INTERFERENCE

Interference is the superposition of two orf p p fmore coherent waves resulting in a new

ttwave pattern.

There are two types of interference:

Constructive InterferenceConstructive InterferenceDestructive Interference

combinedwaveform

wave 1

wave 2

Two waves in phase Two waves 180° outof phasep

Constructive Destructive

Condition for constructive interference

,.......4,2,0differencePhase 2 πππϕ n±±=

;..........2,,0differencepath or λλλ nx =

)12(

Condition for destructive interference

)12(3differencePath

,.......3,differencePhase )12(λλλ

ππϕ π+

=

±±= +nx

n

2.........

2,

2differencePath =x

Note: In each case n = 0,1,2….

Superposition of two waves: Intensity distributionP

S1

Let y and y are the

S

Let y1 and y2 are the displacements of two waves coming from S1S2

g 1and S2

ω= tay cos1

( )ϕω += tayy

cos2

1

ϕ is the phase difference between two waves reaching At P from S1 and S2, a be the amplitude of wave.

)cos(cos21 ϕωω ++=+= tatayyy

2cos

2cos2 ϕϕ −−++

=wtwtwtwta

)2

cos(2

cos2 ϕϕ+= wta

22waveresultant theof amplitude2cos2 =ϕa 2

intensity2cos4 22 ϕaI =Therefore Intensity

amplitude of square is

)maxima(4 2,..4,2,0 2aInif =±±= πππϕ

)minima(0)12,..(3, =+±±= Inif πππϕ

2cos4 22aI = ϕ

)(214

2

2222 aaaa +==×= )(22

4 aaaa +==×=

= sum of intensityFor incoherent light

yof constituent waves.

differentarephasesandamplitudesIf

( )111 )(cos θω += tay

( ) 21222 ~,cos θθϕθω =+= tay

22 )(cos2 2122

21 ϕaaaaI ++=

?max =I?min

max

=I

Young’s Double slit experimentg p

(The experiment can also be performed with a beam of electrons or atoms, showing similar interference patterns.

THE INTERFERENCE FRINGESP

dDwidthfringethatShow λβ =

S1 yn

d/2d O

d/2

d/2

S

O d/2

S2D

At point P for maxima we must haveS2P – S1P = nλ n = 0 1 2 3

LS2P S1P nλ, n 0,1,2,3…

])([)( 2222

dyDPS n ++=)()( 2

12

2dd

PSPS =−])

2([)( 2 yn

])([)( 2221

dyDPS n −+=])

2([])

2([ 2222 dyDdyD nn −+−++

2222 dyyDdyyD +−−++=])2

([)( 1 yn

2)()(thus2

22 dyPSPSdy

dyyDdyyD

n

nnnn

=+++

)()(2)()(

2)()(thus,

12

12

PSPSdyPSPS

dyPSPS

n

n

=−

=−

)()( 12

12 PSPS +

If d<<D then S P≈S P≈DIf d<<D then S2P≈S1P≈D.Thus S2P+S1P=2D and S2P – S1P = nλ

dd2Ddy

Ddyn nn ==λ

22 therefore,

Ddynor n=λ ,

dDnyn

λ=

Thus position of nth bright fringeon the screen d

ynon the screen

fringes,dark for Similarly

..3,2,1,0,2

)12(12 =+=− nnPSPS λ2

So we get, position of nth darkfringe on the screen

dDnyn 2

)12( λ+=

Distance between any two consecutive bright or

d2

Dyy nnλβ ==−+1

Distance between any two consecutive bright or dark fringes (Fringe width)

dyy nn β+1

Proportional to the wavelength of source.Fringe width……..

Proportional to the distance of the screen from plane of S1 and S2.Inversely proportional to the distance between S1 and S2.

Separation between dark and bright fringes

3 βλλλ DDD222

3 βλλλ==−

dD

dD

dD

• O is equidistant from S1 and S2 li ht d t O

S1so light waves superposed at O are in phase so light intensity at O ill b i

OO will be maximum.

• At O, we observe the central S2bright fringe. For this fringe n=0 y = 0. so central bright fringe will be referred as zeroth order bright fringe.

Fresnel’s BiprismIt consists of two thin acute angled prisms joined atthe bases It is constructed as a single prism of obtuse

p

the bases. It is constructed as a single prism of obtuseangle of 179º. The acute angle α on both side is about30´30 .

α α179º

αc

aSThe prism is placed with the refracting edge in a way such

b

that Sa is normal to the face bc of prism.

b

ScreenA c

E

Sd

Fringes of equal width

Oda

F

O

B b

FZ1 Z2

D

1. S is source2. A and B are coherent sources

(Virtual) obtained through ‘ab’

Z1 = distance of biprism from sourceZ2 = distance of screen from biprismD= Z1+Z2(Virtual) obtained through ‘ab’

and ‘ac’ surface of biprismD Z1+Z2

Screen dDλβ =

A cE

Observer

Fringes of equal widthS

d O ObserverEyepiece

da

F

O

B b

FZ1 Z2 Ques: Can we determine

Wavelength of source D through this setup ?

Ans: Yes dβλ =D= distance of screen from source.d = distance between the sources A and B.

Dλ

Determination of the distance between the two sources (d)

dM

d

A convex lens (L1) is placed between the prism and eyepiece (M), suchL1

that the image of the virtual sources A and B are seen in the field ofview of the eyepiece.S th di t b t th i f A d BSuppose the distance between the images of A and B asseen by the eyepiece is d1.

nvdSo ,

mn

uv

dd

==1 ……..(1)

Now move the lens towards eyepiece and bring it to other positon L2

So that again images of A and B are seen clearly in the the field of view

Of eyepiece. Again if the distance between the two images be d2

L2

y p g g 2

mvdnm

uv

dd

==2 ……..(2)

M lti l i (1) d (2) t

Substituting the values of β, d and D we can calculate the

Multiplying (1) and (2) we get

21 1 ddddd⇒

value of wavelength (λ) of given monochromatic light.

dβ21221 1 ddd

d=⇒=

Ddβλ =

Theoretical Calculation of d:(Alternative way) D

dβλ =(Alternative way)

12 zd αμ )( −= Dλβ =or

112 zd αμ )(Where µ is refractive index of biprism. α is angle of biprism and Z1 is distance of biprism

dβ

angle of biprism and Z1 is distance of biprism from source. Assignment: Can you prove it?

Dnλ

1)1(2 zyn αμ −

=∴1

)1(2 ZDand λβ =

1)1(2 Zαμ −

Fringes with white lightWhen white light is used the center fringe at O is white sinceall waves will constructively interfere here while the fringesall waves will constructively interfere here while the fringeson the both side of O are colored because the fringewidth (β) depends on wavelength of lightwidth (β) depends on wavelength of light.

For green light,1)1(2 z

Dny g

g αμλ−

=1)1(2 zg αμ −

For red light, Dny rλ=1)1(2 z

yr

r αμ −=

SummarySummary• Superposition of waves: Interference, Conditions of

constructive and destructive interferenceconstructive and destructive interference.• Coherent sources: division of wavefront: Young’s

double slit experimentdouble slit experiment– Intensity distribution on the screen (Superposition

of two waves).of two waves).– Position of nth order bright (or dark) fringe on the

screen.– Fresnels’ Biprism: Fringe width, determination of

wavelength of light and theoretical approach to calculate distance between the slits ‘d’.

– Use of white light source.

Numerical1. In a certain region of interference 45th order maximum for the

wavelength λ = 5893 Å are obtained. What will be the order of ginterference at the same place for (a) λ = 4820 Å, (b) λ = 7576 Å.

Ans: (a) 55th (b)35th

2. The inclined faces of a glass prism of refractive index 1.5 make an2. The inclined faces of a glass prism of refractive index 1.5 make an angle of 1o with the base of the prism. The slit is 10 cm from the biprism and is illuminated by light of λ = 5900 Å. Find the fringe width observed at a distance of 1 meter from the biprismwidth observed at a distance of 1 meter from the biprism.

Ans: Fringe width= 0.338 mm (Note: Use angle α in radian)

3. In a biprism experiment with sodium light, bands of width 0.1963 mm are observed at 100 cm from the lit O i t d i l 30 fslit. On introducing a convex lens 30 cm away from

the slit, two images of the slit are seen 0.7 cm apart, at 100 cm distance from the slit Calculate theat 100 cm distance from the slit. Calculate the wavelength of sodium light.

Ans: 5889 Å

More numerical on1. Wave representation; Amplitude, phase, frequencyp ; p , p , q y2. Intensity distribution; Imax, Imin, …..3. Biprism, double slit interferencep

Assignment-1Assignment 1

• Show that .2/12

cos2 =⎟⎠⎞

⎜⎝⎛ϕ

2 ⎠⎝

Sol:Sol: ( )1cos

21

2cos

cos

2

0

2

0

2

2

+=

⎟⎠⎞

⎜⎝⎛

=⎟⎞

⎜⎛ ∫∫ ϕϕϕϕϕ

ππ

dd

22cos 2

0

==⎟⎠

⎜⎝

∫π

ϕπ

d

( ) ( ) 12021cos

21 2

0

2

0 =+

=⎟⎟⎠

⎞⎜⎜⎝

⎛+

=∫ ∫ πϕϕϕπ π

dd

222===

ππ

Assignment - 2112 zd αμ )( −=Show that

( )αμδ 1−=For small anglesFrom right angle triangle

……… (1)

A

d

From right angle triangleand equation (1).

d S1

2z

dδ =

d/2 α1

1

2 zdz

δ=→

B δδ

z

Hence

12 zd αμ )( −= z1

δ is angle of deviation112 zd αμ )(=

µµ

(d1+d2)

α1, α2Z1

D

λλ DD)()1()( 21121 ααμ

λλβ+−

=+

=Z

Ddd

D

SSummary• Lecture-1:

• Introduction• Lecture-2:

S iti f t I I i• Superposition of two waves, Imax, Imin.• Double slit experiment (Division of wave front), Expression

for fringe width.• Lecture-3:

– Fresnel’s Biprism- Determine λ of sodium light.• Fringe width D

dβλ =

Fringe width • Determination of ‘d’( using convex Lens).• Dependence of ‘d’ on Biprism.

)1(2 Zd

21ddd =

• Using white light source.• Displacement of fringes.

1)1(2 Zd αμ −=

In Double slit experiment: Insert a thin transparent l h t f thi k t d f ti i d iglass sheet of thickness t and refractive index μ in

the path of one beam.

WHAT WILL HAPPEN? Fringe pattern will remain same

or

St, µ P

or

Fringe pattern ill hiftS1 y will shift

Because S1P ray will travel

Dd

more path. hence effective path difference becomes

d

S2

( )

D

tnDdyn 1−+= μλ

( )[ ]tndDyn 1−+=⇒ μλ

Time required for th li ht t hS

t, µ P

the light to reach from S1 to the point P i

S1 y

is

Ct

CtPST +

−=

0

1Dd

CC0

index refractive since 0 →= μCCS2

where C0 is the velocity of light in air and C its velocity in the medium ( )1 tPS −+ μthe medium ( ) 1).........( 1

0

1

CtPST −+

=∴μ

Time required for the light to reach from S2 to the point P is

)2(2PST ′ )2........(0C

T =

So the path difference between the beams reaching P, f S d S (Δ)from S1 and S2 (Δ) tPSPS )1(12 −−−=Δ μ

dDdyPSPS n=− 12

( )dyn 1Δ

As we know

So the path difference will be ( )tD

yn 1−−=Δ μ

f i h f h h b i h f i hIf P is the centre of the nth bright fringe, then

( )tdyn 1 λ( )

( )[ ]D

ntD

yn

1

1 =−−

λ

λμAt n = 0 the shift y0 of

l b i h f i i( )[ ]tndDyn 1−+=⇒ μλ central bright fringe is

D ( )tdDy 10 −= μ

It means that the introduction of the plate in the pathf f th i t f i b di l th tiof one of the interfering beams displaces the entire

fringe system through a distance( ) tD 1μ( ) t

d1−μ

Note: This displacement is towards the beam in the pathNote: This displacement is towards the beam in the path of which the plate is introduced.

K i th di t th h hi h th t l f i iKnowing the distance through which the central fringe isshifted, D, d and μ the thickness of the material t can be

l l t dcalculated.We have to use white light to determine the

hi k f h i lthickness of the material.For monochromatic light central fringe will similar to

h b i h b i h f i F hi li h l f iother bright bright fringe. For white light central fringe is white.

Displacement of fringesDisplacement of fringes• Determine condition of net path difference.

C’µ1,t1

SCCS1

Sµ2,t2

S2

Screen

( ) ( ) 2211 11 tt −−−=Δ μμCase 1: If µ1= µ2=µ and t1>t2 ,Δ is positive ( upward shift)

or t2>t1 , Δ is negative ( downward shift) Case 2: If t1=t2= t .

THE LLOYD’S MIRROR ARRANGEMENTLight directly coming from the slit S interferesLight directly coming from the slit S1 interfereswith the light reflected from the mirror formingan Interference pattern in the region BC of the

Real source

p gscreen.

Mirror

LVirtual source

L’Note: The central fringe will be dark.

Displacement of fringesDisplacement of fringes• Determine condition of net path difference.

µ t

CC’

µ1,t1

S1C

S2

( ) ( ) 2211 11 tt −−−=Δ μμ

µ2,t2 Screen( ) ( ) 2211 tt μμ

Case1: If µ1= µ2=µ and t1>t2 ,Δ is positive ( upward shift) or t2>t1 , Δ is negative ( downward shift)

Case 2: If t1=t2= t .

SSummary• Lecture-1:

• Introduction• Lecture-2:

• Superposition of two waves, Imax, Imin.• Double slit experiment (Division of wave front),

Expression for fringe width.• Lecture-3:

– Fresnel’s Biprism- Determine λ of sodium light.• Displacement of fringes due to one transparent glass

sheet.• Displacement of fringes due to two transparent

l h tglass sheets.

Phase change on Reflection, Refraction