1

Electromagnetic waves

Hecht, Chapter 2

Wednesday October 23, 2002

2

Electromagnetic waves: Phase relations

Thus E and B are in phase since,

requires that

trkio

trkio ecBeE

,...2,01 ie

EE

BBkk

3

Irradiance (energy per unit volume)

Energy density stored in an electric field

Energy density stored in a magnetic field

32 ,

2

1

m

JEu oE

32 ,

2

1

m

JBu

oB

c

EB 3

22

2 2

1

2

1

m

JE

c

Eu o

oB

4

Energy density

2Euuu oBE Now if Now if E = EE = Eoosin(sin(ωωt+t+φφ)) and and ωω is very large is very large

We will see only a time average of EWe will see only a time average of E

2

1sin

1sin 22

dttT

tTt

t

2

2

1ooEu

5

Intensity or Irradiance

k

In free space, wave propagates with speed cIn free space, wave propagates with speed c

c c ΔΔtt

AA

In time In time ΔΔt, all energy in this volume passes through A.t, all energy in this volume passes through A.Thus, the total energy passing through A is,Thus, the total energy passing through A is,

JoulestAcu

6

Intensity or Irradiance

cAut

P

Power passing through A is,Power passing through A is,

Define: Define: IntensityIntensity or or IrradianceIrradiance as the as the power per unit areapower per unit area

2

2

1ooEcI

cuI

7

Intensity in a dielectric medium

n

cvando

In a dielectric medium,In a dielectric medium,

Consequently, the irradiance or intensity is,Consequently, the irradiance or intensity is,

2

2

1oEvI

8

Poynting vector

DefineDefine

HEcS

cBEcS

EcS

oo

o

o

2

2

9

Poynting vector

BES

VectorPoyntingHES

o

1

For an isotropic media energy flows in the direction of propagation, soFor an isotropic media energy flows in the direction of propagation, soboth the magnitude and direction of this flow is given by,both the magnitude and direction of this flow is given by,

HEStII

The corresponding intensity or irradiance is then,The corresponding intensity or irradiance is then,

10

Example: Lasers

Spot diameter Intensity (W/m2) Electric field magnitude (V/m)

2 mm 1.6 X 103 1.1 X 103

20 µm (e.g. focus by lens of eye)

1.6 X 107 1.1 X 105

2 µm 1.6 X 109 1.1 X 106

Laser Power = 5mWLaser Power = 5mW

nb. Colossal dielectric constant material CaCunb. Colossal dielectric constant material CaCu33TiTi44OO1212 , , = 10,000 at 300K = 10,000 at 300K

Subramanian et al. J. Solid State Chem. 151, 323 (2000)Subramanian et al. J. Solid State Chem. 151, 323 (2000)

Near breakdownNear breakdownvoltage in watervoltage in water

c

IE

EcI

o

o

2

2

1 2

Same as Same as sunlight at earthsunlight at earth

oo = 8.854 X 10 = 8.854 X 10-12-12 CV CV-1-1mm-1 -1 (SI units)(SI units)

11

Reflection, Transmission and Interference of EM waves

12

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

13

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

14

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,

15

Reflection and Transmission at an interface

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

22

111 fv

vgf

16

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

17

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

18

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

19

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

20

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

21

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

22

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)

23

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

24

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

25

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

26

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))

27

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

28

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

29

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

30

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,

31

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

32

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