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Interference

1 Q. Discuss the theoretical details of Superposition principle to de-

rive the condition of interference maxima and minima

1.1 Principle of Superposition

When a number of waves traveling through a medium super-

impose on each other, the resultant displacement at any point

at a given instant is equal to the vector sum of all the displace-

ments due to individual waves at that point.

If y1,y2,y3,....ynbe the displacement due to different waves

acting separately then according to the principle of superposi-

tion, the resultant, when all the waves act together is given by

the vector sum,

~y=~y1+~y2+~y3+.... +~yn

Let us consider two waves of same wavelength i.e., the same

value of ωtravelling in the same direction given by

y1=a1sinωt(1)

y2=a2sin(ωt+φ)(2)

where a1and a2are the amplitudes of the two waves, φis the

constant phase difference between the two waves given by

φ=2π

λx,x is the path difference between two waves

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By the principle of superposition, the resultant displacement

at any point M is

y=y1+y2

=a1sinωt+a2sin(ωt+φ)

=a1sinωt+a2sinωt.cosφ+a2cosωt.sin φ

or, y= (a1+a2cosφ)sinωt+a2sinφ.cosωt(3)

Put

a1+a2cosφ=Acosθ

and a2sinφ=Asinθ

Then,

y=Acosθ.sinωt+Asinθ.cosωt

or, y=Asin(ωt+θ)(4)

Thus, the resulting wave is also a harmonic wave of amplitude

A and initial phase θ. The amplitude can be determined as:

A2=A2cos2θ+A2sin2θ

= (a1+a2cosφ)2+ (a2sinφ)2

or, A2=a2

1+a2

2+2a1a2cosφ(5)

∴A=qa2

1+a2

2+2a1a2cosφ

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We know,

Intensity of light ∝amplitude

Thus, intensity of resultant wave is

I=kA2

and intensity of individual waves are

I1=ka2

I2=ka2

where k is a proportionality constant.

So, eqn (5) can be written as

kA2=ka2

1+ka2

2+2ka1a2cosφ

or, I=I1+I2+2√I1I2cos φ(6)

This gives the total intensity a point M, where the phase dif-

ference is φ

1.2 Maxima

The point having maximum intensity is called maxima. For

maximum resultant intensity

cosφ=1

or, φ=0,2π,4π,....

or, φ=2nπwhere n = 0,1,2,...

Also, φ=2π

λx

or, 2nπ=2π

λx

or, x=nλ

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Hence, the resultant intensity at a point is maximum when the

phase difference between the two superposing wave is an even

multiple of πor path difference is an integral multiple of wave-

length λ. This is the condition of constructive interference.

Here

Imax =I1+I2+2√I1I2

i.e., Imax >(I1+I2)

1.3 Minima

The point having minimum intensity is called minima. For

minimum resultant intensity

cosφ=−1

or, φ=0,3π,5π,....

or, φ= (2n+1)πwhere n = 0,1,2,...

Also, φ=2π

λx

or, (2n+1)π=2π

λx

or, x= (2n+1)λ

Hence, the resultant intensity at a point is minimum when the

phase difference between the two superposing waves is an odd

multiple of πor path difference is an odd multiple of λ

2. This

is the condition of destructive interference.

Here,

Imin =I1+I2+2√I1I2

i.e., Imin <(I1+I2)

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Comparison of Intensities at maxima and minima

amplitude at maxima =a1+a2[from (5)]

amplitude at minima =a1−a2

Therefore, ratio of intensities at maxima and minima is

Imax

Imin =(a1+a2)2

(a1−a2)2=(a1/a2+1)2

(a1/a2−1)2

=r+1

r−12

where r=a1

a2=qI1

2 Q. What do you mean by coherence? Explain the difference be-

tween spatial and temperal coherence

A wave is said to be perfectly coherent if it is purely sinusoidal

for inﬁnitely large space as well as time. In such a wave there

is a deﬁnite relationship between the phase of the wave be-

tween two points of time or two points of space.

In practice, no source emits a perfectly coherent light wave.

The wave are partially coherent i.e., they are pure sine wave

only in a limited space or for a limited period of time. There

are two criteria of coherence of waves.

1) The criteria of time gives rise to temporal coherence

2) The criteria of space gives rise to spatial coherence

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2.1 Temporal Coherence

In a perfect coherent light wave, the oscillating electric ﬁeld

E has a constant amplitude but its phase varies linearly with

time. It is an ideal sinusoidal wave.

The light emitted by actual source is sinusoidal for a short in-

terval of time (0.1 ns), after which the phase changes rapidly.

The average tme interval for which the ﬁeld remains sinu-

soidal, is called temporal coherence. It is denoted by τ.

If c be the speed of light, then

cτ=L,coherence length of the light beam

2.2 Spacial Coherence

The spacial coherence is the phase relationship between the

electric ﬁeld and different points of space.

Let S be the source emitting light waves. The phase relation-

ship between P and Q depends on the distance PQ and also on

the temporal coherence of the beam.

There will be high coherence between P and Q if PQ << L

and no coherence if PQ >> L.

Also, if point R and P be two points equidistance of S. Then,

the waves from point source S reach R and P exactly in same

phase and hence will have perfect spatial coherence. if the

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source S is an extended source, R and P will not have spatial

coherence.

3 Purity of Spectral line

The width of a spectral line is given by

λ=λ2

t=λ2

cτ

For a perfect pure line 4λ=0, i.e., τ=∞

4 Discuss the characteristics of interference by the division of wave-

front and amplitude

The phase relationship between the waves emitted by two in-

dependent sources rapidly changes with time and therefore

they can never be coherent, though the sources are identical

in all aspects. However, if two sources are derived from a sin-

gle source by some device, then the phase difference between

the waves emerging from the two sources remains constant

and the sources are coherent. The techniques used for cre-

ating sources of light can be divided into the following two

broad classes.

4.1 Division of Wavefront

One of the method consists in dividing a light wavefront, emerg-

ing from a narrow slit. This can be done by illuminating two

slits from a single slit or using two virtual sources of a narrow

source or a narrow source with its virtual source to produce

interference. This is known as interference due to division of

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wavefront.

In this case, a wavefront is divided into two parts by utiliz-

ing the phenomenon of reﬂection, refraction or diffraction in

such a way that after travelling slightly different optical path,

the light from two coherent sources so produced are super-

imposed to produce interference pattern.Young’s double slit,

Fresnel’s Biprism, Lloyd’s mirror etc employ this technique.

These techniques requires a narrow source.

4.2 Division of Amplitude

The amplitude (intensity) of a light wave can be divided into

two parts by partial reﬂection at a surface. The two reﬂected

and transmitted components (beam) travel through different

paths and reunite to produce interference patterns. This is

known as interference due to division of amplitude. Interfer-

ence in thin ﬁlm, Michelson’s interferometer etc utilizes this

techniques. This techniques required extended sources.

5 Interference by division of Wavefront

5.1 Q. What are the characteristics of coherent sources? Describe Young’s double

Slit method to determine the wavelength of light source

5.2 Fresnel’s Biprism: Q. How are coherent sources created in Fresnel’s Bi-prism

experiment? Describe with necessary theory, how the wavelength of light is

determined by such bi-prism experiment

Fresnel used a biprism to show interference phenomenon. The

biprism consist of two prisms of very small refracting angles

joined base to base. In practice, it is constructed as a prism

with an obtuse angle of about 179◦and two sides of the angle

of the order of 300. A biprism creates two virtual sources S1

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and S2as shown in the ﬁgure. These two sources are images

of the source S produced by refraction and are hence coherent.

A vertical slit S is taken. A monochromatic light source such

as sodium vapour lamp illuminates the slit S. Therefore, the

slit S acts as a narrow linear monochromatic light source. The

biprism is placed in such a way that its refracting edge is par-

ralel to the length of the slits S. A single cylindrical wavefront

impinges on both prisms.

The top portion of the wavefront is refracted downward and

appears to come from the virtual source S1. The lower seg-

ment, falling on the lower part of biprism, is refracted upward

and appears to come from the virtual source S2. The virtual

sources S1and S2are coherent, and hence interference pattern

is formed on the screen that can be observed using a microm-

eter (eyepiece).

Suppose the distance between S1and S2be d. If the screen

is placed at O, interference fringes of equal width are formed

between E and F. But beyond E and F, fringes of large width

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is produced due to diffraction. MN is a stop to limit the rays.

The point O is equidistant from the sources S1and S2. There-

fore, the central bright fringe of maximum intensity is formed

at O. On both sides of O, alternate bright and dark fringes are

formed.

The width of dark or bright fringe is given by

β=λD

d(1)

where D= (a+b)is the distance of the source from the eye-

piece.

The bright fringes are formed on the screen at distances from

O given by

x=nλD

dwhere n = 0,1,2,3, .....

and the dark fringes are formed on the screen at a distance

from O given by

x=(2n+1)λD

2dwhere n = 0,1,2,3, .....

5.2.1 Determination of wavelength of light

The wavelength of light can be determined by measuring β, D

and d in relation (1)

1) Measurement of β: The vertical cross-wire of eyepiece is

set to the ceentre of any bright fringe. The eyepiece is moved

slowly along the screen and the number of the bright fringes

N that passes across the cross-wire is counted. Micrometer is

used to calculate the distance covered by the eyepiece to pass

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through N bright fringes. Thus, fringe width is given by

β=distance covered

no. of fringes(N)

2) Measurement of distance (d): A convex lens of short focal

length is placed between the slit and the eyepiece without dis-

turbing their positions. The lens is moved back and forth near

the biprism till a sharp pair of images of the slit is obtained

in the ﬁeld of view of the eyepiece. Let the distance between

two images be d1

If u is the distance of the slit from lens and v is the distance of

the eyepiece from lens, then,

u=d1

d(2)

The lens is then moved to a position nearer to the eyepiece,

where again a pair of images of the slit is seen. Let the dis-

tance between the two images be d2. Then,

v=d2

d(3)

Multiplying (2) and (3), we get

d2d1

d2=1

or, d=pd1d2(4)

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Now,the distance D betwen the slit S and the screen is mea-

sured. Using the value of β, d and D in the equation

β=λD

the wavelength λcan be determined.

5.3 Q. Explain what happens when white light is used instead of monochromatic

light in producing fringes in Fresnel’s biprism

If the slit is illuminated by white light, the interference pat-

tern consists of a central white fringe ﬂanked on its both sides

by a few coloured fringes. Moreover, the fringes obtained in

the case of biprism using white light are different from fringes

obtained with monochromatic light. The two coherent sources

in biprism is produced by refraction and the distance between

the two sources depends upon refractive index and thus wave-

length of light. So, distance between two sources is different

for blue light (db)to red light (dr). The distance of nth bright

fringe from the central fringe is

x=nλD

So, the nth bright fringe for blue and red light will be different

xb=nλbD

and xr=nλrD

5.4 Lloyd’s Mirror: Q. How interference pattern can be obtained through a com-

bination of a real and virtual sources

The Lloyd’s mirror consists of a plane mirror about 30cm in

length and 6 to 8 cm in breadth. it is polished on the front sur-

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face and blackened at the back to aviod multiple reﬂections.

A cylindrical wavefront coming from a narrow slits S1falls on

the mirror which reﬂects a portion of the incident wavefront,

giving rise to a virtal image of the slits S2.

The light directly coming from S1interfere with the light re-

ﬂected from the mirror. The slit S1and S2acts as two coher-

ent sources. Interference between direct and reﬂected waves

occurs within the region of overlapping of the two beams and

fringes are produced on the screen placed at a distance D from

S1in the shaded portion as shown below:

The point O is equidistance from S1and S2. So, central bright

fringe is expected to lie at O. But it is not usually seen as the

point O lies outside the region of interference.

But the point O can be brought in the region of interference

by moving the screen nearer to the mirror such that it comes

into contact with the mirror. The central fringe at point O is

then expected to be bright but it is observed dark.

The occurence of dark fringe at O is due to the phase change of

πthat light suffers when reﬂected from the mirror. The phase

change leads to a path difference of λ/2 and hence destructive

interference occurs there.

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Consequently, if the path difference at any point P on the

screen is nλ

i.e., S2P−S1P=nλwhere n=0,1,2,3,...

dark fringes are formed at that point (destructive interference)

and if the path difference at point P on the screen is (2n+

1)λ/2

i.e., S2P−S1P= (2n+1)λ/2 where n=0,1,2,3,...

bright fringes are formed (constructive interference)

The fringe width is given by the relation

β=λD

Measuring β, D and d, the wavelength λcan be determined.

6 Interference in Thin Films

Q. Develop the theory of interference due to reﬂected wave-

fronts from a rectangular slab (thin ﬁlm). Also explain

that the intensity distribution of interference pattern due

to reﬂected & transmitted waves.

When light is incident on a parallel thin ﬁlm, a small portion

of it gets reﬂected from the top surface and a major portion is

transmitted into the ﬁlm. Again, a small part of the transmit-

ted components is reﬂected back into the ﬁlm by the bottom

surface and the rest of it is transmitted from the lower surface

of the ﬁlm.

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The set of reﬂected rays R1,R2,R3,.... are derived from the

same incident ray but appears to come from two sources lo-

cated below the ﬁlm. The sources are virtual coherent sources.

The reﬂected waves travel along parallel paths and interfere at

inﬁnity. Similarly, the transmitted rays T1,T2,T3,.... also inter-

fere in the same way.

6.1 Interference due to Reﬂected light

Let us consider a transparent ﬁlm of thickness t and refractive

index µ. Let a ray SA incident on the upper surface of the ﬁlm

is partly reﬂected along AT and partly refracted along AB.

At B, part of it is reﬂected along BC and ﬁnally emerges out

along CQ.

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The angle of incidence is iand angle of refraction is r. Draw

CN normal to AT and AM normal to BC. From points C and

A onwards, the rays NT and CQ travel equal path. So, the

geometrical path difference between the two rays is

L=AB +BC −AN

Now, the optical path difference = µL

∴X=µ(AB +BC)−1.(AN)

=µ(AB +BC)−µ(CM)

∵µ=sini

sinr=AN/AC

CM/AC =AN

So, x=µ(AB +BC −CM)

or, x=µ(PC −CM)

or, x=µ(PM)

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In 4APM

cosr =PM

or, PM =APcosr

=2tcosr

∴x=µPM =2µtcosr ...(1)

This equn (1), in case of reﬂected light doesnot represent the

correct path difference. On the basis of electromagnetic the-

ory, when a ray of light is reﬂected at the boundary of a rarer

to denser medium, a phase change of πor λ

2occurs. There is

no path difference due to transmission at C.

Therefore, the true path diff is

x=2µtcosr −λ

(i) If the path difference x=nλ, the rays meet each other in

phase and constructive interference takes places and the ﬁlm

appears bright. Thus, for maxima,

2µtcosr −λ

2=nλ

where n= 0,1,2, ..

or, 2µtcosr = (2n+1)λ

(ii) If the path difference x= (2n+1)λ

2, the rays meet each

other in opposite phase and destructive interference takes place

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and the ﬁlm appears dark. Thus, for minima,

2µtcosr −λ

2= (2n+1)λ

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

or, 2µtcosr = (n+1)λ

In can be written as,

2µtcosr =nλ

6.2 Interference due to transmitted light

Consider a thin ﬁlm transparent ﬁlm of thickness t and refrac-

tive index µ. A ray SA after refraction goes along AB. AT B

it is partly reﬂected along BC and partly refracted along BR.

The ray after reﬂection at C, ﬁnally emerges along DQ. Here

at B and C reﬂection take spalce at the rarer medium, so no

phase change occurs.

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Draw BM normal to CD and DN normal to BR. The geomet-

rical path difference between two transmitted rays is

L=BC +CD −BN

The optical path difference between the two rays is

x=µ(BC +CD)−1.(BN)...(1)

From Snell’s law

µ=sini

sinr=BN/BD

MD/BD

or, µ=BN

or, BN =µMD ...(2)

In the above ﬁgure, ∠BPC =rand CP =BC =CD

∴BC +CD =PC +CD =PD ...(3)

So, x=µ(PD)−µ(MD)

or, x=µ(PD −MD)

or, x=µPM

In 4BPM,cosr =PM

or, PM =BPcosr

But BP =2t

∴PM =2tcosr

Thus, the optical path difference between the transmitted rays

x=2µt cosr

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(i) If the path difference x=nλ, bright fringes occurs.

i.e., for maxima 2µt cosr=nλ

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

(ii) If the path difference x= (2n+1)λ

2, dark fringes occors

i.e., for minima 2µt cosr= (2n+1)λ

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

6.3 Intensity Distribution of Interference due to Reﬂected and Refracted beams in

thin ﬁlms

From Fabry-Perot Interference, the intensity of transmitted

beam is given by,

It=0.8521Io,Io=maximum intensity

i.e.,It=85.21% and Ir=100 −85.21 =14.79

(i) In transmitted system, the intensity of the interference max-

ima is 100 % and the intensity of the minima will be 85.21%.

(ii) In reﬂected system, the intensity of the interference max-

ima will be 14.79% of the incident intensity and the intensity

of the minima will be zero.

The condition of interference in the reﬂected and transmit-

ted system for fringes are quite opposite to each other. For a

certain path difference, if a bright fringe is produced in the re-

ﬂected system, a dark fringes will be produced in the transmit-

ted system for the same path difference. hence, intensity dis-

tribution of interference due to reﬂected and refracted beams

are complementary to each other.

Figure

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6.4 Production of Colours

We have path difference =2µtcosr

If t and r are constant, the path difference varies with µor the

wavelength of light. When a white light is incident on the thin

transparent ﬁlm, the fringes are coloured as the path difference

depends upon λ, wavelength of light.

The fringes colour will depend upon the thickness and the an-

gle of incidence of initial ray.

7 Wedge shaped thin ﬁlm

Let us consider two plane surface OA and OB inclined at a

very small θand enclosing a wedge shaped air ﬁlm. The thick-

ness of the air ﬁlm increases from O to A. When the light is

incident on the wedge from above, it gets partly reﬂected from

the glass-air boundary at the top of the ﬁlm. Part of the light

is transmitted through the air ﬁlm and gets reﬂected partly at

the air - glass boundary as shown in ﬁgure. The reﬂected rays

from the top and bottom of the ﬁlm, are coherent and interfere

producing bright and dark fringes depnding upon the path dif-

ference.

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The optical path difference between the two reﬂected wave is

x=2µt cosr−λ

Suppose the nth bright fringes occurs at Pn

The thickness of the air ﬁlm at Pnis PnQN.

For small angle of incidence, cosr =1

So, for bright fringe,

2µt cosr−λ

2=nλ

or, 2µt cosr= (2n+1)λ

Here, µ=1 (for air)

cosr=1

t=PnQn

∴2PnQn= (2n+1)λ

2...(1)

The (n+1)th bright fringe will occur at Pn+1, such that

2Pn+1Qn+1= (2(n+1) + 1)λ

or, 2Pn+1Qn+1= (2n+3)λ

2...(2)

subtracting (1) from (2)

Pn+1Qn+1−PnQn=λ

2...(3)

This shows that the next bright fringe wil occur at the point

where the thickness of air ﬁlm increase by λ

Let (n+m)th bright fringe be at Pn+m. Then, the no. of bright

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fringes between Pnand Pn+mwill be msuch that

Pn+mQn+m−PnQn=mλ

2...(4)

If the distance OQn+m−OQn=x, Then

tanθ=Pn+mQn+m

OQn+m−OQn

or, tanθ=mλ

or, θ=mλ

2x[θ≈tanθ,for small θ]

∴x=mλ

2θ

Hence, fringe width =distance

no. of fringes

∴β=λ

2θ

8 Newton’s Ring

Q. Give with necessary theory Newton’s ring method for

the determination of wavelength of monochromatic light.

Why is the centre of the rays dark and how can we get a

bright cenre?

Circular interference fringes can be produced by elcosing a

very thin ﬁlm of air or any other transparent medium of vary-

ing thickness between a plane glass plate and a convex lens of

large radius of curvature. Such fringes were ﬁrst obtained by

Newton and are known as Newton’s ring.

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Monochromatic light from an extended source S is rendered

parallel by a lens L. It is incident on a glass plate inclined at

45oto the horizontal, and is reﬂected normally down onto a

plano-convex lens N placed on a ﬂat glass plate P.

Light rays reﬂected upward from the top and bottom surfaces

of the air ﬁlm formed between the lens and glass plate, super-

impose each other to give dark and bright circular rings.

Let AB be the plane surface of plano-convex lens of radius of

curvature R placed over the glass plate MN such that a thin

ﬁlm of air having thickness from zero to tis formed between

them.

∴AM =BN =t

Let AOBE be the complete circle to curve part of the lens and

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OE be the diameter. The path difference between two reﬂected

rays one from B and another from N is 2tfor light incident

normally on the air ﬁlm.

∴path diff =2t

The rays reﬂected from B has a phase change of πcorrespond-

ing to path diff λ/2.

So, if a dark ring is formed at B, then

2t=nλ...(1)

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

and λ=wavelength of incident light

Similarly, if a bright ring is formed at B, then

2t=nλ.λ

2t= (2n−1)λ

2...(2)

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

Now, from the geometry of circle in the ﬁgure, we have

AD ×DB =OD ×DE

=OD ×(OE −OD)

=t(2r−t)

∴AD ×DB =2Rt −t2...(3)

or, r×r=2Rt −t2

or, r2=2Rt −t2...(4)

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where rbe the radius of ring at B.

Since, the thickness of air ﬁlm is very small, t2is neglected

∴r2=2Rt

or, 2t=r2

R...(5)

And from (5) and (2) , for bright rings, we have

R= (2n−1)λ

or, r2= (2n−1)λR

∴Radius of nth bright ring is

rn=r(2n−1)λR

2...(6)

and from (5) and (1), for dark ring, we have

R=nλ

or, r2=nλR

∴Radius of nth dark ring is

rn=√nλR...(7)

Let Dnbe the diameter of nth dark ring measured by travelling

microscope, Then

Dn=2rn=2√nλR

or, D2

n=4nλR...(8)

and Dn+mbe the diameter of (n+m)th dark ring

n+m=4(n+m)λR...(9)

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Substracting (8) from (9) , we have,

n+m−D2

n=4mλR

or, λ=D2

n+m−D2

4mR ...(10)

Hence λcan be measured.

8.1 Newton’s ring with bright centre

The central spot is dark as seen by reﬂection. Newton’s ri

x=2t−λ

At centre, t=0

∴x=λ

The wave reﬂected from the lower surface of the air ﬁlm suf-

fers a phase change of πwhile the wave reﬂected from the

upper surface of the ﬁlm doesnot suffer such change. Thus,

the superimposing waves are out of phase by λ

2and hence in-

terfere to produce a dark spot.

But if a transparent liquid of refractive index µis trapped be-

tween lens and plane surface in contact such that

refractive index of lens >µ>refractive index of plate

The reﬂection in both cases (from top and bottom) will be

from denser medium to rarer medium and the two reﬂected

rays will interfere in same phase producing a bright spot.

9 Michelson Interferometer

In a Michelson Interferometer, a beam of light from an ex-

tended source is divided into two parts of equal intensities by

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partial reﬂection and refraction. These beams travel in two

mutually perpendicular direction and come together after re-

ﬂection from plane mirrors. The beams overlap on each other

and produce interference fringes.

It consist of two plane mirrors M1and M2, a beam splitter G1

and a compensating plate G2. The two plates are held parallel

and are inclined 45oto the mirror M2. The plane mirros M1

and M2are made exactly perpendicular to each other. The in-

terference bands can be observed through the telescope T.

Monochromatic light from extended source is rendered par-

allel through lens L and is made incident on beam splitter

G1. The beam is partly reﬂected back along AC and partly

transmitted along AB. The beam AC travels normally towards

plane mirror M1and is reﬂectd back along the same path and

comes out along AT.

The transmitted beam travels towards plane mirror M2and is

reﬂected along the same path. it is reﬂected at G1and Pro-

ceed along AT. The two beams received along AT are pro-

duced from a single source and hence are coherent. They su-

perimpose to produce interference fringes.

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The reﬂected light passes throuigh G1three times whereas the

transmitted light passes through it one time. So, a compen-

sating plane G2of the same thickness is placed in the path of

transmitted light.

The two mirrors form virtual images of source S at S1and S2.

The virtual images are sepaarted by a distance 2d. The path

difference between the two beams will be 2d. The transmitted

blight undergoes rare to denser reﬂection at G1and therefore

aπ-change occurs.

∴optical path diff, x=2d+λ

for bright fringes, 2d+λ

2=nλ

or, 2d= (2n−1)λ

2where n=1,2,3,...

for dark fringes, 2d+λ

2=nλ

or, 2d=nλwhere n=0,1,2,3,...

9.1 Determination of wavelength using Michelson Interferometer

Michelson interferometer can be used to determine the wave-

length of monochromatic light of the source. If the mirror M1

and M2are equidistant from G1, the ﬁeld of view will be per-

fectly dark. The mirror is kept ﬁxed, the mirror M1is moved

with the help of screw. The fringes appears to move across the

ﬁeld of view.

If at a distance d1, the bright fringes is of order m1

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and at distance d2, the bright fringe is of the order mn. Then,

2d1=m1λ

2d2=mnλ

subtracting 2(d2−d1) = (mn−m1)λ

or, 2d=Nλ

∴λ=2d

where dis the distance through which the mirror M1is moved

and Nis the no. of fringes that crosses the ﬁeld of view.

9.2 Measurement of diff in wavelength of Sodium D1and D2lines.

Let the wavelength of D1line is λ1and wavelength of D2line

is λ2. The fringes due to each spectral lines are produced.

The mirror M2is kept ﬁxed and M1is so adjusted that the

bright fringe due to D1coincides with bright fringes due to D2.

This is possible when the nth order of the longer wavelength

coincides with (n+1)th order of shorter wavelength. Then,

2d=nλ1= (n+1)λ2[assume λ1>λ2]

or, n=λ2

λ1−λ2

∴2d=λ1λ2

λ1−λ2

or, λ1−λ2=λ1λ2

Taking λas mean of λ1and λ2, the difference in wavelength is

λ1−λ2=λ2

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9.3 Standarization of Meter

It can be used to standarize a meter. It is not possible to count

the no. of fringes dor a whole length of meter. So, in prac-

tice, nine etalons were used, each being twice the length of

the preceding atalon. The shotest etalon is 0.390625 mm and

the longest is 10cm.

The experiment is divided into two main parts.

i) The no. of wavelength of the monochromatic cadium light

is counted for the shortest etalon.

ii) The length of 2nd etalon is compared with shortest etalon

and the process is repeated until the no. of wavelength for

longest etalon is known. The no. of wavelength for longest

etalon is known which acts as a standard meter.

9.4 Determination of thickness of thin transparent sheet.

When a thin transparent sheet of thickness tis inserted in the

path of one interfereing beam, the optical path diff changes by

2(µ−1)t.

If mbe the no. of fringes displaced due to sheet

2(µ−1)t=mλ

∴t=mλ

2(µ−1)

10 Fabry-Perot Interferometer

The Fabry-Perot interferometer is a high resolving power in-

struments, which makes use of the fringes of equal inclination,

produced by the transmitted light after multiple reﬂections in

an air ﬁlm between two parallel highly reﬂecting glass plates.

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It consistes of two optical plane glass plates A and B with their

inner surface silvered, placed parallel to each other. S is the

broad source of monochromatic light and L1is the lens which

makes the rays parallel. An incident ray sufferes a large no of

internal reﬂections successively at the two silvered surfaces as

shown in ﬁgure. At each reﬂection, a small fucntion of light is

transmitted, producing a group of coherent and parallel trans-

mitted rays. The lens L brings these rays together to focus at

a point where they interfere to give interference pattern.

10.1 Formation of fringes

Let dbe the separation between twosilvered surfaces and θ

be the inclination of a particular incident ray. So, the path

difference between two successive rays is x=2dcosθ.

The condition for maximum intensity is given by

2dcosθ=nλ...(1)where n=0,1,2,3,.....

The interference pattern consist of concentric rings, each ring

corresponding to a particular value of θ.

10.2 Resolving Power

Let us consider two consequtive rings due to light rays of

wavelengths λand λ+dλ.

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The path diff between two waves is

x=2dcosθ

and phase φis given by

φ=2π

λx

or, φ=2π

λ2dcosθ

or, φ=4πdcosθ

λ...(2)

differentiating above equation

dφ=−4πd

λsinθdθ...(3)

If the maximum for a wavelength λ+dλoccurs at some an-

gular separation dθ

−2dsinθdθ=ndλ[∴differentiatinf equn (1)]

equation (3) becomes

dφ=2πndλ

λ...(4)

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Thus, the resolving power is given by

dλ=2πn

dφ...(5)

For a phase change dφgoing from one maximum to the posi-

tion of other in Fabry-Perot Interferometer.

Again, sin dφ

4=1−r2

2rr=reﬂection coefﬁcient

But sin dφ

4=dφ

4[∴dφs small]

or, dφ

4=1−r2

or, dφ=2(1−r2)

So, equn (5) becomes

dλ=2πnr

2(1−r2)

there f ore λ

dλ=πnr

(1−r2)

Hence, the resolving power increases with increase in the value

of ni.e., order of the fringes.

11 Numericals

Q. A light source emits visible light of two wavelengths

λ1=430nm and λ2=570nm. The source is used in a double-

slit interference experiment in which L=1.5 m and d =

0.025 mm. Find the separation between the third -order

bright fringes.

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Q. A viewing screen is separated from a double-slit source

by 1.2m. The distance between the two slits is 0.030mm.

The secoond order bright fringe (n=2) is 4.5 cmfrom the

centre line. (a) Determine the wavelength of the light (b)

Calculate the distance between adjacent bright fringes.

Q. In Fresnel’s birism experiment, on inserting a thin glass

plate in the path of one of the interfereing beams, it is

found that the actual bright fringe shift into the position

previously occupied by the sixth bright fringe. if the wave-

length of light used is 5890 ˚

Aand refractive index of glass

plate is 1.5 for the wavelength. Find the thickness of the

plate.

Q. The inclined face of a glass prism of refractive index 1.5

makes an angle of 1owith the base of the prism. The slit is

10cm from the prism and is illuminated by light of wave-

length 5800 ˚

A. Find the fringe width at a distance of 1.2 m

from the bi-prism.

Q. A glass wedge of an angle 0.01 radian is illuminated

by monochromatic light of wavelength 5500 ˚

Afalling nor-

mally on it. At what distance from the edge of the wedge

with the 10th fringe be observed by reﬂected light.

Q. Light of wavelength 650nm falls normally on a thin

wedge shaped ﬁlm of refractive index 1.4 forming fringes

that are 2mm apart. Find the angle of wedge in degree.

Q. The diameter of 6th bright ring with air and liquid ﬁlms

are respectively 2.5 mm while performing a Newton’s ring

experiment. What is the refractive index of liquid?

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