I-I, PHYSICS QUESTION BANK WITH ANSWERS UNIT I … · 2017-11-23 · nawab shah alam khan college of engineering & technology 1 prof. s.m.asadullah nsakcet i-i, physics – question

NAWAB SHAH ALAM KHAN COLLEGE OF ENGINEERING & TECHNOLOGY 1

Prof. S.M.ASADULLAH NSAKCET

I-I, PHYSICS – QUESTION BANK WITH ANSWERS

UNIT I

INTERFERENCE IN THIN FILMS

1. Define 1) physical thickness 2) apparent thickness 3) optical thickness of a transparent

medium.

1) physical thickness: it is actual thickness of a transparent medium. It is also called

geometric length

2) Apparent thickness: a denser medium appears to be less thick than its actual thickness

where as a rarer medium appears to be thicker than its actual thickness. If ‘t’ is the geometric

or actual thickness and ‘μ’ is the refractive index then its apparent thickness is given by t/μ.

3) Optical thickness: it is defined as the thickness of vacuum which contains the same

number of waves as the medium under consideration. If ‘t’ is the geometric or actual thickness

and ‘μ’ is the refractive index then its optical thickness is given by ‘μt’. The concept of optical

thickness is used to find the path difference between two waves traveling in different media.

Problem 1: A glass slab has thickness 6 cm. Find its apparent & optical thickness (μ = 3/2)

2. What is phase difference and path difference?

Phase difference: In a wave the phase changes gradually. If two points are separated by a

distance ‘λ’ the phase difference between them is 2π. If two points are separated by a distance

‘∆x’ then the phase difference between them ‘δ’ is given as x

2

Problem 2: A wave has wave length 6 m. Find the phase difference between two particles

separated by 1) 2 m 2)20 cm.

a) Path difference: suppose two rays reach a point taking two different paths. The difference

between optical paths of the two rays travelling in different directions is known as the optical

path difference.

If two rays of light travel distances t1 and t2 in air or vacuum their path difference ∆ = t2 – t1

If a ray travels a distance t1 in air and t2 in a medium of refractive index μ then the path

difference ∆ = μt2 – t1

If a ray travels a distance t1 in a medium of refractive index μ1 and t2 in a medium of refractive

index μ2 then the path difference ∆ = μ2t2 – μ1t1

Problem 3: Three rays R1,R2 and R3 travel distances 6 cm each, but R1 in air,R2 in water and

R3 in glass. Find the path difference in R1&R2,R1&R3 and R2&R3 respectively.

b) Phase change due to reflection at boundaries of optical interfaces: Light waves may

undergo phase change due to reflection at some point in their path. If the waves are reflected at

a rarer-to-denser medium boundary, the reflected waves suffer a phase change of π rad or

180° or a path difference of λ/2. Therefore, we must add or subtract λ/2 in the calculation of

true optical path difference whenever a reflection occurs at a denser medium.

3. What are coherent waves?

Two light waves having same frequency, same wave length and same phase or constant phase

difference are called coherent waves. They may or may not have same amplitude.

4. What are coherent sources? OR

Explain the term coherent sources. Describe the general methods to obtain coherent

sources.

Coherent sources: If two sources produce light of same frequency, same wave length and

same phase or constant phase difference they are called coherent sources. The respective

waves are called coherent waves. Two independent monochromatic sources cannot be



coherent. Hence coherent sources can be obtained by either 1) division of wave front or

2) division of amplitude

5. What is division of wave front (Dec 2016)

1 Division of wave front: in this method a narrow slit is used as source. The wave front is

spherical. This wave front is subsequently divided into two parts by two narrow slits S1 and S2.

This method is used in Young double slit experiment. Fresnel’s bi prism, Lloyd’s mirror, etc

are the other examples where the division of wave front is used.

6. What is division of amplitude

Division of amplitude: In this method a broad source is used. The wave

front is plane. The amplitude of the light beam is divided by partial reflection

into two or more beams. Thin films (parallel sided, wedge shaped, Newton’s

ring, etc), interferometers such as Michelson’s interferometer etc utilize this

method in producing interference.

7. Discuss why two independent sources of light of the same wavelength cannot produce

interference fringes.

Two independent sources of light cannot maintain phase difference constant. Hence they

cannot produce sustained interference fringes.

8. What is Stokes’s law of change of phase on reflection

According to Stokes, when a light wave is reflected at the surface of an optically denser

medium, it suffers a phase change of π i.e., a path difference of λ/ 2. No such phase change is

introduced if the reflection takes place from the surface of rarer medium

9. What is superposition principle? (June 2005)

When two waves travel simultaneously in a medium the resultant displacement at any point is

the algebraic sum of the displacements due to the individual waves. Suppose y1 and y2 are the

displacements of two waves at a point. The resultant displacement due to the principle of

superposition is Y = y1 ± y2.

If y1 = a sin(ωt) and y2 = a sin(ωt + δ) then Y = A sin(ωt + θ).

The amplitude A = 2a cos (δ/2) and the resultant intensity I = 4a2 cos2 (δ/2) = 4Io cos2 (δ/2)

where Io is the intensity of each source.

10. What is meant by interference of light? State the fundamental conditions for the

production of interference fringes. (June 2017)

The phenomenon of redistribution of light energy due to superposition of two coherent waves

is called interference. Suppose y1 and y2 are the displacements of two waves at a point. The

resultant displacement due to the principle of superposition is Y = y1 ± y2.

If y1 = a sin(ωt) and y2 = a sin(ωt + δ) then Y = A sin(ωt + θ).

The amplitude A = 2a cos (δ/2) and the resultant intensity I = 4a2 cos2 (δ/2) = 4Io cos2 (δ/2)

where Io is the intensity due to each source.

For interference maximum cos (δ/2) = ±1. i.e., the phase difference δ = 0,2π, 4π,………2nπ

Or Path difference ∆ = nλ

For interference minimum cos (δ/2) = 0. i.e., the phase difference δ = π, 3π, 5π,………(2n±1)π

Or Path difference ∆ = (2n±1)λ/2

Conditions for observing sustained interference.

1) The waves must be coherent.

2) Their amplitudes should be equal

3) If the waves are polarized their planes of polarization should be parallel

4) The sources must be close.



11. Explain the interference of light due to thin films. (June 2006, May 2007)

With ray diagram discuss the theory of thin films and the condition for constructive and

destructive interference in the case of reflected system. (2008)

Interference in reflected light.

Let us consider a thin transparent film of uniform thickness t

and refractive index μ. Let GH and G1H1 be the two surfaces of

the film as shown in Fig.

Suppose a ray AB of monochromatic light be incident on its

upper surface GH. This ray is partly reflected along BR and

refracted along BC. After one internal reflection at C, it travels

along CD and finally emerges out along DR1 in air.

DR1 is parallel to BR. To find out the effective path difference

between the rays BR and DR1we draw a normal DE on BR and

normal BF on CD. Extend DC and BQ to meet at P. In the figure ABN = i, (the angle of

incidence) and QBC = r, (the angle of refraction). From the geometry of the figure BDE =

i and QPC = r.

The optical path difference between the two reflected light rays (BR and DR1) is given by

= Path (BC + CD) in film - Path BE in air = (BC + CD) - BE …………(1)

We know that = FD

BE

FD/BD

BE/BD

sinr

siniμ

BE = (FD) ……….(2)

From equations (1) and (2) = (BC+CD)-(FD) = (BC+CF+FD) - (FD)

= (BC + CF)

= (PF) (As BC = PC) ………..(3)

From triangle BPF, cos r = PF/BP or PF = BP cos r = 2t cos r ...(4)

Substituting the value of PF from equation (4) in equation (3), we have

= x2t cos r = 2t cosr ...(5)

It should be remembered that a ray reflected at a surface backed by a denser medium suffers

an abrupt phase change of which is equivalent to a path difference /2. Thus the effective

path difference between the two reflected rays is (2 t cos r / 2). We know that maxima

occurs when effective

Path difference = n .

i.e., 2 t cos r / 2 = n

or 2 t cos r = (2n 1) /2 ...(6)

2 t cos r = (2n + 1) /2 where n = 0,1,2,3,4 …..

2 t cos r = (2n - 1) /2 where n = 1,2,3,4 …..

If this condition is fulfilled, the film will appear bright in the reflected light.

The minima occurs when the effective path difference

2 t cos r / 2 = (2n 1) /2

Or 2 t cos r = n

Here n = 0, 1, 2, 3 etc.

When this condition is fulfilled the film will appear dark in the reflected light.



12. With ray diagram discuss the theory of thin films and the condition for constructive

and destructive interference in the case of transmitted system.(May 2007, Dec 2016)

Interference in transmitted light.

Let us consider a thin transparent film of uniform thickness t and

refractive index μ. Let GH and G1H1 be the two surfaces of the

film as shown in Fig.

Suppose a ray AB of monochromatic light be incident on its

upper surface GH.

Due to multiple reflection and refraction at the two surfaces we

obtain two transmitted rays CT and ET1. Since these rays have

originated from the same point source they are coherent. Hence

they have a constant phase difference and are in a position to

produce sustained interference when combined. In order to calculate the path difference

between the two transmitted rays, we draw normal CQ and EP on DE and CT respectively. We

produce ED and CF to meet at I

The effective path difference is given by = (CD + DE) - CP . ...(1)

We know that QE

CP

QE/CE

CP/CE

sinr

siniμ

Therefore CP = (QE) ……(2)

From equations (1) and (2) = (CD + DQ + QE) – (QE)

= (CD+DQ) = (QI) = 2 t cos r

Here it should be remembered that inside the film, reflection at different points takes place at

the surface backed by rarer medium (air) thus no abrupt change of takes place in this case.

The interference maxima occur when effective path difference = n

i.e. 2 t cos r = n

If this condition is fulfilled, the film will appear bright in transmitted light.

The minima occur when the effective path difference

2 t cos r = (2n 1) /2 where n = 1, 2, 3, etc. When this condition is fulfilled, the film will

appear dark.

Thus the conditions of maxima and minima in transmitted light are just reverse of the

conditions for reflected light.

13. Show that with monochromatic light, the interference patterns of reflected and

transmitted light are complementary. Thus the effective path difference between the two reflected rays from a thin film

∆ = 2 t cos r / 2.

We know that maxima occur when effective Path difference = n .

i.e., 2 t cos r / 2 = n

or 2 t cos r = (2n 1) /2 …....(1)

2 t cos r = (2n + 1) /2 where n = 0,1,2,3,4 …..

2 t cos r = (2n - 1) /2 where n = 1,2,3,4 …..

If this condition is fulfilled, the film will appear bright in the reflected light.

The minima occurs when the effective path difference 2 t cos r / 2 = (2n 1) /2

Or 2 t cos r = n ……(2) where n = 0, 1, 2, 3 etc.

When this condition is fulfilled the film will appear dark in the reflected light.

The effective path difference between the two transmitted rays from a thin film

∆ = 2 t cos r .

The maxima occur when effective path difference = n



i.e. 2 t cos r = n …………..(3)

If this condition is fulfilled, the film will appear bright in transmitted light.

The minima occur when the effective path difference

2 t cos r = (2n 1) /2 ………(4) where n = 1, 2, 3, etc.

When this condition is fulfilled, the film will appear dark.

From equations (1) and (4) it is clear that the condition for maximum in reflected light is

same that for minimum in transmitted light.

Similarly from equations (2) and (3) it is clear that the condition for minimum in reflected

light is same as that for maximum for transmitted light.

Thus the conditions of maxima and minima in transmitted light are just reverse of the

conditions for reflected light.

14. Explain the colors in a thin film when exposed to sun light. (June 2005)

Colours in thin films: when a thin film (of soap solution or oil on the surface of water) is

exposed to a source of white light beautiful colours are observed. This is due to the

phenomenon of interference in thin films.

Explanation: In thin films the path difference between the light reflected from the top

and the bottom surface is 2μt cosr – λ/2. At a particular point of the film and for a particular

position of the eye, the interfering rays of only certain wavelengths satisfy the conditions of

maxima. As we change the position of the maximum condition is satisfied by different

colours. Hence we see multicolours.

15. Why the colours seen in reflected light are not observed in transmitted light.

for thin films the path difference for reflected light is 2 t cos r / 2 whereas for transmitted

light the path difference is 2 t cos r.

The conditions for maxima and minima in transmitted light are inverse that of reflected light.

Hence the colours suppressed or absent in reflected light appear as intense colours in

transmitted light or colours of reflected and transmitted light are complementary.

16. Why do soap bubbles appear multicoloured when viewed under sun light.

In thin films the condition for brightness is given by 2 μt cos r = (2n + 1)λ/2 .

Since the bubble is spherical in shape, the light incident will have different values of r at

different points on the soap bubble. Hence we get multicolours.

17. Why does a film appear black in reflected light when it is excessively thin?

The path difference in thin films is given by 2μt cos r – λ/2

If the thickness of the film is extremely small when compared to λ, then 2μt cos r

term can be neglected and the path difference is λ/2. Hence destructive interference occurs. So

the film appears dark.

18. Why a thick film seen by reflected light shows no colours but appears white ?

Why a thick film does not show colors when seen in white light. Light waves can maintain phase up to a certain length only. This is called coherence length.

For visible light it is of the order of fraction of a millimeter. In thick films or glass plates of

thickness millimeter or more interference is not observed as the light waves are no longer

coherent.

Problem 4: A parallel beam of light (λ = 5890x10-8 cm) is incident on a thin glass plate

(μ = 1.5) such that the angle of refraction onto the plate is 60o. Calculate the smallest thickness

of the glass plate which will appear dark by reflection [3.26 x 10-5 cm]

Hint: rcosμ2

λt



Problem 5: White light falls normally on a film of soapy water whose thickness is 5 x 10-5 cm

and μ = 1.33. Which wavelength in the visible region will be reflected most strongly?

[ 5320 x 10-8 cm, for n = 2]

Hint: The condition of maxima for reflected light is 2 t cos r = (2n 1) /2

Find λ for different values of n

Problem 6: White light is incident on a soap film at an angle sin -1(4/5) and the reflected light

on examination by a spectroscope shows dark bands. Two consecutive dark bands correspond

to wavelengths 6.1 x10 -5 and 6.0 x 10-5 cm. If the refractive index of the film 4/3, calculate the

thickness.[0.0017 cm]

Hint: 21

221

λλ

λnorλ1)(nλncosrt2μ

or 21

21

λλ

λλcosrt2μ

INTERFERENCE IN WEDGE SHAPED FILM

19. Discuss the formation of interference fringes in a thin wedge shaped film.(May 2017)

problem: Find the thickness of wedge-shaped air film at a point where fourth bright fringe is

situated. Wave length of light is 589.3 nm

A thin wedge of air film can be obtained by two glass slides

resting on each other and separated by a thin spacer at one end.

The experimental arrangement is shown. When light is incident

from above, it gets partially reflected from the top and bottom

surfaces of the air film. The two rays BC and FE are coherent as

they are derived from the same ray AB through the division of

amplitude. For small film thickness the rays interfere producing

dark and bright fringes depending on the phase difference. Since

the glass plates are thick the interference is mainly due to the air

film.

At a point of air film of thickness t the optical path

difference

between the two rays BC and FE is given by

∆ = 2μt cos r ± λ/2 ………….(1)

the term λ/2 is due to reflection at the air glass boundary at

the bottom.

For interference maximum the path difference ∆ = mλ ..(2)

or 2μt cos r = (2m ± 1)λ/2 ………….(3)

For interference minimum the path difference ∆ = (2m ± 1)λ/2 …..(3)

or 2μt cos r = m λ ………….(4)

for normal incidence cos r = 1. For mth dark fringe suppose the thickness of the air film is t1.

Then 2μt1 = m λ …………(4)

Suppose the next dark fringe occurs at a thickness t2. Then 2μt2 = (m + 1) λ …..(5)

From the above equations we get 2μ(t2 – t1) = λ …..(6)

But (t2 – t1) = BC

2μ(BC) = λ

BC = λ/2μ

From the triangle ABC, BC = AB tanθ

hence AB tanθ = λ/2μ

the fringe width AB, β = θ2μ

λ

tanθ2μ

λ



Salient features of interference pattern

1. Fringe at the apex is dark.

2. Fringes are straight and parallel.

3. Fringes are of equal thickness.

4. Fringes are localized.

Problem Given that wave length λ = 589.3 nm

number of bright fringe m = 4

thickness of the film t = ?

Applying 2μt cos r = (2m - 1 )λ/2 we get

2t = 3.5x589.3 nm

or thickness t = 1.75x589.3 nm = 1031.3 nm

problem 7: Fringes of equal thickness are observed in a thin glass wedge of refractive index

1.52. The fringe spacing is 1 mm and wavelength of light is 5893Å. Calculate the angle of the

wedge.[39.96 seconds of an arc]

Hint: β θ2μ

λ

Problem 8:If the angle of wedge is 0.25o and the wavelengths of sodium lines are 5890Å and

5896Å, find the distance from the apex at which the maxima due to two wavelengths first

coincide when observed in reflected light [n = 981]

Hint: 2t = (2n + 1)λ1/2 = (2n + 3)λ2/2

NEWTON’S RINGS

20. Discuss the theory of Newton’s rings with relevant diagram (b) Explain how

Newton’s rings are formed in the reflected light.

21. Derive the expressions for the diameters of dark and bright rings.(June 2005 ) 22. Prove that in reflected light, (i) Diameters of the dark rings are proportional to the

square roots of natural numbers and (ii) diameters of bright rings are proportional to

the square roots of odd numbers. (iii) the difference in the

squares of the diameters is constant When a plano-convex lens with its convex surface is placed on a

plane glass plate, an air film of gradually increasing thickness is

formed between the two. The thickness of the film at the point of

contact is zero. If monochromatic light is allowed to fall normally,

and the film is viewed in reflected light, alternate dark and bright

rings concentric around the point of contact between the lens and

glass plate are seen. These are called Newton’s rings

Experimental arrangement.

The experimental arrangement of obtaining Newton’s rings is

shown in figure1. L is a plano-convex lens of large radius of

curvature R. This lens with its convex surface is placed on a plane

glass plate P. The lens makes contact with the plate at C. Light

from an extended monochromatic source such as sodium lamp

falls on a glass plate G held at an angle 45° with the vertical. The

glass plate G reflects normally a part of the incident light towards

the air film enclosed by the lens L and the glass plate P. A part of

the incident light is reflected by the curved surface of the lens L and a part is transmitted

which is reflected back from the plane surface of the plate. These two reflected rays interfere



and give rise to an interference pattern in the form of circular rings. These rings are localized

in the air film, and can be seen with a microscope focussed on the film.

Explanation of the formation of Newton’s rings. Newton’s rings are formed due to interference between the waves reflected from the top and

bottom surfaces of the air film formed between the plate and the lens. The formation of

Newton’s rings can be explained with the help of Fig. 2. AB is a monochromatic ray of light

which falls on the system. A part is reflected at C (glass-air boundary) which goes out in the

form of ray 1 without any phase reversal. The other part is refracted along C D. At point D it is

again reflected and goes out in the form of ray 2 with a phase reversal of . The reflected rays

1 and 2 can produce interference fringes as they have been derived from the same ray AB. As

the rings are observed in the reflected light, the path difference between them is (2 t cos r +

/2); For air film = 1 and for normal incidence r = 0.

Hence in this case, the path difference is (2 t + / 2).

At the point of contact, t = 0, and the path difference is / 2, which is the condition of

minimum intensity. Thus the central spot is dark. For nth maximum, we have 2t + /2 = n

This expression shows that a maximum of a particular order n will occur for a constant value

of ‘t’. Since ‘t’ remains constant along a circle the maximum is in the form of a circle. For

different values of t, different maxima will occur. In a similar way, this can be shown that

minima are also in the circular form.

Theory : Newton’s rings by reflected light Now we shall calculate the diameters of dark and bright rings. Let

LOL be the lens placed on a glass plate AB. The curved surface

LOL is a part of spherical surface (shown dotted in figure) with

centre at C. Let R be the radius of curvature and r be the radius of

Newton’s ring-corresponding to the constant film thickness t.

As discussed above 2 t + /2 = n ,

or 2 t = (2n -1) /2 for the bright ring where n = 1, 2, 3, etc.

And 2t = n for dark ring where n = 1, 2, 3, etc.

From the property of the circle

NP x NQ = NO x ND

Substituting the values

r x r = t x (2R - t) = 2Rt – t2 2Rt (approximately)

r2 = 2Rt or t = r2/2R

Thus for bright ring

2

λ12n

2R

r2

2

2

Rλ12nr2

Replacing r by D/2 we get the diameter of the nth bright ring as

2

Rλ12n

4

2

D

or D = 12nRλ2

Or D 12n

Thus the diameters of the bright rings are proportional to the square roots of odd natural

numbers as (2n - 1) is an odd number.

Similarly for a dark ring

D n

Thus diameters of dark rings are proportional to the square roots of natural numbers.



With the increase in ‘n’ the fringe width decreases but the area of the ring remains same.

(iii) the diameter of the n th bright ring Dn = 12nRλ2

Square of the diameter of the n th bright ring Dn2 = 12nRλ2

Square of the diameter of the (n+1) th bright ring Dn+12 = 12nRλ2

The difference in the diameters of the successive rings Dn+12 - Dn

2 = 4λR

since the above is independent of the number n, it is constant.

Newton’s rings by transmitted light

In case of transmitted light, the rings are just opposite to the rings in reflected light. The

central ring is bright .

The diameter of the nth bright ring D = nRλn2

The diameter of the nth dark ring D = 1n21n2Rλ2

23. Describe Newton’s rings experiment to determine the radius of curvature of a plano -

convex lens (June 2014, Dec 2016) or determination of wavelength of light.

The experimental arrangement is shown in figure. The microscope is

adjusted such that Newton’s rings are clearly seen. The centre of the

cross-wires is adjusted at the centre of the fringe pattern. The

microscope is moved to the extreme left of the pattern and the cross

wire is adjusted tangentially in the middle of a clearly nth bright or

dark fringe. The reading of micrometer is noted. The microscope is

now moved to the right and the readings of micrometer screw are

noted successively at (n -2)th , (n - 4) th rings etc. till we are very near

to the central dark spot. Again crossing the Central dark spot in the

same direction the readings corresponding to... (n -4)th , (n-2)th ,nth

rings are noted on other side. The diameters of different rings are found out. Now a graph is

plotted between number of rings n and the square of the corresponding diameter. The graph is

a straight line. From the graph we find the slope or nm

DD nm

22

If R is the radius of curvature of the plano convex lens then the wavelength of the

monochromatic light can be found using Rnm

DD nm

4

22

If we know the wave length of the light we can find the radius of curvature of the lens using

nm

DDR

nm

4

22

24. Explain what happens when the air in the inter space is replaced by a transparent

liquid in Newton’s rings method (June 2017)

25. Determine the refractive index of transparent liquid by using Newton’s ring

method. (June 2005)

When air in the inter-space is replaced by a liquid the diameters of the Newton’s rings

decrease. When air is present between the lens and a glass plate the diameter of the nth

bright ring Rλ12n2D

when a liquid of refractive index μ is introduced in the space the diameter of the nth

bright ring

Rλ12n2D`



Experiment to determine the refractive index of the given liquid.

The experimental arrangement is as shown. The diameters of the mth and nth dark rings are

measured as Dm and Dn. Now the given liquid is introduced between the lens and the glass

plate. The rings shrink. The new diameters D1m and D1n are measured. The refractive index of

the liquid can be calculated using the formula DD

DDμ

21n

21m

2n

2m

NEWTON’S RINGS WITH WHITE LIGHT

26. What will happen to Newton’s rings if (a) white light is used (b) the lens is lifted

slowly from the flat plate.

When monochromatic light is used, Newton’s rings are alternately dark and bright. In case of

white light, first few rings will be colored. Other rings cannot be viewed due overlapping of

rings of different colors.

b) As the lens is lifted slowly the central dark spot becomes bright and the diameters of the

rings increase. Finally the rings disappear.

27.Why the centre of Newton’s ring is black ? How can it be made bright. at the center the lens surface is in contact with the glass plate. Hence the effective path

difference for reflected rays is λ/2. Therefore the centre of Newton’s ring is black. It can be

made bright by providing a space of λ/2 with the help of dust particles.

Problem 9: In a Newton’s ring experiment, the diameter of the 5th ring was 0.3 cm and the

diameter of 25 th ring was 0.8 cm. if the radius of curvature of the plano convex lens is 100 cm,

find the wavelength of light used. [687x10-5 cm]

Hint: Rnm4

DDλ

2n

2m

Problem 10: A Newton’s rings arrangement is used with a source emitting two wave lengths

λ1 = 6x10-5 cm and 4.5 x 10 -5 cm and it is found that nth dark ring due to λ1 coincides with

(n + 1) th dark ring for λ2. If the radius of curvature radius of curvature of the curved surface is

90 cm, find the diameter of nth dark ring for λ1. [0.254 cm]

Hint: nλ1= (n + 1)λ2

Dn2 = 4nRλ1

Problem 11:In Newton's ring experiment the diameters of the 4th and 12th dark rings are

0.400 cm and 0.700 cm respectively. Find the diameter of the 20th dark ring. [0.906 cm]

412

1220

DD

DD24

212

212

220

Problem 12:Newton’s rings formed by sodium light between a flat glass plate and a convex

lens are viewed normally. What will be the order of the dark ring which will have double the

diameter of that of 40th dark ring. [160]

Hint: n

m

D

D2n

2m

Problem 13: In a Newton’s rings experiment the diameter of the 10th ring changes from 1.40

to 1.27 cm when a liquid is introduced between the lens and the plate. Calculate the refractive

index of the liquid. [1.125]

Hint: Dn2 =

μ

Rλn4

Problem 14: Light containing two wavelengths λ1 and λ2falls normally on a plano convex lens

of radius of curvature R resting on a glass plate. If the nth dark ring due to λ1 coincides with



(n + 1) th dark ring for λ2. If the radius of curvature radius of curvature of the curved surface is

R, find the radius of nth dark ring for λ1.

HINT: the radius of the nth dark ring r = Rn 1

but nλ1 = (n + 1)λ2

Or n = 21

2

Hence r = 21

21

R

Problem 15: If the diameter of two consecutive Newton’s rings in reflected light of

wavelength 5890 Ǻ are 2.0 and 2.02 cm respectively, what is the radius of curvature of the

lens surface in contact with plane glass surface ? [Ans. 341.2 cm]

Problem 16: In a Newton’s rings experiment, the diameter of the 5th ring was 0.336 cm and

the diameter of the 15th ring was 0.590 cm. Find the radius of curvature of the plano-convex

lens if the wavelength of light used is 5890 x 10 -8 cm. [Ans. 99.82 cm]

Problem 17: In a Newton’s rings experiment the diameter of the 15th ring was found to be

0.59 cm and that of 5th ring is 0.336 cm. if the radius of curvature of the lens is 100 cm find the

wavelength of the light (June 2005)[Ans. 588 nm]

Problem 18: calculate the thickness of air film at 10th dark ring when viewed in reflected light

of wavelength 500 nm. The diameter of the 10th ring is 2 mm. (2004)[2.5 µm]

Problem 19: In Newton’s rings experiment the diameter of the 10th ring is 0.5 cm. If

λ= 5900 Ǻ find the radius of curvature of the lens.[1.059 m](2005,2011)

Problem 20: Newton’s rings are formed by light reflected normally from a plano-convex lens

and a plane glass plate with a liquid between them. The diameter of the nth ring is 2.18 mm and

that of (n +10) th ring is 4.51 mm. Calculate the refractive index of the liquid, given that the

radius of curvature of the lens is 90 cm and the wavelength of light used is 5893 A. [Ans. 1.7]

Problem 21: In a Newton’s ring experiment the diameter of the 10th ring changes from 1.40

cm to 1.27 cm when a liquid is introduced between the lens and the plate. Calculate the

refractive index of the liquid. [Ans. 1.215]

MULTIPLE CHOICE QUESTIONS

Wave nature of light is evidenced

a) photoelectric effect b) interference

c) black body radiation d) nuclear emission ‘

1. Two sources are said to be coherent if their emitted waves have

a) same wavelength J b) same amplitude

c) constant phase difference d) all the three

2. When the light wave is reflected at the glass-air interface, the change of phase of the

reflected wave is equal to

a) 0 b)π/4 c) π/2 d)π

3. When the light wave is reflected at the air-glass interface, the change of phase of the

reflected wave is equal to .

a) 0 b)π/4 c) π/2 d)π

4. Two waves having their intensities in the ratio 9:1 produce interference. In the

interference pattern the ratio of maximum to minimum intensity is equal to

a) 2:1 b) 9:1 c) 3:1 d) 4:1



5. Two beams interfere have their amplitudes ratio 2: 1. Then the intensity ratio of bright

and dark fringes is

a) 2:1 b) 1:2 c) 9:1 d) 4:1

6. When a thin film of oil or soap bubble is illuminated with white light, multiple colours

appear. This is due to

a) diffraction . b) polarization c) total internal reflection d) interference

7. In Newton’s rings interference is due to light rays reflected from

a) lower surface of lens and upper surface of glass plate

b) lower surface of glass plate and upper surface of lens

c) lower surface of lens and lower surface of glass plate ‘

d) upper surface of lens and upper surface of glass plate

8. Choose the correct statement

a) In Newton’s rings experiment, when viewed through reflected light

the centre of the rings pattern is bright

b) In Newton’s rings experiment, when viewed through transmitted

light the centre of the rings pattern appears dark

c) In Newton’s rings experiment, when viewed through reflected light

the centre of the rings pattern is dark ,

d) In Newton’s rings experiment, when liquid of higher refractive index

is introduced in between the lens and the glass plate, the radius of the rings increases.

1. The resultant displacement due to the presence of both the waves of displacements yl and

y2 is given by y = y1 + y2. This is known as principle of ……………….(principle of

superposition)

2. Two waves are said to be………….. if their waves have same wavelength, same amplitude

and constant phase difference. (coherent)

3. If 2 : 1 is the amplitude ratio of the sources, the intensity ratio of bright fringe to dark fringe

is…………(9:1)

4. If the film thickness is extremely small when compared to λ, the film will appear

………… (dark)

5. Newton’s ring experiment, the radius of nth dark ring is given by ……….. where λ is the

wavelength of light and R is the radius of curvature of the lens. Rn

6. On introducing a liquid between the lens and the glass plate in Newton’s rings experiment,

the diameter of the rings observed ……(decreases)



2.

3. DIFFRACTION

1. What is meant by diffraction of light?

The phenomenon of bending of light into the geometrical shadow region is called

diffraction. This is shown by all kinds of waves, irrespective of their nature.

2. What are the types of diffraction and give the differences between them? (June

2005, June 2011)

Explain Fresnel and fraunhofer diffraction (May 2017) Diffraction phenomenon can be divided into following two general classes:

(1) Fresnel’s diffraction: In this class of diffraction, source and screen are placed at finite

distances from the aperture of obstacle having sharp edges. In this case no lenses are used for

making the rays parallel or convergent. The incident wave front are either spherical or

cylindrical.

(2) Fraunhofer’s diffraction: In this class of diffraction source and the screen or telescope

(through which the image is viewed) are placed at infinity or effectively at infinity. In this case

the wave front which is incident on the aperture or obstacle is plane. OR

Fresnel diffraction Fraunhofer diffraction

1 In this class of diffraction, source and

screen are placed at finite distances from

the aperture of obstacle having sharp

edges.

In this class of diffraction source and the

screen or telescope (through which the

image is viewed) are placed at infinity or

effectively at infinity.

2 The incident wavefront are either

spherical or cylindrical.

In this case the wave front which is

incident on the aperture or obstacle is

plane.

3 Source used is small in size Big or extended source is used

4 Lenses are not used Lenses are used

3. What is the difference between interference and diffraction? (June 2005)

4. Differentiate between interference and diffraction. (May 2007, June 2011)

5. Explain what is meant by diffraction of light. How diffraction is different from

interference? (June 2011)

DIFFERENCE BETWEEN INTERFERENCE AND DIFFRACTION Following are the differences between interference and diffraction phenomena:

interference diffraction

Interference is due to the interaction

between two separate wave fronts

originating from two coherent sources

Diffraction is due to the interaction

between the secondary wavelets

originating from different points of same

wave front.

In an interference pattern all the

maxima are of same intensity

In a diffraction pattern the intensity

decreases on either side of the central

maximum

In interference all fringes have equal In diffraction the central maximum is very



widths. wide. The width decreases on either side

In the interference pattern the regions

of minimum intensity are usually

almost perfectly dark

In the in diffraction pattern the regions of

minimum intensity are not perfectly dark.

6. a) Give the theory of Fraunhofer diffraction due to single slit and hence b)

obtain the condition for primary and secondary maxima. Using this c) obtain

intensity distribution curve (June 2006)

7. Obtain the condition for primary maxima in Fraunhofer diffraction due to a

single slit and d) derive an expression for width of central maximum (2005, 2011) briefly explain fraunhofer diffraction at single slit (May 2017)

a) FRAUNHOFER DIFFRACTION AT SINGLE SLIT

suppose AB is a narrow slit width ‘a’ and

perpendicular to the plane of the paper. Let a

plane wave front WW΄ of monochromatic light

of wave length λ propagating normally to the

slit be incident on it. Let the diffracted light be

focussed by means of a convex lens on a screen

placed in the focal plane of the lens. The

secondary wavelets travelling normally to the slit, i.e., along the direction OPo are brought to

focus at Po by the lens. Thus Po is a bright central image. This is called zero order maximum.

The secondary wavelets travelling at an angle θ with the normal are

focussed at a point P1 on the screen. In order to find out intensity at

P1, draw a perpendicular AM to the diffracted rays.

b) Suppose the slit is divided into two halves AO and OB. Suppose

the waves starting from the top of the two halves have a path

difference λ/2. These two waves interfere destructively producing

minimum. For every point in the first half there is a corresponding

point in the second half producing waves having a path difference

λ/2.

2

λθsin

2

a or a sinθ = λ ………………(1)

The path difference between secondary wavelets from A and B direction θ.

The rays travelling in this direction interfere destructively producing minimum called first

order minimum. The higher order minima can be obtained in the direction d sinθ = m λ

where m = 2,3,4,..

In addition to the central maximum there are secondary maxima which lie in between

the secondary minima on either side. Hence for secondary maxima a sinθ = (2m + 1) λ/2

It should be noted that secondary maxima do not fall exactly mid-way between two minima,

but they are displaced towards the centre of the pattern, of course, the displacement decreases

as the order of maximum increases.

Thus the diffraction pattern due to single slit consists of a

central bright maximum flanked by secondary maxima and

minima on both the sides.

c) Intensity distribution in diffraction pattern



Suppose the width of the slit is divided into n equal parts and the amplitude of the wave

from each part is ‘a’ (because width of each part is same). Suppose the phase difference

between any two consecutive waves from these parts would be ‘δ’

n

1

sinθ

λ

2πa = δ =

n

12α (say)

Using the method of vector addition of amplitudes, the resultant amplitude Aθ is given by

n

αsin

αsinAθ a

for large value of n , α/n is very small.

Hence

α

αsinAθ an

α

αsinAA oθ

The intensity in any direction is given by 2

oθα

αsinII

where Io is the intensity of the principal maximum at θ = 0.

Figure represents the intensity distribution. It is a graph of 2

oα

αsinI

(along Y-axis) as a

function of α or sin θ (along X-axis).

It can be seen that most of the light is confined to the central maximum. The intensity of the

secondary maxima falls off rapidly Io/22, Io/61 ..

d) Linear width of the principal maximum

Linear width of the principal maximum is distance between

the first order secondary minima on either side of the central

maximum. If x is the distance of the first secondary minimum

from the center then the width of the central maximum

W = 2x.

If ‘f ’ is the focal length of the lens used to focus the

diffraction pattern we have sinθ = x/f

But sinθ = λ/a.

Therefore a

λfxor

a

λ

f

x

Hence, the width of the central maximum is given by a

λf2x2 W

From the above it is clear that as the slit narrows the width of the central maximum increases

as shown.

8. How do you measure the slit width.

Measurement of slit width

The slit is illuminated with a laser beam and the diffraction pattern is obtained on a

screen placed at a distance one meter. The distance between the first minima on either

side 2x is measured. The slit width a can be measured using x

λfa

where f is the focal length of the lens or distance of the screen from the slit.

Problem 1: A slit is illuminated with light of wave length λ. find the angular width of

the central maximum if the width of the slit is 1) 4λ 2) 3λ 3) 2λ 4) λ.

Hint: a sin θ = λ

in the above if the screen is at a distance 2 m find the width of the central maximum.



Problem 2: A slit of width 1.5 mm is illuminated by a light of wavelength 500 nm and

diffraction pattern is observed on a screen 2 m away. Calculate the width of the central

maximum. [1.33 mm]

Hint: a

λD22x

Problem 3: A screen is placed 2 m away from a narrow slit. If the first minima lie at 5

mm on either side of the central maximum when light of wavelength 500 nm is used.

Find the width of the slit.[0.2 mm] Hint: a = λD/x.

Problem 4: A lens whose focal length is 40 cm forms a Fraunhofer diffraction pattern

of a slit 0.3 mm wide. Calculate the distances of the first dark band and of the next

bright band from the axis (wavelength of light used is 5890 Å [0.785 mm, 1.178 mm]

Hint: a

fx

1 and

a

fx

2

32

Problem 5: find the half angular width of the central bright maximum in the

Fraunhofer diffraction pattern of a slit of width 12x10-5 cm when the slit is illuminated

by light of wave length 6000Å.[30o]

Hint: a sinθ = nλ

Problem 6: Find the angular width of the central maximum in the Fraunhofer

diffraction using a slit of Width 1 μm when the slit is illuminated by light of wavelength

600 nm.(June 2006 ) [θ = 36.87o][2θ = 73.74o]

Hint: a sinθ = λ Problem 7: Calculate the angular separation between the first order minima on either side of

central maximum when the slit is 6 x10-4 cm width and light illuminating it has a wavelength

6000 Å

9. Explain with theory the Fraunhofer diffraction due to ‘N’ slits. (May 2003, June

2004, June 2011)

Fraunhofer Diffraction due to N slits (Diffraction grating)

Diffraction grating consists of very large number of narrow slits side by side and

separated by opaque spaces. The incident light is transmitted through the slits and

blocked by opaque spaces. Such a grating is called transmission grating. When light

passes through the grating, each one of the slit diffracts the waves. All the diffracted

waves reinforce one another producing sharper and intense maxima on the screen. In

practice a plane transmission grating is a plane sheet of transparent material on which

opaque rulings are made with a diamond point. The spaces between the rulings are

equal and transparent and constitute the parallel slits. The rulings are opaque and are of

equal width. The combined width of a ruling and a slit is called grating element.

Theory of plane transmission grating

Let ABC….H represent the section of grating normal to the plane of the paper. Let the

width of each slit be ‘a’ and that of opaque ruling be ‘b’. Now (a + b) which is the

combined width of a ruling and a slit is called grating element. It is also the distance

between two successive slits. Any two points on successive slits separated by a distance

(a + b) are called corresponding points. Let a plane wave front be incident normally on

the grating. The points in the slits act as secondary sources of light giving rise to

secondary waves. These waves spread in all directions on the other side of the grating.

These waves are brought to focus on a screen with the help of a lens. The secondary



waves travelling in the same direction as that of the incident wave are focused at Po.

Since all these secondary waves have travelled equal distance to reach Po, they

reinforce constructively and hence the point Po is the position of central bright

maximum.

Now let us consider secondary waves travelling at an angle θ with the direction of

incidence and reaching P1. The intensity at P1 depends on the path difference between

the secondary waves originating from the corresponding points of two adjacent slits.

Since the distance between corresponding points is (a + b) the path difference is

(a + b) sinθ.

The intensity at P1 will be maximum if (a + b) sinθ = nλ

Intensity distribution

Suppose the waves from the grating reach a point on the screen.

Suppose the phase difference between two waves form the edges of a slit is given by

sinθ

λ

2πa = 2α (say)

and suppose the phase difference between two waves form the adjacent slits is given by

sinθ)(

λ

2πba = 2β (say)

The intensity at any point is due to diffraction as well as interference. It is given as 22

βsin

sinNβ

α

sinαAI

The factor α

sinαA2

gives the distribution of intensity due to diffraction at single slit

while the factor 2

βsin

sinNβ

gives the distribution of intensity as a combined effect of all

slits.

Principal maxima: the most intense maxima are called principal maxima. They are

obtained for (a + b) sinθ = ± nλ where n = 0,1,2,3…..

The intensity of principal maxima is 22

Nα

sinαAI

Minima: in between any two principle maxima there will be (N - 1) minima.

Secondary maxima: as there are (N - 1) minima there will be (N - 2)

9. Calculate the maximum number of orders possible for a plane diffraction grating.

(June 2011)

The principal maxima in a grating satisfy the relation (a + b) sinθ = nλ or

λ

θsinban

The maximum angle of diffraction can be 90o.

Hence the maximum possible order is given by

λ

ba

λ

sin90ba(n)

o

max

11. Find the missing order in a grating spectrum

In a grating spectrum the total intensity is controlled by diffraction at a single slit

and interference due to two adjacent slits.

If the intensity due to any one cause is zero the total intensity at that point will be

zero.



Suppose at a point we nth principal maximum satisfying the relation

(a + b) sinθ = nλ …….(1)

Suppose at the same point we get mth diffraction minimum satisfying the relation

a sinθ = mλ …….(2)

If both the conditions (1) and (2) are satisfied simultaneously then nth diffraction order

will be missing. Dividing equation (1) by (2) we get

m

n

sinθa

sinθb)(a

or

m

n

a

b)(a

…….(3)

12. Explain experimental method of determination of wavelength of Spectral lines

of a given source of light using plane transmission grating.

13. Explain with theory how wavelength of spectral line is determined using Plane

diffraction grating (M 2003, J 2004)

what is resolving power. What is its formula?

The resolving power of a grating is defined as the capacity to form separate diffraction

maxima of two wavelengths λ1 and λ2which are very close to each other. It is given by

λ/dλ where λ is the average wavelength and dλ is the difference in wavelengths. For a

grating it is given as Nndλ

λ where N is the total number of lines on the grating.

The resolving power of a grating depends on the total number of lines on the grating

and the order of the spectrum. Examine if two spectral lines of wavelengths 5890Å and 5896Å can be clearly resolved in

the 1) first order 2) second order by a diffraction grating 2 cm wide and having 425

lines/cm (May 2017)

Hint: Nndλ

λ [the lines will be clearly resolved in second order]

What is dispersive power. What is its formula?

MULTIPLE CHOICE QUESTIONS

10. The Penetration of waves into the regions of the geometrical shadow Is

a) interference b) diffraction c) polarization (1) dispersion

11. In a Single slit diffraction, the first diffraction minima is observed at an an

angle of 30o, when the light of wavelength 500 nm is used. The Width of the slit is

a) 5x10-5cm b) 2.5x10-5 cm . c) 10 x 10-5cm d) 1.5x10-5 cm

12. In Fraunhofer diffraction the wavefront undergoing diffraction has to be

a) spherical b) cylindrical c) elliptical d) plane

13. In a single slit experiment if the slit width is reduced

a) the fringes becomes brighter

b) the fringes become narrower

c) the fringes become wider

d) the colour of the fringes change

14. Instead of red colour source, if blue colour source is used in single slit experi

ment

a) the diffraction pattern does not change

b) the diffraction bands become wider



c) the diffraction pattern becomes narrower and crowded together

d) the diffraction pattern disappears .

15. The diffraction pattern of a single slit consists of

a) wider dark band at the center with alternate bright and dark bands on either side.

b) narrow bright band at the center with alternate dark and bright bands of equal intensity on

either side .

c) wider bright band at the center with alternate dark and bright bands of equal intensity on

either side .

d) wider and brighter band at the center with alternate dark and bright bands of decreasing

intensity on either side.

16. A parallel beam of monochormatic light falls normally on a plane diffraction grating

having 5000 lines/ cm. A second order Spectral line is diffracted through an angle of 30°. The

wavelength of light is

a) 5 x 10 -7 cm b) 5 x 10-6 cm C) 5 x10-5 cm d) 5 x 10-4 cm

/

. . . urn lies 5 : Th W d . m e 1 th 0f Slit Is on elthel' Side Of the central maxrmum % < a)

0.024 mm b) 0 x c) 2.4 mm ‘ d) 242:1? r

a) 1 pm b) 10 ”In C) 2 5 pm d) 25 um 1‘

20. When white light is incident on a diffraction grating, the light diffracted more

a) blue b) yellow c) violet (1) red

21. Monochromatic light falling normally on a grating gives rise to diffracted second order

beam at angle 30°. If the grating has 5000 lines / cm, the wavelength of light is

a) 600 nm ' b) 400 nm c) 500 nm d) 650nm

22. Maximum number of orders possible with a grating is

a) independent of grating element

b) directly proportional to grating element.

c) inversely proportional to grating element.

d) directly proportional to wavelength.

23. Polarization of light conclusively proves that

a) light waves are longitudinal

b) light waves are transverse

c) light waves are longitudinal as well as transverse.

d) light travels with velocity3x108 m/s

24. The phenomenon causing polarization of light is a) interference b) diffraction c) double

refraction d) refraction

25. A calcite crystal-is placed over a dot through the calcite one finds

a) one dot. b) two dots which are stationary

c) one dot rotating about the other. d) two rotating dots.

26. In a doubly refracting crystal, along optic axis

a)μo > μe b) μo < μe c) μo = μe d) none

Fill in the Blanks

12, With decrease of slit width in single slit, the width of the fringes We



13 In a grating, the combined Width of a ruling and a slit is called ,4“:

14 Points on successive slits separated by a distance equal to the grating element.

are called

15. In Fraunhofer diffraction at a plane transmission grating, when white light source is used,

the angle of diffraction for violet is , , than that of red.

16. The phenomenon which confirms transverse nature of light is

17. If the intensity of polarized light passing through rotating analyser falls to zero, . then the

light is polarized.

18. The angle of incidence of light for which the reflected beam 1s completely plane polarized

IS known as

19. The polarizing angle for glass is,

20. When unpolarized light passes through certain crystals, two refracted beams are produced.

This phenomenon is called

21In Nicol prism for A 2 5893A, the refractive index of Canada balsam cement is

22. In Nicol prism the refractive index for e-ray varies between ' and

slate True or False

1Two monochromatic sources of same amplitude and some wavelength ”can act as coherent

sources. ‘ _ True / False

2In Young’ 5 double slit experiment, if one of the slits lS closed the fringe contrast decreases.

. ' ' True/False

3If the path difference between the two interfering waves is integral multiple of i v\,

constructive interference occurs. . True/ False

1In Newton’s ring experiment with reflected light the point of contact of lens With the glass

plate where thickness of air is zero, appears dark. True/ False

i. Radius of 4“ dark ring in Newton’s ruing experiment is 4 EA. True/ False

RThinfilmswhenviewedund hit li t, ul et ." persion. er W e gh appears m l 3 _ 06 l

7. For diffraction to occur, the size of the obstacle must be comparable with wave length. True

/ False

B. Converging lens is used to focus parallel rays in Fresnel diffraction. True / False

9. The source and screen are at infinite distance from the obstacle, producin



Fraunhofer diffraction. True / FalsN 10. Diffraction is due to superposition of secondary

waves originating from diffek ent parts of the same wavefront. True/PaleQ 11. In diffraction

all bright fringes have the same intensity. True/False 12. In diffraction, intensity of dark

fringes is not zero. ~ True/Fallse

13. In single slit Fraunhofer diffraction, the fringe width is inversely proportional to the slit

width. . True/False

14. In Praunhofer diffraction at a double slit, the angular separation between two consecutive

maxima decreases if the slits are narrower and closer. True / False

15. Dispersion spectra obtained in prisms are called standard spectra. True/False

16. According to Brewster’s law the refractive index of the material medium is

equal to the sine of the angle of polarization. True / False 17. Nicol prism is used to produce

dispersion of White light. True/ False

18. On rotating the analyser, if the intensity of transmitted beam does not vary, I . then the

light analyzed may be circularly polarized or unpolarized. True / False

19. The three dimensional movies seen in the theaters are due to' polarization of light. _ True

/Fa1se

20. In the path of light, when an analyzer is inserted and rotated, if the intensity of transmitted

beam varies between maximum and minimum but not zero then the light analyzed may be

unpolar‘ized or circularly polarized. True / False

Pan‘ A Questions and Answers

1, State principle of superposition of waves.

When two or more waves travel simultaneously in a medium the resultant displacement at any

point is due to the algebric sum of the displacements due to individual waves. This is the

principle of superposition.

i.e. y 2 3/1 i .712

+ve sign has to be taken when both the displacement y] and 3/2 are iIl Phase with each other;

~ve sign has to be taken whey they are out of phase.

Optics 7.6’l‘

Define coherent waves or what are coherent sources? ' 5; 2 /I‘wo waves are said to be coherent

if they emit waves with i) same wavelength ' ii) same amplitude and ' , . .v iii) constant phase

difference Can two independent monochromatic light sources be coherent?

' Two independent monochromatic light sources cannot be coherent. E‘ven two different parts



of a single monochromatic source cannot act as’ coherent sources. Instead either waves from

two points on the same wavefront or two rays split from a single parent ray can act as coherent

sources. .

4 What is an interference pattern? Or Explain the phenomenon of interference of light.

If two waves of same amplitude a interfere, at regions of constructive interference the intensity

is 4a2 while at regions of destructive interference the intensity is zero. Such a modification of

intensity due to superposition of coherent waves is known as interference pattern. ‘

5_ Mention the condition for sustained interference pattern.

When the waves from two coherent sources interfere and if their path difference is less than

the coherent length, we get sustained interference pattern.

‘6. State the condition for formation of bright and dark interference fringes.

When the path difference between the two interfering beams is either zero or ‘ kintegral

multiple of X (i.e. even multiple of A/2) constructive interference occurs. ' ‘

Hence for brightness, the path. difference

_6=n}\ or 271(52

When the path difference between the two interfering beams 1s odd multiple of A / 2

destruc'tive intereference occurs. '

) where n = 0,1,2,3 etc.

Hence for darkness, the path difference

6 = (2n + 1))./2 where n :2 0,1,2,3 etc. . (or) = (271 1))\/2 where n =2 1,2,3 etc.

7‘ in Young's double slit experiment, what is the expression for bandwidth?

I Bandwidth or fringe spacing d~ é;

)e is the wavelength of light

0 is the diStanCE between the double slit and screen and g d is the separation between the two

slits. I 3 .3 0 CI) J l 7

8' What type of fringes are obtained in Young’s double slit experiment Whe

white light is need as source?

At the center of the fringe pattern, Where the path difference is zero Whita bright fringe is

seen. On either side a system of coloured bands are seen.



In air-wedge experiment, straight line interference fringes are seen while in Newton ring

experiment circular rings areseen. Why?

In air-wedge experiment the air film is formed between two optically flat glass plates. Hence

the loci of points of same thickness of air-film form straight lines and hence straight line

fringes are seen.

In Newton’sring experiment; the air-film‘is formed between a convex lens and a glass plate.

Hence the loci of points of same thickness of air-film form circular rings and hence circular

fringes are seen.

Extremely thin film appear dark, when viewed under monochromatic/white light. Why? ' .

i The phenomenon of bending of Ii h ding of light ane ~ g t round the spfea , . . 5 Into the .

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tag. zero and‘ also increases with in“i; ii

L . ‘ . I l: Slty1 . m crease Of order. __NL&‘fd&idg as” $4”; iwigngg i l 1” g :03 a. W MW?“

“M“ W Wmmyemwiwawww “1W? . ft W W I . ? 16. What is plane diffractlon gratmg h t of

transparent material on which . . ee . . ' S a lane 5 ' A plane transmisswn grating; a clliamond

point. The rulings are opaque and W1

opaque rulings are made are of equal width. The spage , . and ”Mime the Parallel SW" fine

linearly polarized light.



What is meant by polarization 0 . . i atio ire ction

332706330114:

5 between the rulings are e

The intensity of dark fringes is' not Q.

1 .

Define diffraction.

13' The phenomenon of bending of light round the sharp corners of obstacles and Preading of

light waves into the geometrical shadow of the obstacle placed in

313 path of light IS called diffraction.

pistinguish between Fresnel and Fraunhofer diffraction.

14 F 1 ff ‘_ W; w ~.. V” m m ~ :1" h $~mkszikui .1 W“??? , ‘ WM resne Di raction my 1 .1».

gm“. _ Fraunhofer Diffraction A?“ g

1 1. Either a point source or an illumi3 Extended source at infinite disi“

1 nated narrow slit is used. 1 tance is used. it?

1

l

I 2, :The wavefront undergoing diffracF The wavefront undergoing {. . tion is either spherical

or cylindridiffraction is a plane wavefront. E‘

> ‘, cal. “ 3, The source and the screen are at The source and the screen’are at ‘ 1 finite

distances from the obstacle infinite distances from the obsta'

“ producing diffraction. cle producing diffraction.

l

4. a No lens is used to focus the rays g Converging lens is used to focus .» sarallelravs..

15 DistingUiSh between interference and diffraction fringes.

..>

Interference “m Diffraction '5 1. superposition is due to two sepSuperposition is due to



secondary arate wavefronts originating from wavelets originating from different two coherent

sources. parts of the wavefront from a single psource. ‘

2. The fringes normally have equal 1 The width between fringes is never

widths. equal. ‘ 3. All the bright fringes have the same The intensity of bright fringes usu~

intensity. ally decreases with increase of ' 1 order. 1

l

4, All the dark fringes have zero intenThe intensity of dark fringes is'not . sity. ‘ zero and‘ also

increases with in‘ crease of order. ‘

16. What is plane diffraction grating?

A plane transmission grating is a plane sheet of transparent material on which opaque rulings

are made with a diamond point. The rulings are opaque and are of equal width. The spaces

between the rulings are equal and transparent and constitute the parallel slits.

1'7 What' 18 meant by polarization of light (or) Define linearly polarized light.

in em. wave the electric vector executes periodic oscillations in a direction perpendicular to

the direction of propagation. If such oscillations are confined to a single plane h then it is

called plane polarized light or linearly polarized

light. What is circularly polarized light?

In em. wave if the electric vector of constant amplitude no longer oscillateg but rotates While

proceeding in the form of a helix, then it is called circularly polarized light. '

What is elliptically polarized light?

In em. wave if the electric vector no longer oscillates but rotates while proceeding in the form

of a helix and if the amplitude of the electric vector is not a constant but varies periodically

then it is called elliptically polarized light.

What is meant by partially polarized light?

If the linearly polarized light contains small additional components of unpolarized light, then it

is called partially plane polarized light. I

Define Brewster’s angle.

When unpolarized light is reflected at the surface of some transparent medium, the reflected

light is found to be partially plane polarized. The degree of polarization changes with angle of

incidence. The angle of incidence for which the reflected beam is completely plane polarized

is knoWn as polarizing angle or angle of polarization. This angle is known as Brewster’s

angle. ‘



State Brewster’s law. 9.,

The refractive index of the material medium a is equal to the tangent of the angle of

polarization z'p.

p = tan 2p

. What is pile of plates?

When unpolarized light is incident at polarizing angle the reflected right is completely plane

polarized and transmitted light is partially plane polarized. The transmitted light contains a

greater proportion of light vibrating parallel to the plane of incidence. If the process of

reflection at polarized angle is repeated using number of plates all inclined at polarizing angle,

finally the transmitted light becomes purely plane polarized. Such an arrang 1) t i0 kno V7 p]

A of plates. . ‘ (x

State Malus law.

Malus stated that the intensity of the polarized light transmitted through the analyser varies as

the square of cosine of the angle between the plane of transmission of the analyser and the

plane of polarizer.

i.e. I1 : Io cos2 0

. Write note on polarization by scattering.

When light beam passes through a medium consisting of very small particles, light is

scattered. If the dimensions of the particles are smaller than wavelength of light, then the

intensity of scattered light is found to be inversely proportional to the fourth power of

wavelength i. e., I or 1/ A4 In a direction perpendicular to the direction of propagation of light

the scattered light 15 found to be plane polarized.

Write a note on polarization by selective absorption.

’ . m nlight passes 1aroug‘h crystals such as tourmaline it is split into two components, the

horizontal and vertical. The horizontal component is being com

\‘ pletely absorbed by the crystal and hence vertical component alone comes out.

27.

The light emerging from crystal, therefore, is linearly polarized. This 15 called

polarization by selective absorption.

What are polaroids?



Crystals of quinine sulfate per iodide called herapathite also exhibit polarization by selective

absorption but as these crystals are very small, they cannot be used for polarization as such.

However, these crystals are aligned in a solid matrix and efficient and economical polarizing

material Polaroid is produced. The most common type of a polarizer used today is H-sheet.

This has a matrix of parallel chains of the synthetic material polyvinyl alcohol (PVA, -C'H2 -C

H OH --), impregnated with ioome. The conduction electrons of iodine move along the chains

and absdrb the electric field oscillating parallel to them; "the plane of transmission is normal.

K-sheet is a polyvinylene pol arizer. This is similar to H-sheet but more resistant to high

temperature and humidity.

28. What is meant by double refraction?

We know that unpolarized light has two components one vertical and another horizontal.

When unpolarized light passes through certain anisotropic crystals such as Calcite or quartz,

velocity of propagation of these two components vary. This means that the material exhibits

two different refractive indices. Since μ = (sin i/ sin r), though both the components have the

same angle of incidence, they have different angles of refraction. This is called double

refraction.

Part -B Questions

12. (a) Discuss the theory of Newton’s rings with relevant diagram (b) Explain how Newton’s

rings are formed in the reflected light.



(c) Derive the expressions for the diameters of dark and bright rings. (June 2005)

13. Discuss about Newton’s rings experiment and deduce the equation for the calculation of

radius of curvature of plano-convex lens. (June, 2014)

14. Determine the refractive index of transparent liquid by using Newton’s ring method. 1

(June 2005)

15. What is meant by diffraction of light? Explain it on the basis of Huygen's wave theory

(June 2005)

16. What are the types of diffraction and give the differences between them? (June 2005, June

2011)

17. How the fringes are obtained in diffraction and why they are unequally spaced? (June

2004 )

18.What is the difference between interference and diffraction?

(June 2005)

19 Differentiate between interference and diffraction. (May 2007, June 2011)

20 Explain what is meant by diffraction of light. How diffraction is different from

interference? (June 2011)

21. Obtain the condition for primary maxima in Fraunhofer diffraction due to a single slit and

derive an expression for width of central maximum(June 2005, June 2011)

22. Give the theory of Fraunhofer diffraction due to single slit and hence obtain the condition

for primary and secondary maxima. Using this obtain intensity distribution curve.(June 2006)

23. Explain with theory how the wave length of spectral line is determined using Plane

diffraction grating (May 2003, June 2004)

24. Explain with theory the Fraunhofer diffraction due to ’n’ slits.

(May 2003, June 2004, June 2011)

25. Calculate the maximum number of orders possible for a plane diffraction grating

. (June 2011)

26. Explain experimental method of determination of wavelength of Spectral lines of a given

source of light using plane transmission grating. ‘

Exercise Problems

1. Two coherent sources whose intensity ratio is 36:1 produce interference fringesDeduce the

ratio of maximum intensity to minimum intensity.

(June 2008) (49: 25)

3. A thin equiconvex lens of focal length 4 m and refractive index 1.50 rests on and in contact

with an optical flat surface, and using light of wavelength 546 nm Newton’s rings are viewed

normally by reflection. What is the diameter of the 5th bright ring? (0.627x10-2 m)

4. In Fraunhofer diffraction pattern due to a single slit, the screen is at a distance of 100 cm

from the slit and the slit is illuminated by monochromatic light of wavelength 5893Å. The



width of the slit is 0.1 mm. Calculate the separation between the central maximum and the first

secondary minimum. (5. 893 mm)

5. In double slit Fraunhofer diffraction, calculate the fringe spacing on a screen 50 cm away

from the slits, if they are illuminated with blue light (λ = 480 nm). Slit separation b = 0.1mm

and slit width a = 0.020 mm. (2mm).

6. A grating has 6000 lines per cm. Find the angular separation of two yellow lines of mercury

of wavelengths 5770 Å and 5791 Å in second order (11΄)

7. A beam of monochromatic light of wavelength 546.1 nm incident normally on a grating

produces first order diffracted image 18°50΄ from the direction of incidence. Calculate the

number of lines per unit length of the grating.

(5911 lines/cm)

8. For a glass plate of refractive index μ = 1.54, calculate the angle of polarization and also

angle of refraction. (57o,33΄)

Coherence wedge

Message body

'

Interference: Coherence, division of amplitude and division of wave front, interference in

thin films (transmitted and reflected light), Newton’s rings experiment.

Diffraction: Distinction between Fresnel and Fraunhoffer diffraction, diffraction due to

single slit, N-slits, Diffraction grating experiment.

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I-I, PHYSICS QUESTION BANK WITH ANSWERS UNIT I … · 2017-11-23 · nawab shah alam khan college of engineering & technology 1 prof. s.m.asadullah nsakcet i-i, physics – question