (ii) potential energy of

the electron in stationary state of hydrogen atom. Draw the energy level diagram showing how the transitions between energy levels result in the

appearance of Lyman Series. [CBSE Delhi 2013] Ans. Suppose m be the mass of an electron and v be its speed in nth orbit of radius r. The centripetal

force for revolution is produced by electrostatic attraction between electron and nucleus.

( ) ( )

rmv

r

Z e e4

12

02rf

= … (i)

or, mv rZ e

412

0

2

rf=

So, Kinetic energy [ ]K mv21 2=

K rZ e

41

20

2

rf=

Potential energy ( ) ( )

rZ e e

41 –

0rf=

rZe

41

–0

2

rf=


Total energy, E KE PE= +

rZ e

rZ e

41

2 41

–0

2

0

2

rf rf= + f p

E rZ e

41

2–0

2

rf=

Negative sign indicates that the electron remains bound with the nucleus (or electron-nucleus form an attractive system).

Lyman series

–0.85 eV

–1.51 eV

–3.4 eV

–13.6 eV

n = ∞0 eV

= 4n

= 3n

= 2n

= 1n

7. Using Bohr’s postulates, derive the expression for the frequency of radiation emitted when electron in hydrogen atom undergoes transition from higher energy state (quantum number ni) to the lower state, (nf).

When electron in hydrogen atom jumps from energy state ni = 4 to nf = 3, 2, 1 identify the spectral series to which the emission lines belong. [CBSE (AI) 2013]

Ans. Let m be the mass of an electron and v be its speed in nth orbit of radius r. The centripetal force for revolution is produced by electrostatic attraction between electron and nucleus.

( ) ( )

rmv

r

Ze e4

12

02rf

= …(i)

For nth orbit, E can be written as En

so, –E rZe

41

2nn0

2

rf= …(ii)

From Bohr's postulate for quantization of angular momentum

mvrnh

v mrnh

2 2&r r

= =

Substituting this value of v in equation (i), we get

orrm

mrnh

rZe

rmZe

h n

2 412

02

2

20

2 2

r rf r

f= =; E

or, rmZe

h nn 2

02 2

r

f= …(iii)

Substituting value of rn in equation (ii), we get

– –E

mZe

h nZe

h nmZ e

41

28n

0

20

2 2

2

02 2 2

2 4

rf

r

f f= =

f p

or, – , whereEn

Z RhcR

chme

8n 2

2

02 3

4

f= =

R is called Rydberg constant.


For hydrogen atom Z=1, EnRhc–

n 2=

If ni and nf are the quantum numbers of initial and final states and Ei & Ef are energies of electron in H-atom in initial and final state, we have

andEnRhc

EnRch– –

ii

ff

2 2= =

If ν is the frequency of emitted radiation, we get

h

E E–i fo =

–nRc

nRc

Rcn n1 1– –

–i f f i2 2 2 2&o o= =f p > H

When, electron jump from ni = 4 to nf = 3, then Spectral lines belongs to Paschen series Similarly, ni = 4 to nf = 2 ⇒ Balmer series ni = 4 to nf = 1 ⇒ Lyman series 8. (a) Using Bohr’s postulates, derive the expression for the total energy of the electron in the

stationary states of the hydrogen atom. (b) Using Rydberg formula, calculate the wavelengths of the spectral lines of the first member

of the Lyman series and of the Balmer series. [CBSE (F) 2014] Ans. (a) Suppose m be the mass of an electron and v be its speed in nth orbit of radius r. The centripetal

force for revolution is produced by electrostatic attraction between electron and nucleus.

( ) ( )

rmv

r

Z e e4

12

02rf

= … (i)

or, mv rZ e

412

0

2

rf=

So, Kinetic energy [ ]K mv21 2=

K rZ e

41

20

2

rf=

Potential energy ( ) ( )

rZ e e

41 –

0rf=

rZe

41

–0

2

rf=

Total energy, E KE PE= +

rZ e

rZ e

41

2 41

–0

2

0

2

rf rf= + f p

E rZ e

41

2–0

2

rf=

Negative sign indicates that the electron remains bound with the nucleus (or electron-nucleus form an attractive system).

(b) Rydberg formula for the first member of Lyman series

R1

11

21

–2 2m= e o ⇒ R

1 1 41

–m

= c m


R1

43

×m

= ⇒ R34

m =

Rydberg formula for the first member of Balmer series

R1

21

31

–2 2m= e o ⇒ R

141

91

–m

= c m

R1

365

×m

= ⇒ R5

36m =

Numerical Questions 1. When electron in hydrogen atom jumps from energy state ni =4 to nf =3, 2, 1, identify the

spectral series to which the emission lines belong. [CBSE (AI) 2013] Ans. When electron jump from ni = 4 to nf = 3 then spectral lines belongs to Paschen series Similarly, ni = 4 to nf = 2 ⇒ Balmer series ni = 4 to nf = 1 ⇒ Lyman series

2. The energy of electron in nth orbit of H-atom is –.

En

13 6n 2

= eV. What is the energy required for

transition from ground state to first excited state? Ans. Energy in ground state (n = 1),

– . – . eVE1

13 6 13 61 2= =

Energy in first excited state (n = 2)

– . – . eVE2

13 6 3 42 2= =

Required energy = E2 – E1

= –3.4 – (–13.6) = 10.2 eV 3. A 12.1 eV beam is used to bombard gaseous hydrogen at room temperature. What series of

wavelengths will be emitted? Ans. For hydrogen atom

.

En

13 6n 2

= eV

For n = 1, ( )

.E

113 6–

1 2= = –13.6 eV

For n = 2, ( )

.E

213 6–

2 2= = –3.4 eV

For n = 3, ( )

.E

313 6–

3 2= = –1.51 eV

For n = 4, ( )

.E

413 6–

4 2= = –0.85 eV

Now, E3 – E1 = –1.5 – (–13.6) = 12.1 eV Since applied beam has energy = 12.1 eV, the excitation will be upto E3 and possible number of

transitions = 3. There will be two spectral lines in the Lyman series and one in the Balmer series.


4. A 12.9 eV beam of electrons is used to bombard gaseous hydrogen at room temperature. Upto which energy level the hydrogen atoms would be excited?

Calculate the wavelengths of the first member of Paschen series and first member of Balmer series. [CBSE Delhi 2014]

Ans. Energy of an electron, .

En

13 6–n 2

= eV

For n = 1, .

.E1

13 613 6– –1 2

= = eV

For n = 2, –.

– .E2

13 63 42 2

= = eV

For n = 3, –.

– .E3

13 61 513 2

= = eV

For n = 4, –.

– .E4

13 60 854 2

= = eV

Energy required to excite hydrogen atoms from ground state to excited state = Ef – Ei

= 0.85 – (–13.6) = 12.75 eV

Thus, hydrogen atom would be excited upto level n = 4.

For Paschen series

R n n1 1 1

–f i2 2

m= > H

.1

1 097 1031

41

× –72 2m

= = G

.1

1 097 10 91

161

× –7

m= ; E

.1

1 097 10 1447

× ×7

m=

.

.7 679 10

14418 75 10

×× –

77m = = m = 1875 nm

For Balmer Series

R1

21

31

–2 2m= = G

.1

1 097 10 41

91

× × –7

m= ; E

.1

1 097 10 365

× ×7

m=

.

.5 485 10

366 56 10

×× –

77m = = m = 656 nm

5. Calculate the shortest wavelength in the Balmer series of hydrogen atom. In which region (infra-red, visible, ultraviolet) of hydrogen spectrum does this wavelength lie? [CBSE Allahabad 2015]

Ans. For Shortest wavelength (or series limit) of Balmer series ni → ∞

∴ –RR1

21 1

4min

2 3m= =d n

.

.mR4

1 097 104 3 646 10min 7

7–

##m = = = m = 3646 Å

Obviously the lines of Balmer series are found in the visible region and first, second, third… lines are called Hα, Hβ, Hγ..., lines respectively.


6. Calculate the shortest wavelength of the spectral lines emitted in Balmer series. [Given Rydberg constant, R = 107 m–1] [CBSE Central 2016] Ans. For shortest wavelength (or series limit) of Balmer series, ni → ∞

∴ –RR1

21 1

4min

2 3m= =d n

.

.mR4

1 097 104 3 646 10min 7

7–

–

##m = = = m =3646 Å

Obviously the lines of Balmer series are found in the visible region and first, second, third… lines are called Hα, Hβ, Hγ..., lines respectively.

7. Use Bohr model of hydrogen atom to calculate the speed of the electron in the first excited state. [CBSE East 2016]

Ans. The speed of electron in stable orbit of H-atom is

.. .

( . )v

he

n n21

2 8 85 10 6 63 10

1 6 10 1

0

2

12 34

19 2

– –

–

# # # #

#

f= = a k

.

/m sn2 18 106#

=

For n=1, v1 = 2.18 × 106 m/s.

For n=2, .

. /m sv 22 18 10

1 09 102

66#

#= =

For n=3, .

. /m sv 32 18 10

7 27 103

65#

#= =

Obviously the speed of electron goes on decreasing with increasing n. 8. Calculate the wavelength of radiation emitted when electron in a hydrogen atom jumps from

n=∞ to n= 1. [CBSE North 2016] Ans. For shortest wavelength of Lyman series, ni = ∞

∴ –R R1

11 1

min2 3m

= =d n

.

. . Åm mR1

1 097 101

0 9116 10 911 6min 77–

##m = = = =

This is called series limit of Lyman series λlimit = 911.6 Å Obviously the lines of Lyman series are found in ultraviolet region. 9. The ground state energy of hydrogen atom is – 13.6 eV. If an electron makes a transition from

an energy level – 0.85 eV to –3.4 eV, calculate the wavelength of the spectral line emitted. To which series of hydrogen spectrum does this wavelength belong? [CBSE (AI) 2012]

Ans. As we know:

.

eVEn13 6–

n 2= ...(i)

For n =1, E1 = –13.6 eV

When electron undergoes transitions from – . – .eV eVE E0 85 3 4toA B= =

Then, from equation (i)

– ..

nn0 85

13 64

–

AA2 &= =

Similarly, – ..

nn3 4

13 62

–

BB2 &= =


Hence electron transits from n = 4 to n = 2. It corresponds to Balmer series.

We know, R n n1 1 1

–B A2 2

m= f p

Here, , , . × mn n R4 2 1 097 10 –A B

7 1= = =

Then, . ×1

1 097 1021

41

4862– Å72 2 &m

m= =e o

10. In a Geiger–Marsden experiment, calculate the distance of closest approach to the nucleus of Z = 75, when an α-particle of 5 MeV energy impinges on it before it comes momentarily to rest and reverses its direction.

How will the distance of closest approach be affected when the kinetic energy of the α-particle is doubled? [CBSE (AI) 2012]

Ans. Let r0 be the distance of closest approach where the K.E. of α-particle is converted into its potential energy.

Given, Z = 75, Ek = 5 MeV

( ) ( )

K rZe e

41 2

0 0rf=

⇒ ( ) ( )

r KZe e

KZe

41 2

42

00 0

2

rf rf= =

( . )

( . )r

5 10 1 6 10

9 10 2 75 1 6 10

× × ×

× × × × ×0 6 19

9 19 2

–

–

=

.

5 1018 75 1 6 10

×× × ×

6

10–= m = 4.32 × 10–14 m

As r K1

0 ?

If kinetic energy (K) of α-particle is doubled, the distance of closest approach will become half. 11. Using Rydberg formula, calculate the longest wavelengths belonging to Lyman and Balmer

series. In which region of hydrogen spectrum do these transitions lie? [Given R = 1.1 × 107 m–1] [CBSE (F) 2015] Ans. Rydberg formula for electron transition is given by

1m

= ( )R n n n n1 1

– >f i

i f2 2f p

For longest wavelength belonging to Lyman series, nf = 1 and ni = 2

1m

= .1 1 1011

21

× × –72 2e o

= . ×.

×1 1 10 1 41

43 3

10–7 7=c m

and m = .3 34 10× –7 = 121.2 × 10–9 m = 121.2 nm

For longest wavelength of Balmer series, nf = 2 and ni = 3

1m

= .1 1 1021

31

× × –72 2e o

= .1 1 10 365

× 7 c m


m = .5 536 10× –7

= 654.5 × 10–9 = 654.5 nm Hence, Lymanm = 121.2 nm

and Balmerm = 654.5 nm First transition lies in ultraviolet region. Second transition lies in visible region. 12. The ground state energy of hydrogen atom is – 13.6 eV. (i) What is the kinetic energy of an electron in the second excited state? (ii) If the electron jumps to the ground state from the second excited state, calculate the

wavelength of the spectral line emitted.

Ans. (i) The energy of electron in H-atom (Z = 1) in nth orbit, .

En

13 6–n 2

= eV

For, n = 1 (ground state), .

.E 113 6

13 6– –1 = = eV

For, n = 3 (second exited state), ( )

. ..E

313 6

913 6

1 51– –

–3 2= = = eV

Now, K = – E = – (–1.51) ∴ Kinetic energy, K = 1.51 eV (ii) Given, ni = 3, nf = 1

So, ( )

R RR1

11

31

99 1

98– –

2 2m= = => ;H E

∴ .R8

98 1 097 10

9× × 7

m = = = 102.7 nm.

13. In the ground state of hydrogen atom, its Bohr radius is given as 5.3 × 10–11 m. The atom is excited such that the radius becomes 21.2 × 10–11 m. Find (i) the value of the principal quantum number and (ii) the total energy of the atom in this excited state. [CBSE (South) 2016]

Ans. (i) The radii of Bohr's orbits are given by

rme

h nn 2

02 2

r

f= ⇒ rn = n2

For, n = 1 (ground state), r1 = 5.3 × 10–11 m (given)

Now, ( )

( )rr

n

1

n

12

2

= ⇒ .

.n21 2 10

5 3 10 1×

×–

–

11

11

2=

⇒ ..

n 5 321 2

42 = =

∴ n 4 2!= = Here, –ve value of (n = –2) is not taken for quantum number. Hence, required quantum number is n = 2.

(ii) For n = 2, .

En13 6–

n 2=

⇒ ( )

.E

213 6–

2 2= = –3.4 eV.


14. Calculate the de-Broglie wavelength associated with the electron revolving in the first excited state of hydrogen atom. The ground state energy of the hydrogen atom is –13.6 eV.

[CBSE 2020(55/5/1)]

Ans. The energy of the electron in the first excited state is ( )

.– . eV

213 6

3 4–

2=

Kinetic energy of electron = – K = + 3.4 eV

( . )m K Eh

2m=

. . .

.

2 9 1 10 3 4 1 6 10

6 63 10

× × × × ×

× –

31 19

34

– –m=

.

.

99 008 10

6 63 10

×

×50

34

–

–=

= 6.66 Å 15. In a hydrogen atom, find the ratio of the time taken by the electron to complete one revolution

in the first exited and in the second excited states. [CBSE 2020(55/2/3)] Ans. rn ∝ n2

vn ∝ n1

T vr2

n n

nr=

⇒ Tn ∝ n3

⇒ T

T

278

3

2 =

z z z


CHAPTER-6: NUCLEI

1. (a) Write three characteristic properties of nuclear force. (b) Draw a plot of potential energy of a pair of nucleons as a function of their separation. Write

two important conclusions that can be drawn from the graph. [CBSE Allahabad 2015] Ans. (a) The nuclear forces are found to possess the following properties: (i) Nuclear force are strong attractive force: The nuclear force is the strongest attractive

force in nature. It is much stronger than the coulomb force acting between the charges or the gravitational force between the masses. As nucleons (protons and neutrons) exist inside the nucleus, therefore for stability of nucleus the nuclear force between protons inside the nucleus must dominate over coulomb repulsive force. Actually the gravitational force is the weakest force in nature.

(ii) Nuclear forces are short range forces: The range of nuclear forces is very small of the order of 2 – 3 fm (1 fm = 10–15 m). These forces act between nucleons only when they are within the nucleus. The range of nuclear forces is taken as 1.5 fermi where the magnitude of nuclear force becomes one-third of its maximum value.

(iii) Nuclear forces are charge independent forces: The nuclear force between proton-proton (p – p), neutron-proton (n – p), neutron-neutron (n – n) is charge independent. Thus the existence of nuclear force explains the stability of nucleus. In lighter nuclei the nuclear forces between protons are much stronger than electrical repulsive forces, therefore these nuclei remains stable. As the nucleus becomes heavier, the number of protons and neutrons in nucleus goes on increasing. As the electrical repulsive force acts between each pair of protons and is inversely proportional to the square of distance

between them . .,i e Fr12?d n , while nuclear force decreases much more rapidly with

increase of distance; this results in a larger increase of electric force as compared to nuclear force and consequently the stability of nucleus begins to decrease. That is why, all the elements with Z > 83 are unstable and show the property of radioactivity.

(b)

–100

+100

1

Attractive

2 3 4

MeV

Repulsive

r (fm)

r°

A

B0

D

Conclusions: (i) The potential energy is minimum at a distance r0 of about 0.8 fm. (ii) Nuclear force is attractive for distance larger than r0. (iii) Nuclear force is repulsive if two are separated by distance less than r0. (iv) Nuclear force decreases very rapidly at r0/equilibrium position. (Any two)


2. (a) In a typical nuclear reaction, e.g.,

. ,H H He n MeV3 2712

12

23+ + +

although number of nucleons is conserved, yet energy is released. How? Explain. (b) Show that nuclear density in a given nucleus is independent of mass number A. Ans. (a) In nuclear reaction,

.H H He n MeV3 2712

12

23+ + +

Cause of the energy released: (i) Binding energy for nucleon of He2

3 becomes more than the binding energy of H12 .

(ii) Mass defect between the reactant and product nuclei DE = Dmc2

= H Hem m M c2 – nucleus12

23 2+` `j j8 B

(b) The radius of nucleus of mass number A is given by, R = R0 A

1/3

Volume of the nucleus, V R R A34

343

03r r= =

Density of the matter in the nucleus,

VolumeMass

R A

A

34

03

tr

= =

R R

34

14

3

03 0

3tr r

= =

The density is independent of mass number A.

Numerical Questions

1. Two nuclei have mass numbers in the ratio 2 : 5. What is the ratio of their nuclear densities? [CBSE Delhi 2009]

Ans. Nuclear density is independent of mass number, so ratio is 1 : 1. 2. Two nuclei have mass numbers in the ratio 8 : 125. What is the ratio of their nuclear radii?

[CBSE (AI) 2009] Ans. Nuclear radius R = R0 A

1/3

∴ R

R

A

A

1258

52

/ /

2

1

2

11 3 1 3

= = =f dp n

3. Two nuclei have mass numbers in the ratio 27 : 125. What is the ratio of their nuclear radii? [CBSE (AI) 2009]

Ans. R

R

A

A

12527

53

/ /

2

1

2

11 3 1 3

= = =f dp n

4. Obtain approximate value of the radius of (a) a nucleus of He24 (b) a nucleus of U92

238 (c) what is

the ratio of these radii? (Assume R0 = 1.2 × 10–15 m) Ans. (a) R0 = 1.2 × 10–15 m, A = 4 u Ra = R0 A

1/3 = 1.2 × 10–15 × (4)1/3 = 1.9 × 10–15 m. (b) R0 = R0 A

1/3, A = 238 u Rb = 1.2 × 10–15 × (238)1/3 = 7.4 × 10–15 m.


(c) Ratio, . ( )

. ( )R

R

1 2 10 238

1 2 10 4238

4× ×

× ×/

/ /

b

a15 1 3

15 1 3 1 3

–

–

= = c m

So, Ra : Rb = 1 : 4. 5. In the fission of a piece of uranium 1.0 g mass is decayed, then how much kilowatt hour (kWh)

energy will be obtained by it? Ans. According to mass-energy equivalence relation DE = Dmc2 Here Dm = 1.0 g = 1.0 × 10–3 kg ∴ Energy produced, DE = 1.0 × 10–3 × (3 × 108)2 = 9 × 1013 joule As 1 kWh = 3.6 × 106 joule

∴ 1 joule = .3 6 10

1× 6 kWh

∴ .

E3 6 109 10

××

6

13T = kWh = 2.5 × 107 kWh

6. Rest mass of a proton is 1.67 × 10–27 kg. What may be the maximum wavelength of a photon which can produce proton-antiproton pair?

Ans. Mass of proton = 1.67 × 10–27 kg According to Einstein’s mass-energy equivalence relation E = mc2

Energy of a photon, Ehcm

=

and mass m = mass of proton + mass of anti - proton = 1.67 × 10–27 + 1.67 × 10–27 = 2 × 1.67 × 10–27 kg

∴ hc

mc2

m=

∴ Wavelength of photon, mchc

mch

2m = = = .

.2 1 67 10 3 10

6 6 10× × × ×

×–

–

27 8

34

= 6.6 × 10–16 m = 6.6 × 10–6 Å 7. In the fission of one (U235) nucleus 200 MeV energy is released. If 2 MW power is available from

a reactor, then how many fissions per second are taking place? What is mass loss per hour? Ans. Power of reactor 2 MW = 2 × 106 W = 2 × 106 J / s

= .1 6 102 10

××

–19

6

eV / s = 12.5 × 1024 eV / s = 12.5 × 1018 MeV / s

The number of fissions per second = .200

12 5 10× 18

= 6.25 × 1016

Energy obtained per second = 2 × 106 J Energy obtained per hour, DE = 2 × 106 × 60 × 60 J = 72 × 108 J

∴ Mass-loss per hour = ( )c

E3 1072 10

××

2 8 2

8T = = 8 × 10–8 kg

8. Consider one of the fission reactions of U235 by thermal neutrons: U n Sr Xe n292

2350

138

9454

140$+ + +

The fission fragments are, however, not stable. They undergo successive b-decays until 38Sr94 becomes 40Zr94 and 54Xe140 becomes 58Ce140. Estimate the total energy released in the process. Is all that energy available as kinetic energy of the fission products (Zr and Ce)? Atomic masses are given to be

m (U235) = 235.0439 u, mn = 1.00866 u m (Zr94) = 93.9065 u, m (Ce140) = 139.9055 u


Ans. The given reaction is

U n Sr Xe n292235

01

3894

54140$+ + +

Successive b-decays of 38Sr94 and 54Ze140 are expressed as

38Sr94 → 40Zr94 + 2 e–

54Xe140 → 58Ce140 + 4 e– Therefore, net reaction is 94U

235 + n → 40Zr94 + 58Ce140 + 6 e– + Q The reaction involves nuclear masses; therefore, from given atomic masses, masses of electrons

are to be subtracted. So Q = {m (U235) – 92me} – [m (Zr94) – 40me + m (Ce140) – 58me + mn + 6me] × c2 = {m (U235) – m (Zr94) – m (Ce140) – mn} × c2 = {235.0439 – 93.9065 – 139.9055 – 1.00866} u × 931 MeV = (235.0439 – 234.82066) u × 931 MeV = {0.22324 u} × 931 MeV = 207.8 MeV The whole energy is not available in the form of kinetic energy of fission products, but it is partly

shared by neutrons produced and partly by b-particles produced. 9. A deuterium reaction that occurs in experimental fusion reactor is 1H

2 + 1H2 → 1H

3 + 1H1 + Q1

and then it follows 1H

3 + 1H2 → 2He4 + 0n

1 + Q2

(a) Calculate the energy released in each of these reactions. (b) Calculate the total energy released per gram of deuteron used in fusion. Given masses of 1H

2 = 2.014740 amu, 1H3 = 3.017005 amu

1H1 = 1.008145 amu, 0n

1 = 1.008986 amu 2He4 = 4.003179 amu, 1 amu = 931 MeV Ans. (a) Mass defect in reaction (1) Dm1 = 2.014740 + 2.014740 – 3.017005 – 1.008145 = 0.004330 amu As 1 amu = 931 MeV ∴ Energy released, Q1 = 0.004330 × 931 MeV = 4.031 MeV. Mass defect in reaction (2) Dm2 = 3.017005 + 2.014740 – 4.003179 – 1.008986 = 0.019580 amu ∴ Energy released Q2 = 0.019580 × 931 MeV = 18.23 MeV. (b) Total energy released Q = Q1 + Q2 = 4.031 + 18.23 = 22.26 MeV. This total released energy is from the fusion of 3 (1H

2) nuclei.

∴ Energy released per 1H2-nucleus =

.3

22 26 = 7.42 MeV.

Number of nuclei in 1 g of 1H2 =

..2 01474

6 02 10× 23

Total energy released from 1 g of 1H2 =

..2 01474

6 02 10× 23

× 7.42 MeV.

= 2.217 × 1024 MeV.

z z z


CHAPTER-7: ELECTRONIC DEVICES

1. (i) What is forbidden energy gap (or band gap)?

(ii) What is the order of band gap Eg in Ge and Si?

(iii) What happens to its band gap width when a semiconductor material is doped?

Ans. (i) Energy Bands: In a solid, the energy of electrons lie within certain range. The energy levels of allowed energy are in the form of bands, these bands are separated by regions of forbidden energy called band gaps.

(ii) Energy gap or band gap for Ge, Eg = 0.7 eV

and Si, Eg = 1.1 eV.

(iii) When semiconductor material is doped then its band gap width decreases and conductivity of material increases.

2. (i) What is meant by intrinsic and extrinsic semiconductors?

(ii) What are n-type and p-type semiconductors?

(iii) What do you mean by drift and diffusion in relation to with p-n junction?

Ans. (i) Intrinsic semiconductors: Pure semiconductors in which the conductivity is caused due to charge carriers made available from within the material are called intrinsic semiconductors. There are no free charge carriers available under normal conditions. However, when the temperature is raised slightly, some of the covalent bonds in the material get broken due to thermal agitation and few electrons become free. In order to fill the vacancy created by absence of electron at a particular location, electron from other position move to this location and create a vacancy (absence of electron) at another place called hole. The movement/shifting of electrons and holes within the material results in conduction.

An intrinsic semiconductors behaves as a perfect insulator at temperature 0 K.

Extrinsic semiconductors: The semiconductors in which the conductivity is caused due to charge carriers made available from external source by adding impurity from outside are called extrinsic semiconductor. The process of adding impurity is called doping. The impurity added is generally from third group or fifth group. There are two types of extrinsic semiconductors: n–type or p–type.

(ii) n-type semiconductors: When a pentavalent impurity like phosphorus, antimony, arsenic is doped in pure-germanium (or silicon), then the conductivity of crystal increases due to surplus electrons and such a crystal is said to be n-type semiconductor, while the impurity atoms are called donors atoms. Thus, in n-type semiconductors the charge carriers are negatively charged electrons and the donor level lies near the bottom of the conduction band.

p-type semiconductors: When a trivalent impurity like aluminium, indium, boron, gallium, etc., is doped in pure germanium (or silicon), then the conductivity of the crystal increases due to deficiency of electrons i.e., holes and such a crystal is said to be p–type semiconductor while the impurity atoms are called acceptors. Thus in p–type semiconductors the charge carriers are holes. Acceptor level lies near the top of the valence band.

Conduction bandDonor level

Valence band(a) n-type semiconductor

Conduction band

Acceptor levelValence band

(b) p-type semiconductor


(iii) Two important processes occurring during the formation of a p-n junction are (a) diffusion and (b) drift.

(a) Diffusion: In n-type semiconductor, the concentration of electrons is much greater as compared to concentration of holes; while in p-type semiconductor, the concentration of holes is much greater than the concentration of electrons. When a p-n junction is formed, then due to concentration gradient, the holes diffuse from p-side to n-side (p → n) and electrons diffuse from n-side to p-side (n → p). This motion of charge carriers gives rise to diffusion current across the junction.

(b) Drift: The drift of charge carriers occurs due to electric field. Due to built in potential barrier, an electric field directed from n-region to p-region is developed across the junction. This field causes motion of electrons on p-side of the junction to n-side and motion of holes on n-side of junction to p-side. Thus a drift current starts. This current is opposite to the direction of diffusion current.

With passage of time, diffusion current decreases whereas drift current increases and they balance each other. This creates a potential barrier.

3. (i) On what factors does the conductivity of intrinsic semiconductor depend? (ii) How does the width of the depletion layer of a p-n junction diode change with decrease in

reverse bias? [CBSE (AI) 2008] Ans. (i) Conductivity of a pure semiconductor depends upon: (a) The width of band gap or energy band (b) Intrinsic charge carrier concentration (i.e., electrons or holes). (ii) If the reverse bias across a p-n junction is decreased, the depletion region of p-n junction

decreases. 4. State the factor, which controls: (i) wavelength of light, and (ii) intensity of light emitted by an LED. Ans. (i) Wavelength of light (photons) emitted from LED depends upon band gap or band energy

which control the wavelength of LED. (ii) The forward current increases as the intensity of light (photons) increases and reaches a

maximum value. But further increase in forward current results in decrease of light intensity. Note: LEDs are biased such that the light emitting efficiency is maximum. 5. Two semiconductor materials X and Y shown in the alongside figure, are

made by doping a germanium crystal with indium and arsenic respectively. The two are joined end to end and connected to a battery as shown.

(i) Will the junction be forward biased or reverse biased? (ii) Sketch a V-I graph for this arrangement. Ans. (i) Y → n-type semiconductor and X → p-type semiconductor. Hence, p-n junction is reverse biased.

Electron diffusionElectron drift

Hole diffusion

Electron

Hole

p n– +

– +

– +

Hole drift

Vo Potential barrier

Depletionlayer

w

N

– +

– +

– +P

X Y


(ii)

Reverse bias

mA

V

µA

6. Describe, with the help of a circuit diagram, the working of a photo diode. [CBSE (F) 2017] Ans. The characteristic curves of a photodiode for two different illuminations I1 and I2 (I2 > I1) are shown.

Reverse bias

mA

Volt

(I2 > I1)I2

I1

µA

7. Draw a circuit diagram of an illuminated photodiode in reverse bias. How is a photodiode used to measure the light intensity?

Ans.

p-side n-side

R

A

E

h

The photocurrent from photodiode depends on the intensity of incident light i.e., photocurrent is proportional to incident light intensity. A change in the light intensity, is observed when a reverse bias is applied.

8. Plot V-I characteristics for an illuminated photodiode under reverse bias for three different illumination intensities I1 > I2 > I3. [CBSE 2020 (55/2/2)]

Ans.

µA

I3I2I1

I1 > I2 > I3

mA

VoltsReverse bias


9. Distinguish between, n-type and p-type semiconductors on the basis of energy band diagrams. Compare their conductivities at absolute zero temperature and at room temperature. [CBSE Delhi 2015(C)]

Ans.

In n-type semiconductors, most of the electrons come from the donor impurity while in p-type semiconductor, the density of holes in the valence band is predominant due to impurity in the extrinsic semiconductors.

At absolute zero temperature conductivities of both type (n-type and p-type) will be zero. For equal doping, an n-type semiconductor will have more conductivity than a p-type

semiconductor at room temperature. 10. Write the two processes that take place in the formation of p-n junction. Explain with the help of

a diagram, the formation of depletion region and barrier potential in a p-n junction. [CBSE Delhi 2017]

Ans.

Two processes which occur during the formation of a p-n junction are diffusion and drift. Due to the concentration gradient across p and n-sides of the junction, holes diffuse from p-side to n-side (p → n) and electrons diffuse from n-side to p-side (n → p). This movement of charge carriers leaves behind ionised acceptors (negative charge immobile) on the p-side and donors (positive charge immobile) on the n-side of the junction. This space charge region on either side of the junction together is known as depletion region.

The loss of electrons from the n-region and the gain of electron by the p-region causes a difference of potential across the junction of the two regions. The polarity of this potential is such as to oppose further flow of charge carriers across the junction so that a condition of equilibrium exists. Since this potential tends to prevent the movement of electron from the n-region into the p-region, it is called a barrier potential.

11. Draw a labelled circuit diagram of a full wave rectifier. Explain its underlying principle and working. Depict the input and output waveforms. Describe briefly the role of a capacitor in filtering. [CBSE Chennai 2015]

Ans. Rectification: Rectification means conversion of ac into dc. A p-n diode acts as a rectifier because an ac changes polarity periodically and a p-n diode allows the current to pass only when it is forward biased. This makes the diode suitable for rectification.

Ele

ctro

n en

ergy

n type p type

EC

Eg

EV

{

{

ECED

EV

EA

=0.01 - 0.05 eV

≈0.01eVEg

(a) T > O K(b) T > O KOne thermally generated electron-hole

pair + 9 electrons from donor atoms

_ _

_ _

_ _

_ _

_ _

+ +

+ +

+ +

+ +

+ +

n

Depletion regionHole diffusion Hole drift

Electron diffusionElectron drift

p


Working: The ac input voltage across secondary S1 and S2 changes polarity after each half cycle. Suppose during the first half cycle of input ac signal, the terminal S1 is positive relative to centre tap O and S2 is negative relative to O. Then diode D1 is forward biased and diode D2 is reverse biased. Therefore, diode D1 conducts while diode D2 does not. The direction of current (i1) due to diode D1 in load resistance RL is directed from A to B In next half cycle, the terminal S1 is negative and S2 is positive relative to centre tap O. The diode D1 is reverse biased and diode D2 is forward biased. Therefore, diode D2 conducts while D1 does not. The direction of current (i2) due to diode D2 in load resistance RL is still from A to B. Thus, the current in load resistance RL is in the same direction for both half cycles of input ac voltage. Thus for input ac signal the output current is a continuous series of unidirectional pulses.

p1 n1

RLP

P

2

1

Inpu

tAC

sign

alto

berectified S1

S2p2 n2

D1

D2

+

–

A

B

i1

i2 Output

OCentre tap

32TTT

2t

2TO

Waveform

Waveform

at p 1

32TTT

2t2T

Oat p 2

32TTT

2t2TO

Outputwaveform

(across R )L

Due toD1

Due toD1

Due toD2

Due toD2

In a full wave rectifier, if input frequency is f hertz, then output frequency will be 2f hertz because for each cycle of input, two positive half cycles of output are obtained.

In full wave rectifier both positive and negative half cycles of the input ac current will charge the capacitor. The main role of the capacitor filter is to short the ripples to the ground and to block the pure dc components, so that it flows through alternate path and reaches to output load resistor (RL).

12. (a) In the following diagram, is the junction diode forward biased or reverse biased?

+ 5 V

(b) Draw the circuit diagram of a full wave rectifier and state how it works. [CBSE (AI) 2017] Ans. (a) The junction shown in figure is reverse biased. (b) Rectification: Rectification means conversion of ac into dc. A p-n diode acts as a rectifier

because an ac changes polarity periodically and a p-n diode allows the current to pass only when it is forward biased. This makes the diode suitable for rectification.



p1 n1

RLP

P

2

1In

putA

Csi

gnal

tobe

rectified S1

S2p2 n2

D1

D2

+

–

A

B

i1

i2 Output

OCentre tap

32TTT

2t

2TO

Waveform

Waveform

at p 1

32TTT

2t2T

Oat p 2

32TTT

2t2TO

Outputwaveform

(across R )L

Due toD1

Due toD1

Due toD2

Due toD2


13. (a) Draw the circuit diagrams of a p-n junction diode in (i) forward bias, (ii) reverse bias. How are these circuits used to study the V–I characteristics of a silicon diode? Draw the typical V–I characteristics?

(b) What is a light emitting diode (LED)? Mention two important advantages of LEDs over conventional lamps. [CBSE 2020 (55/4/1)]

Ans. (a)

The following are the basic features of forward biasing: (i) Within the junction diode the current is due to both types of majority charge carriers

but in external circuit it is due to electrons only. (ii) The current is due to diffusion of majority charge carriers through the junction and is

of the order of milliamperes. The basic features of reverse bias are:

(i) Within the junction diode the current is due to both types of minority charge carriers but in external circuit it is due to electrons only.

(ii) The current is due to leakage of minority charge carriers through the junction and is very small of the order of µA.

Characteristics of a p–n junction diode: The graph of voltage V versus current I in forward bias and

reverse bias of a p–n junction is shown in the figure. (b) A light emitting diode (LED) is a semiconductor device which emits light when current flow

through it. Electrons in the semiconductor recombine with electron-holes, releasing energy in the form of photon (light energy).

Advantages: (i) Low operational voltage and less power consumption. (ii) Fast action and no warm-up time required. (iii) Long life and ruggedness. (iv) Fast on-off switching capability. (Any two)

+–Reverse biasingForward biasing

nn pp

Forwardbias

F

V

I ( A)Reverse

biasR

OV(–) (+)

I (mA)

Avalanchebreakdown

�


14. What is a rectifier? Explain the working of a full wave rectifier giving a labelled circuit diagram.

[CBSE 2020 (55/5/1)]

Ans. Rectifier is a semiconductor device which convert ac into dc.

Rectification: Rectification means conversion of ac into dc. A p-n diode acts as a rectifier because an ac changes polarity periodically and a p-n diode allows the current to pass only when it is forward biased. This makes the diode suitable for rectification.


p1 n1

RLP

P

2

1

Inpu

tAC

sign

alto

berectified S1

S2p2 n2

D1

D2

+

–

A

B

i1

i2 Output

OCentre tap

32TTT

2t

2TO

Waveform

Waveform

at p 1

32TTT

2t2T

Oat p 2

32TTT

2t2TO

Outputwaveform

(across R )L

Due toD1

Due toD1

Due toD2

Due toD2


15. What is the function of a solar cell? Briefly explain its working and draw its I-V characteristic curve. [CBSE 2020 (55/2/1)]

Ans. Solar cell is a device which converts solar energy into electrical energy.

Working: When photons of energy greater than band gap energy (hν>Eg) are made to incident on the junction, electron-hole pairs are created which move in opposite directions due to junction field. These are collected at two sides of junction, thus producing photo-voltage; this gives rise to photocurrent. The characteristic curve of solar cell is shown below.

The V- I characteristics of solar cell:

I

VOCV

ISC

Short circuit current

(Open circuit voltage)


16. Name the optoelectronic device used for detecting optical signals and mention the biasing in which it is operated. Draw its I-V characteristics. [CBSE Sample Paper 2018]

Ans. Photo diode is an opto electric device which is used for detecting optical signals. It is easier to observe the change in the current with change in light intensity if a reverse bias is

applied. Hence, it is operated in reverse biased.

Reverse bias

mA

Volt

(I2 > I1)I2

I1

µA

17. (i) Can a slab of p-type semiconductor be physically joined to another n-type semiconductor slab to form p-n junction? Justify your answer.

(ii) In a p-n junction diode the forward bias resistance is low as compared to the reverse bias resistance. Give reason. [CBSE 2020 (55/4/1)]

Ans. (i) No, slab will have roughness much larger than the inter atoms crystal spacing and hence continuous contact at the atomic level will not be possible. The junction will behave as a discontinuity for the flowing charge carriers.

(ii) In a p-n junction diode the forward bias has a low resistance because the barrier potential decreases and when the diode is reverse biased then the barrier potential increases hence it has a high resistance.

18. Why is an intrinsic semiconductor deliberately converted into an extrinsic semiconductor by adding impurity atoms? [CBSE 2020 (55/3/1)]

Ans. The conductivity of intrinsic semiconductor is very low at room temperature hence it cannot be used to develop any electronic device from it so its electrical conductivity is increased by adding impurity to intrinsic semiconductor.

z z z

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