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Electromagnetic waves: Reflection, Transmission and Interference

Monday October 28, 2002

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Amplitude Transmission & Reflection

022

21

1

21

212

nn

n

vv

v

12

1212 nn

nn

For For normalnormal incidence incidence

Amplitude Amplitude reflectionreflection Amplitude Amplitude transmissiontransmission

Suppose these are plane wavesSuppose these are plane waves

itxkioR

txkioT

txkio eeEgeEfeEf 121121

3

Intensity reflection

12

1212 nn

nnee

E

E iio

oR

Amplitude reflection co-efficientAmplitude reflection co-efficient

and intensity reflectionand intensity reflection

2

12

12212

2

11

2

11

12

2121

nn

nn

I

I

Ev

EvR

o

R

o

oR

4

Intensity transmission

2

1

2

2

11

2

22

12

2121

o

oT

o

oT

o

T

E

E

n

n

Ev

Ev

I

IT

Intensity transmissionIntensity transmission

and in generaland in general

21122

121

212

n

nT

R + T = 1R + T = 1

(conservation of energy)(conservation of energy)

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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

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Young’s Double slit experiment

MonochromaticMonochromatic, , planeplane wave wave Incident on slits (or pin hole), SIncident on slits (or pin hole), S11, S, S22

separated by distance separated by distance aa (centre to centre) (centre to centre) Observed on screen Observed on screen L >> a L >> a ((LL- meters, - meters, aa – –

mm)mm) Two sources (STwo sources (S11 and S and S22) are ) are coherentcoherent and and in in

phasephase (since same wave front produces both (since same wave front produces both as all times)as all times)

Assume slits are very narrow (width Assume slits are very narrow (width b ~ b ~ ) ) so radiation from each slit alone produces so radiation from each slit alone produces

uniformuniform illumination across the screen illumination across the screen

AssumptionsAssumptions

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Young’s double slit experiment

slits at x = 0 The fields at S1 and S2 are

Assume that the slits might have different width Assume that the slits might have different width and therefore and therefore EEo1o1 E Eo2o2

tio

tio eEEeEE 2211

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Young’s double slit experiment

What are the corresponding E-fields at P?What are the corresponding E-fields at P?

trkioP

trkioP e

r

EEe

r

EE 21

2

22

1

11

Since Since L >> a (L >> a ( small) small) we can put we can put r = |rr = |r11| = |r| = |r22||

We can also put We can also put |k|k11| = |k| = |k22| = 2| = 2// ( (monochromatic sourcemonochromatic source))

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Young’s Double slit experiment

PPP EEE 21

22

2

1oPEvEvI PP

The total amplitude at PThe total amplitude at P

Intensity at PIntensity at P

**22

**21

*2

122121

21

PPPPPP

PP

P

EEEEEE

EEEE

EEE

PP

PP

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Interference Effects

Are represented by the last two terms

If the fields are perpendicular

then,

and,

**

1221 PPPPEEEE

222

21 PPPEEE

PPPIII

21

222

21 PPPEvEvEv

In the absence of interference, the total intensity is a simple sumIn the absence of interference, the total intensity is a simple sum

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Interference effects

Interference requires at least parallel components of E1P and E2P

We will assume the two sources are polarized parallel to one another (i.e.

PPPPPPEEEEEE

212121cos

12

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

21

Other forms of two source interference

Fresnel BiprismFresnel Biprism

ss22

SS11

SS

dd ss

22

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”