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Electromagnetic waves: Interference

Wednesday October 30, 2002

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Two-source interference

1r

What is the nature of the superposition of radiation What is the nature of the superposition of radiation from two from two coherentcoherent sources. The classic example of sources. The classic example of this phenomenon is this phenomenon is Young’s Double Slit ExperimentYoung’s Double Slit Experiment

aa

SS11

SS22

xx

LL

Plane wave (Plane wave ())

2r

PP

yy

3

Interference terms

12

2

221

*1 ____________________________

rrkioo

P

er

EE

EEP

12

21

221

* ____________________________

rrkioo er

EE

EEPP

12221

**

cos2

_________________2121

r

EE

EEEE

oo

PPPP

1212 rrk

where,where,

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Intensity – Young’s double slit diffraction

122121 cos2 PPPPP IIIII

1212 rrk

PhasePhase difference of beams occurs because of a difference of beams occurs because of a pathpath difference difference!!

12222 cos2

2121

PoPoPoPooPEEEEE

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Young’s Double slit diffraction

I1P = intensity of source 1 (S1) alone

I2P = intensity of source 2 (S2) alone

Thus IP can be greater or less than I1+I2 depending on the values of 2 - 1

In Young’s experiment r1 ~|| r2 ~|| k Hence

Thus r2 – r1 = a sin

122121 cos2 PPPPP IIIII

2112 rrkrkrk

rr22-r-r11

aa

rr11

rr22

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Intensity maxima and minima

m2...4,2,012

2

12...3,12 m

am

2

1sin

Maxima for,Maxima for,

Minima for,Minima for,

...2,1,0sin2sin ma

mormka

If IIf I1P1P=I=I2P2P=I=Ioo

If IIf I1P1P=I=I2P2P=I=Ioo 0cos2 122121 PPPPMIN IIIII

oPPPPMAX IIIIII 4cos2 122121

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Fringe Visibility or Fringe Contrast

MINMAX

MINMAX

II

IIV

To measure the contrast or visibility of these fringes, one To measure the contrast or visibility of these fringes, one may define a useful quantity, the fringe visibility:may define a useful quantity, the fringe visibility:

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Co-ordinates on screen

Use sin ≈ tan = y/L Then

These results are seen in the following Interference pattern

amLy

amLy

2

1min

max

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Phasor Representation of wave addition

Phasor representation of a wave E.g. E = Eosint is represented as a vector of

magnitude Eo, making an angle =t with respect to the y-axis

Projection onto y-axis for sine and x-axis for cosine

Now write,

tEE

as

trkEE

o

o

sin

,

sin

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Phasors

Imagine disturbance given in the form

222

111

sin

sin

tEE

tEE

Po

Po

==φφ22--φφ11

φφ11

φφ22

Carry out addition at t=0Carry out addition at t=0

1212

2121

212

22

12

,

cos2

,

180cos2

rrkwhere

IIIII

or

EEEEE PoPoPoPo

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Other forms of two-source interference

1r

Lloyd’s mirrorLloyd’s mirror

screenscreenSS

S’S’

2r

12

Other forms of two source interference

Fresnel BiprismFresnel Biprism

ss22

SS11

SS

dd ss

13

Other sources of two source interference

nn

Altering path length for rAltering path length for r22 rr11

rr22

With dielectric – thickness dWith dielectric – thickness d

krkr22 = k = kDDd + kd + koo(r(r22-d)-d)

= nk= nkood+ ko(rd+ ko(r22-d)-d)

= k= koorr22 + k + koo(n-1)d(n-1)d

Thus change in path length = k(n-1)dThus change in path length = k(n-1)d

Equivalent to writing, Equivalent to writing, 22 = = 11 + k + koo(n-1)d(n-1)d

Then Then = kr = kr22 – k – koorr1 1 = k= koo(r(r22-r-r11) + k) + koo(n-1)d(n-1)d

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Incidence at an angle

ii

a sin a sin

a sin a sin ii

1r

2r

Before slitsBefore slitsDifference in path lengthDifference in path length

After slitsAfter slitsDifference in path lengthDifference in path length

= = a sin a sin II in r in r11

= = a sin a sin in r in r22

Now k(rNow k(r22-r-r11) = - k a sin ) = - k a sin + k a sin + k a sin ii

Thus Thus = ka (sin = ka (sin - sin - sinii))

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Reflection from dielectric layer

Assume phase of wave at O (x=0, t=0) is 0

Amplitude reflection co-efficient (n1n2) = 12

(n2 n1) ’=21

Amplitude transmission co-efficient (n1n2) = 12

(n2 n1) ’= 21

Path O to O’ introduces a phase change

nn22nn11 nn11

AA

O’O’

OO

ttx = 0x = 0 x = tx = t

A’A’’’

’’

'cos

2

222

tSk

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Reflection from a dielectric layer

At O: Incident amplitude E = Eoe-iωt

Reflected amplitude ER = Eoe-iωt

At O’: Reflected amplitude Transmitted amplitude

At A: Transmitted amplitude Reflected amplitude

tSkioeE

22' tSki

oeE 22'

tSkioeE

222'' tSki

oeE 222''

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Reflection from a dielectric layer

tSkioA eEE 11''

AA

A’A’

z = 2t tan z = 2t tan ’’

and and ΔΔSS11= z sin = z sin = 2t tan = 2t tan ’ sin ’ sin

•At A’

96.0'2.0'12

12

andnn

nn

Since,Since,

The reflected intensities ~ 0.04IThe reflected intensities ~ 0.04Ioo and both beams (A,A’) will have and both beams (A,A’) will have

almost the same intensity.almost the same intensity.Next beam, however, will have ~ |Next beam, however, will have ~ |||33EEoo which is very small which is very small

Thus assume interference at Thus assume interference at , and need only consider the two , and need only consider the two beam problem.beam problem.

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Transmission through a dielectric layer

At O’: Amplitude ~ ’Eo ~ 0.96 Eo

At O”: Amplitude ~ ’(’)2Eo ~ 0.04 Eo

Thus amplitude at O” is very small

O’O’

O”O”

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Reflection from a dielectric layer

Interference pattern should be observed at infinity

By using a lens the pattern can be formed in the focal plane (for fringes localized at )

Path length from A, A’ to screen is the same for both rays

Thus need to find phase difference between two rays at A, A’.

AA

A’A’

z = 2t tan z = 2t tan ’’

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Reflection from a dielectric surface

tSkioA eEE 11''

AA

A’A’

z = 2t tan z = 2t tan ’’

tSkio

iA eEeE 222'

If we assume If we assume ’ ~ 1’ ~ 1

and since and since ’ = |’ = |||

This is just interference between two sources with equal amplitudes This is just interference between two sources with equal amplitudes

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Reflection from a dielectric surface

cos2 2121 IIIII

'cos2

sin'tan2'cos

2

2

2

12

1122

tkn

tknt

kn

skSk

o

oo

tSkio

iA eEeE 11

' tSki

oA eEE 222'

where,where,

Since Since kk22 = n = n22kkoo kk11=n=n11kkoo

and nand n11sinsin = n = n22sinsin’’ (Snells Law)(Snells Law)

Thus, Thus,

112212 2 skSk

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Reflection from a dielectric surface

Since ISince I11 ~ I ~ I22 ~ I ~ Ioo

Then, I = 2IThen, I = 2Ioo(1+cos(1+cos))

Constructive interferenceConstructive interference

Destructive interferenceDestructive interference

= = 2m 2m = 2ktcos = 2ktcos’ - ’ - (here k=n(here k=n22kkoo))

2ktcos2ktcos’ = ’ = (2m+1)(2m+1)

ktcosktcos’ = ’ = (m+1/2)(m+1/2)

2n2n22coscos’ = ’ = (m+1/2) (m+1/2)oo

2n2n22coscos’ = ’ = m moo

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Haidinger’s Bands: Fringes of equal inclinationdd

nn22

nn11

Beam splitterBeam splitter

ExtendedExtendedsourcesource

PPII PP22

PP

xx

ff

FocalFocalplaneplane

11

11

DielectricDielectricslabslab

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Fizeau Fringes: fringes of equal thickness

Now imagine we arrange to keep cos ’ constant We can do this if we keep ’ small That is, view near normal incidence Focus eye near plane of film Fringes are localized near film since rays diverge

from this region Now this is still two beam interference, but whether

we have a maximum or minimum will depend on the value of t

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Fizeau Fringes: fringes of equal thickness

cos2 2121 IIIII

where, where,

kt

kt

2

'cos2

Then if film varies in thickness we will see fringes as we move our eye.Then if film varies in thickness we will see fringes as we move our eye.

These are termed These are termed Fizeau fringesFizeau fringes..

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Fizeau Fringes

Extended sourceExtended source

Beam splitterBeam splitter

xx nn

nn22

nn

kt

kt

2

'cos2

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Wedge between two plates11 22

glassglassglassglass

airair

DDyy

LL

Path difference = 2yPath difference = 2yPhase difference Phase difference = 2ky - = 2ky - (phase change for 2, but not for 1) (phase change for 2, but not for 1)

Maxima 2y = (m + ½) Maxima 2y = (m + ½) oo/n/n

Minima 2y = mMinima 2y = moo/n/n

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Wedge between two plates

Maxima 2y = (m + ½) Maxima 2y = (m + ½) oo/n/n

Minima 2y = mMinima 2y = moo/n/n

Look at p and p + 1 maximaLook at p and p + 1 maxima

yyp+1p+1 – y – ypp = = oo/2n /2n ΔΔxx

where where ΔΔx = distance between adjacent maximax = distance between adjacent maxima

Now if diameter of object = DNow if diameter of object = D

Then LThen L = D = D

And (D/L) And (D/L) ΔΔx= x= oo/2n or /2n or D = D = ooL/2n L/2n ΔΔxx

airair

DDyy

LL

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Wedge between two platesCan be used to test the quality of surfacesCan be used to test the quality of surfaces

Fringes follow contour of constant yFringes follow contour of constant y

Thus a flat bottom plate will give straight fringes, otherwise Thus a flat bottom plate will give straight fringes, otherwise ripples in the fringes will be seen. ripples in the fringes will be seen.