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Electromagnetic waves: Reflection, Refraction and Interference

Friday October 25, 2002

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Optical cooling

mm

Photonsvv

Because of its motion, the atoms “see” an incoming photon with a frequency Doppler-shifted upward by,

Laser frequency (fL) chosen to be just below the resonance frequency of the atom (fo)

fo = fL(1+v/c)c

vf

vkL

2

k

3

Photons of the right frequency will be absorbed by the atom, whose speed is reduced because of the transfer of the momentum of the photon,

Emission occurs when the atom falls back to its ground state. However, the emission is randomly directed

An atom moving in the opposite direction, away from the light source, sees photons with a frequency, ffLL(1-(1-vv/c),/c), far enough from fo that there can be little or no absorption, and therefore no momentum gainno momentum gain.

Radiation pressure force : As v decreases, there must be a corresponding change in the laser frequency…..

One must have three mutually perpendicular laser beamsthree mutually perpendicular laser beams in order to reduce the speed of the atoms in all directions.

Optical Cooling

L

fLi

kvm

mvkmv

0

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Reflection, Transmission and Interference of EM waves

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Reflection and Transmission at an interface

tvxf 11 tvxf 22

tvxg 11

11 22

Normal Incidence – Two media characterized by vNormal Incidence – Two media characterized by v11, v, v22

incidentincident transmittedtransmitted

reflectedreflected

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Reflection and Transmission at an interface

Require continuity of amplitude at interface:f1 + g1 = f2

Require continuity of slope at interface:f1’ + g1’ = f2’

Recall u = x – vt

',' vft

u

u

f

t

f

du

f

x

ff

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Reflection and Transmission at an interface

t

f

vt

g

vt

f

v

2

2

1

1

1

1

111

Continuity of slope requires,Continuity of slope requires,

t

f

v

v

t

g

t

f

2

2

111

or,or,

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Reflection and Transmission at an interface

Integrating from t = - to t = t Assuming f1(t = - ) = 0 Then,

22

111 fv

vgf

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Amplitude transmission co-efficient ()

21

2

1

212

2

vv

v

f

f

Medium 1 to 2Medium 1 to 2

21

1

1

221

2

vv

v

f

f

Medium 2 to 1Medium 2 to 1

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Amplitude reflection co-efficient ()

21

12

1

112 vv

vv

f

g

12

12

12

1212 nn

nn

vv

vv

11 n

cv At a dielectric interfaceAt a dielectric interface

22 n

cv

11

Phase changes on reflection from a dielectric interface

12

1212 nn

nn

120

1212 ie

nn22 > n > n11 nn22<n<n11

Less dense to more denseLess dense to more densee.g. air to glasse.g. air to glass

More dense to less denseMore dense to less densee.g. glass to aire.g. glass to air

ie1212

phase change on reflectionphase change on reflection NoNo phase change on reflection phase change on reflection

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Phase changes on transmission through a dielectric interface

022

21

1

21

212

nn

n

vv

v

Thus there is no phase change on transmissionThus there is no phase change on transmission

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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

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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

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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

21

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

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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