Interference

Superposition Principle

For all linear systems, the net response at a given place and time caused by two

or more stimuli is the sum of the responses which would have been caused by

each stimulus individually. So if input A produces response X and input B

produces Y than input (A+B) produces response (X+Y).

Interference

When the light from two different sources with same frequency and having a

constant phase difference move in the same direction, then these light wave trains

superimpose upon each other. This results in the modification of distribution of

intensity of light. This modification of the intensity of light resulting from the

superposition of wave is called interference.

Coherent Sources

Two sources are said to be coherent if they emit light of same frequency and

always having a constant phase difference between them.

At some points the resultant intensity is greater than the sum of the intensities of

the waves.

Destructive Interference

Constructive Interference

At some points the resultant intensity is smaller than the sum of the intensities of

the waves.

21 III

21 III

This is called Constructive Interference.

This is called Destructive Interference.

The wave front originating from a common source is divided into two parts by

using mirrors, prisms or lenses and the two wave fronts thus separated travels and

finally brought together to produce interference.

In this type sources are small like a point source.

Classification of Interference

1. Division of Wave front

2. Division of Amplitude

The amplitude of the incoming beam is divided into two parts either by partial

reflection or refraction. These two parts travel in different paths and finally brought

together to produce interference.

In this type broad sources are required.

Young’s Double Slit Experiment:

crest

Trough

S

(Coherent Source)

S1

S2

(Slit)

(Screen)

Let S be a narrow slit illuminated by a monochromatic light of wavelength λ. S1 and S2

are two narrow slits close to each other and equidistant from S. Suppose is the

frequency of the waves. Let a1 and a2 be the amplitudes of the two wave coming out of

S1 and S2.

Analytical treatment of Interference

tay sin11

The displacement y1 due to one wave from S1 at any instant t is

The displacement y2 due to other wave from S2 at any instant t is

)sin(22 tay

Where Φ is the constant phase difference between the two waves.

The resultant displacement at P is the algebraic sum of the individual

displacements

21 yyy

)sin(sin 21 tatay

Squaring (i) and (ii) and then adding

222222

221

22

2

2

1 sincossincos2cos AAaaaaa

sincoscossinsin 221 tatatay

tataay cossinsin)cos( 221

coscos21 Aaa

sinsin2 Aa

tAtAy cossinsincos

(i)

(ii)

Let

For maximum intensity

1cos

2

2

2

1max aaI

or

IAaaaa 2

21

2

2

2

1 cos2

n2

2

2121

2

2

2

1max )(2 aaaaaaI

Phase Difference Path Difference

2

Path Difference = nλ

For minimum intensity

1cos

2

2

2

1min aaI

or

IAaaaa 2

21

2

2

2

1 cos2

)12( n

2

2121

2

2

2

1min )(2 aaaaaaI

Phase Difference Path Difference

2

Path Difference

For good contrast

21 aa 2

max 4aI 0minI and

2)12(

n

Average intensity

2

0

2

0

d

Id

Iav

When a1 = a2

2

0

2

0

21

2

2

2

1 )cos2(

d

daaaa

2

0

2

021

2

2

2

1 sin2 aaaa

2

2

2

1 aaIav

22aIav

Conditions for sustained interference:

Two light sources must be coherent.

Two coherent sources must be narrow, otherwise a single source

will act as a multi sources.

The amplitude of two waves should be equal so that we can get

good contrast between bright and dark fringes.

The distance between two coherent sources must be small.

The distance between two coherent sources and screen should be

reasonable. The large distances of screen reduce to intensity.

Calculation of the fringe width:

To determine the spacing between the bands/ fringes and

the intensity at point P.

D

S1

2d

S2

S

(Coherent Source)

Slit

Screen

O

P

d

d

N

M

x

Path difference (Δ) = S2P-S1P

To calculate S2P, consider the ∆S2NP

222

22 NPNSPS

222

2 dxDPS

2

1

2

2

2 1

D

dxDPSor

Expending by binomial theorem

2

2

22

11

D

dxDPS

Therefore, higher power term of D can be neglected. Then we get

2

2

22

1D

dxDPS

D

dxDPS

2

2

2or

dxD Here

Similarly, we can calculate S1P, consider the ∆S1MP

D

dxDPS

2

2

1

Then the path difference is (Δ) = S2P - S1P D

xd2

For the nth fringe the path difference = D

dxn2

(a)Bright Fringes:

The path difference should be equal to n

.

nD

dxn 2

nd

Dxn

2 where n = 0, 1, 2, 3, 4, ----------------

nd

Dn

d

Dxx nn

21

21

nd

D

d

Dn

d

D

222

d

D

2

d

D

2

The distance between two consecutive fringes is also known as fringe width.

2

12

n

2

122

nD

dxn

d

Dnxn

412

(b) Dark Fringes:

The path difference should be equal to

Point P to be dark

where n = 0, 1, 2, 3, 4, --------------

d

D

d

Dxx nn

24

21

d

D

2

Fringe width

D

d

1

From the above equations, it is clear that fringe width β depends on

1. It is directly proportional to the distance between two coherent sources

and screen

2. It is directly proportional to the wavelength of light

3. It is inversely proportional to the spacing between two coherent sources

.

Fresnel’s Biprism:

Fresnel’s biprism is a device to produce two coherent

sources by division of wave front.

D

b a

Overlap

region S

S2

S1

O

H

P

G

Q

Construction:

A biprism consists of a combination of two acute angled prisms placed

base to base.

The obtuse angle of the biprism is 179º and other two acute angles are 30’.

03

03

179°

(b) Determination of the distance between two virtual sources:

Displacement method is one of the methods to calculate the distance

between two virtual coherent sources:

U’

v u

v’

lensandobjectbetweencedis

lensandimagebetweencedis

Oobjecttheofsize

Iimagetheofsize

tan

tan

According to the linear magnification produced by the lens:

Further the lens moves towards the eyepiece and a focused image of virtual

sources S1 and S2 is visible in eyepiece again. This time the image separation of S1 and

S2 should be appear different (d2) so that:

u

v

d

d1

'

'2

u

v

d

d

(1)

(2)

d1 d2

L2 L1

S1

S2

d

1

2

d

d

d

d

212 ddd

21ddd

From equation (1) and (3), we get

or

But

'vu and

'uv

Thus equation (2) becomes

v

u

d

d2 (3)

Applications of Fresnel’s Biprism:

Determination of thickness of thin sheet of transparent

material like glass or mica.

or

How to calculate the displacement of fringes when a mica

sheet is introduced in the path of interfering rays?

S1

2d

S2

t

x

D

P m

O

v

t

c

tPST

1

c

t

c

tPST

m

1

The time taken by light to reach P from S1 is

v

cmBut

ttPScT m 1

ttPSPS m 12

The path difference S2P and S1P will then be given by

The path difference between S2P and S1P is

m nttPSPS 12

m ntPSPS 112or

We have already calculated that PSPS 12 D

dxn2

m ntD

dxn 12

m ntd

Dxn 1

2or

Let the point P is the center of the nth bright fringe if the path difference is equal to nλ

Where xnis the distance of the nth bright fringe from the central fringe in the absence of mica.

The position of the central bright fringe when the mica sheet is placed in

the path S1P is obtained by putting n=0 in equation (1) we get

(1)

td

Dx 1

20 m

m0xSince >1 so that is positive.

(2)

The fringe width is

d

Dxx nn

21

Using equation (1)

It means the fringe width is not affected by introduce of mica sheet.

d

D

2

Put these values in equation (2) we get,

1

2 0

mD

dxt

1

0

m

xtor

Thus we can calculate the thickness of mica sheet.

Light Reflection From Denser Media:

/2 Shift In Position of Wave

Inversion with /2 shift in position No inversion

Change of Phase on Reflection

When a wave of light is reflected at the surface of denser medium, it always gives a phase change

of π or path difference of λ/2

i Air

Glass

ar

at r

N M

D

C

B

A

a

Here r and t are the reflection and the transmission coefficients when wave is travelling from rarer

to denser medium.

If we reverse the direction of reflected and transmitted light then according to the Principle of

reversibility , the original wave of amplitude a is produced, provided that there is no absorption of

energy

i Air

Glass

ar

at r

N M

D

C

B

A

ar2+att’

art+atr’

Here r’ and t’ are the reflection and the transmission coefficients when wave is travelling from

denser to rarer medium.

The reversal of ar and at must reproduce the amplitude a. The sum of components along BE

should be zero.

0' atrart rr '

E

Interference due to Reflection:

Source

t

airinANpathfilminABCPath

i i

r r

R1 R2

T1 T2

A

B

C

D

M

N

(Reflected rays)

(Transmitted rays)

The path difference between the reflected rays

ANBCAB m

rAB

BMcos

r

tAB

cos

r

tBC

cos BCAB

rBM

AMtan

rBMAM tan

(1)

BM = t

and also

Now, for AN

rtAM tan

rtAC tan2

iAC

ANsin

iACAN sin

irtAN sintan2

r

ri

r

rt

sin

sinsin

cos

sin2

rr

rt sincos

sin2 m

ri sinsin m

r

rtAN

cos

sin2

2

m

AC = AM + CM

(because AM = CM)

or

r

rt

r

t

r

t

cos

sin2

coscos

2

mm

r

rt

r

t

cos

sin2

cos

2 2

mm

rr

t 2sin1cos

2

m

rr

t 2coscos

2m

rt cos2m

So that,

So that the actual path difference:

2cos2

m rt

As the ray is reflected from a denser medium, so an addition of

path difference of λ/2 will be there.

So for Maximum Intensity, path difference should be equal to

m nrt cos2

m nrt 2

cos2

n

2)12(cos2

m nrt

Where n = 0,1,2,3,4,5…………………..

So for Minimum Intensity, path difference should be equal to 2

)12(

n

2)12(

2cos2

m nrt

Where n = 0,1,2,3,4,5…………………..

Interference will not be perfect as there is difference in the amplitude of the reflected rays.

Production of colors in thin films:

When a thin film of oil on water, or a soap bubble, exposed to white light (such as

sunlight) is observed under the reflected light. The brilliant colors are seen due to

the following reasons

The path difference depend on the wavelength. It means the path

difference will be different for different colors, so that with the white light

the film shows various colors from violet to red.

The path difference also varies with the thickness of film so that various

colors appear for the same angle of incidence when seen in white light.

The path difference changes with the angle r and angle r changes with

angle i. So that the films assumes various colors when viewed from

different directions with white light.

Newton’s Rings:

Source

2

cos2

m rt

2

cos2

m rt

Actually the path difference between the interfering rays is

The effective path difference for large radius of curvature or for small angle θ

m nt 2

2

2

122

m nt

2

122

2

m nt

m nnt 22

122

m nt 2

For normal incidence cosr =1, then the path difference 2

2

m t

For maxima

For minima

Why Central Ring is dark in Newton’s rings experiment

2

2

m t

At point of contact t = 0

2

Path difference for dark ring 2

12

n

For n = 0, Path difference 2

That’s why Central Ring is dark

How to calculate the radius or diameter of the nth fringe:

n

22 2 tRtn

Rt 2

Rtn 22

Rt n

2

2

Let be the radius of bright ring at point C and t is the thickness of air film at that point.

Let R be the radius of plano-convex lens. In triangle OAC

so that the higher power terms are neglected.

Therefore, we have

ρn

R

R

R-t

O

A

B

C

D

RttRR n 22222

222 tRR n

or

But t

2

122

m nt

2

122

22

m nR

n

m

2122 R

nn

For constructive interference: We have,

m

2

12nRn

m

2

1222

nRD nnDiameter

m

m

122

2

1242

nRnRDn

1m

Now, for the air film the refractive index

Therefore,

Therefore,

1222 nRDn

12 nDn

Diameter of the nth bright ring is proportional to the square root of the odd natural numbers

For Dark rings

m nt 2

m nR

pn 2

22

1mBut for air film

So,

Rnpn 2

Therefore, RnDn 42

nDn

Diameter of the nth dark ring is proportional to the square root of the natural numbers

Rnpn

Spacing between successive rings

This shows that the spacing decreases with increase in the order of the rings.

]1[1 nnkDD nn

]12[12 kDD

]23[23 kDD

]34[34 kDD

1222 pnRD pn

12212222 nRpnRDD npn

RnRRpRnR 24244

pRDD npn422

pR

DD npn

4

22

Determination of wavelength of light by Newton’s Rings method

thpn Now the diameter of bright ring,

Therefore,

1222 nRDn

For Air film,

RnD airn 4][ 2

Determination of refractive index of unknown liquid by Newton’s Rings method

For unknown liquid with refractive index m

m

RnD liquidn

4][ 2

liquidn

airn

D

D

][

][2

2

m