1

+

-

+B

E

Chapter 4 The Propagation of LightSeptember 19,22 Scattering and interference

4.1 IntroductionScattering: The absorption and prompt re-emission of electromagnetic radiation by electrons in atoms and molecules.Scattering plays an essential rule in light propagation.Transmission, reflection and refraction are all macroscopic manifestations of scattering by atoms and molecules.

4.2 Rayleigh scatteringRayleigh scattering: Elastic scattering of light by molecules which are much smallerthan the wavelength of the light.

Re-emission of light from an induced dipole:

2

2

032

420

0

20

sin

32

)()(

)cos(

4

sin)(

rc

pI

r

tkrkpE

Rayleigh’s scattering law:Why is the sky blue? Why is the sunset red?

4/1 I

2

4.2.1 Scattering and interferenceScattering in tenuous media (d >):

Lateral scattering: random phase m

Summation is a random walk (no interference).

nI

nE

nen

eeeS

nS

SeSeSS

eSSeS

n

jmjm

i

n

jm

in

j

in

m

in

n

ni

ni

nn

inn

n

m

in

jm

jmjm

nn

nm

,1,

)(

1,

)(

11

2

2

2*21

11

||

or

||

1||))((||

,

11

1

Forward scattering: in phaseSummation is a constructive interference.

2

1

nI

nE

neeS in

m

in

P

P

Primary light

Lateral scattering

Forward scattering

3

4.2.2 The transmission of light through dense mediaScattering in dense media (d<<):

Lateral scattering: destructive interferenceWherever there is an atom A, there will be an atom A' /2 away laterally.There is little or no light scattered laterally or backwards in a dense homogenous media.

Forward scattering: in phase, constructive interference.

Tyndall and Mie scattering:When the size of the molecules increases, scattering of the longer wavelengths increases proportionally. (Destructive interference starts for the scatter of short wavelength).Why is the cloud (or milk) white?

large.) is if that (Note 2 nnn

P

P

Primary light

Lateral scattering

Forward scattering

4

0

0

220

1

220

1

222220

0

0220

1

222220

220

,2

,2

0 and ,0 ,tan lag phase Dipole

tanexp/

tanexp/

)(/

)(

timEq

eEimq

tEi

mqtx

ee

tieeee

1) Phase lag of the oscillation of the damped oscillator:(It is easier to use the phase factor eit when discussing phase lag or phase lead.)

4.2.3 Transmission and the index of refraction

Why does n >1 lead to v < c?Both the primary and the secondary waves propagate with speed c in the space between atoms. They overlap and generate the transmitted wave.

+

-Primary Primary +secondary

5

Note that the “+ ” comes from our initial definition ofIf we instead define then

.)( 0tieEtE

,)( ,)( 00titi extxeEtE

.lag phase same thehave we,tanexp

/

tanexp/

)(/

)(

220

1

222220

0

0220

1

222220

220

timEq

eEimq

tEi

mqtx

ee

tieeee

2) Additional phase lag of /2 for emitting the secondary wave (Problem4.5):

)]2/(exp[)1()](exp[

)2/exp()1(1)](exp[

])1(1)][(exp[

)0let Then( )1(exp

)(exp

kxtixnkkxti

ixnkkxti

xnikkxti

xxnkkxti

xknxxktiE

P

x

xtieE

6

Total phase lag = + /2, < 3.Note that a phase lag of > is equivalent to a phase lead of 2

Transmitted = primary + secondarySuccessive phase shift at each atom is tantamount to a difference in phase velocity.

Re(n)

Primary

SecondaryTransmitted

7

Read: Ch4: 1-2Homework: Ch4: 2,4,5Due: September 26

8

September 24 Reflection and refraction

4.3 ReflectionReflection: Backward scattering at an interface.Mechanism: Consider the sum of all the backward scattered wavelets at an observation point close to the interface. The backward wavelets from different dipoles inside a uniform dense medium are paired and cancel each other because of their phases. However, at the interface, this cancellation is incomplete because of their amplitudes, which results in a net reflection from a thin layer of about /2 deep on the surface.

External reflection: nincident < ntransmitting

Internal reflection: nincident > ntransmitting

P

9

4.3.1 The law of reflectionRay: A line drawn on the direction of the flow of the radiant energy.Perpendicular to the wavefronts in an isotropic medium.

riADBC

1) The angle of incidence equals the angle of reflection.2) The incident ray, the normal of the surface, and the reflected ray are all in one plane (called the plane of incidence).

Normal incidence: Glancing incidence:

0i90i

The law of reflection:

Specular reflectionDiffuse reflection

A

B

C

D

ir

10

4.4 RefractionRefraction: Forward scattering on an interface + primary beam.

4.4.1 The law of refraction

ttii

tt

ii

t

t

i

i

ti

ti

nn

v

cn

v

cn

vvtvACtvBD

ACADBD

sinsin

,

sinsin

,

sin1sin

Snell’s law:

1)

2) The incident, reflected, and refracted rays all lie in the plane of incidence.

ttii nn sinsin

A

B

C

Di

t

vit

vtt

11

'" '

vt '

4.4.2 Huygens’s principleHuygens’s principle (1690): How can we determine a new wavefront ' from a known wavefront ? 1) Every point on a propagating wavefront acts as the source of spherical secondary wavelets, the wavefront at a later time is the envelope of these wavelets.2) The secondary wavelets have the same frequency and speed as the propagating wave.

• Proposed long before Maxwell• First step toward scattering theory• Interference not included• Primary wave completely scattered• Still useful in inhomogeneous media

12

Read: Ch4: 3-4Homework: Ch4: 8,12,22,23,25Due: October 3

13

4.5 Fermat’s principleFermat’s principle (1657): The actual path between two points taken by a beam of light is the one that is traversed in the least time. Principle of least time.Example: refraction

S

O

P

h

b

a

x

i

t

ni

nt

t

t

i

i

ti

titi

vv

xabv

xa

xhv

x

dx

dt

v

xab

v

xh

vvt

sinsin

0)(

)(

)(OPSO

2222

2222

Optical path length (OPL):

Light traverses the route that have the smallest optical path length.

P

SdssnOPL )(

ni

September 26 Fermat’s principle

c

OPL

c

nds

nv

nds

v

dst

P

SP

S

P

S

14

Optical mirages in the atmosphere:

dz

dT

T

P

dz

dn

T

Pn

TNkPTkNPV

Nni

f

m

Nqn

BBm

j jj

j

e

e

2

2200

22

1

11)(

0dz

dT

Inferior mirage(road, desert)

0dz

dT

Superior mirage(sea, airplane)

15

Modern formulation of Fermat’s principle: The actual light ray going from point S to point P must traverse an optical path length that is stationary with respect to variations of that path.

Stationary path stationary phase constructive interferenceNon-stationary path varying phase destructive interference

Stationary paths:An actual OPL for a ray can be stationary, minimum, or maximum.

S P

S P

OPL×/c=

16

Read: Ch4: 5Homework: Ch4: 32,34Due: October 3

17

September 29,October 1 Electromagnetic approach

4.6 The electromagnetic approachElectromagnetic theory provides a complete description on light propagation in media, regarding its amplitude, polarization, phase, wavelength, direction, and frequency.

4.6.1 Waves at an interfaceBoundary conditions: 1) The tangent component of the electric field E is continuous on the interface.

2) Similarly, the tangent component of H (= B/) is continuous on the interface.

)](exp[0 ti rkΕ

tt

tztz

txtxtxtxAC EEEE

EELEEdt

d21

21

2121

Similarly

0)(

SB

lE

yx

E1tx

E2tx

.02

2

1

1

2

2

1

1

tttt

ACAC

BBL

BB

dt

ddt

d

SE

JlB

SE

JlB

18

)](exp[ˆ)](exp[ˆ)](exp[ˆ

)](exp[ˆ)](exp[ˆ)](exp[ˆ

ˆˆˆ

ˆˆ)(ˆˆ

ˆˆˆ

000

0000

0

0

tan

tzkxkitzkxkitxki

tititi

ttztxtnrrzrxrniixin

ytttnrrrniiin

ytnrnin

ytnnrinn

nnn

ΕuΕuΕu

rkΕurkΕurkΕu

EuEuEu

EuuEEuu

EuuEuEE

Deriving the laws of reflection and refraction using the boundary conditions:Let y =0 be the interface plane, and be the unit vector of the normal of the interface.Let z=0 be the plane-of-incidence, so that and kiz=0.

The three participating waves are:

nu

1) Incident wave:

2) Reflected wave:

3) Refracted wave:

Here E0i, E0r, E0t are complex amplitudes.

The boundary condition (at the interface y = 0) is now

)](exp[

)](exp[

)](exp[

0

0

0

ti

ti

ti

tttt

rrrr

iiii

rkΕE

rkΕE

rkΕE

tan2tan1 EE

ir

t

nu krki

kt

y=0 x

y

)ˆ//(ˆ ni ukz

19

• At the (x=0, z=0) point on the interface, this indicates a phasor sum of a(t)+ b(t) = c(t) at all time with fixed lengths a, b, and c, which is only possible when the three phasor vectors have an identical angular velocity (but may have different initial phases). Therefore r=t=i.

• Similarly, we have krx=ktx=kix by fixing (z=0,t=0) and varying x on the interface.

• Similarly, we have krz=ktz (=kiz)=0 by fixing (x=0,t=0) and varying z on the interface.

The laws of reflection and refraction:

1) krx=kix, krz=kiz =0 a) The incident beam, the reflected beam and the normal are in one plane;b) kisini = krsinr. Since ki= kr, we have i = r.

2) ktx=kix, ktz=kiz =0 a) The incident beam, the refracted beam and the normal are in one plane;b) kisini = ktsint. Since ki/ni= kt/nt, we have ni sini = nt sint.

)](exp[ˆ)](exp[ˆ)](exp[ˆ 000 tzkxkitzkxkitxki ttztxtnrrzrxrniixin ΕuΕuΕu

20

4.6.2 The Fresnel equationsHow much are E0r and E0t for a given E0i? Answer: Solving the two boundary conditions.

s polarization (TE mode) and p polarization (TM mode):

Ei

Bi

ki

i

x

ys polarization

Bi

Ei

ki

i

x

yp polarization

Eis

Bis ki

i r

t

ni

nu

Ers

Brs

kr

Ets

Bts kt

ntx

y

Brp

Erp

Btp

Etp

Question: will an s-polarized incident light produce both p and s polarized reflected and refracted light?

Choosing the positive directions of the fields so that at large angles of incidence they appear to be continuous.

Boundary condition 1, continuity of Etan:

ttprrp EE coscos

Bip

Eip

Boundary condition 2, continuity of Btan/ :

ttptirpittpirp EnEnBB ////

Solution for the p-polarization: .0 tprp EE

Therefore, s-polarized light won’t produce any p-polarized reflection or refraction, and vise versa.

21

I) E perpendicular to the plane-of-incidence (s polarization):

tritri ΕΕΕΕΕΕ 000 Boundary condition 1, continuity of Etan:

Boundary condition 2, continuity of Btan/ :

ttt

tiri

i

i

ririri

tt

tr

r

ri

i

i

En

EEn

vvcnEvEB

BBB

coscos)(

, , ,//

coscoscos

000

000

Ei

Bi

ki

i r

t

ni

nu

Er

Br

kr

Et

Btkt

ntx

y

Solving these two equations for E0r and E0t, using Cramer’s rule, we have

coscos

cos2

coscos

coscos

0

0

0

0

tt

ti

i

i

ii

i

i

t

tt

ti

i

i

tt

ti

i

i

i

r

nn

n

E

Enn

nn

E

E

For dielectrics with i ≈ t ≈ 0, we have theAmplitude reflection (and transmission) coefficients:

coscos

cos2

coscos

coscos

0

0

0

0

ttii

ii

i

t

ttii

ttii

i

r

nn

n

E

Et

nn

nn

E

Er

22

II) E parallel to the plane-of-incidence ( p polarization):

ttiri

ttrrii

ΕΕΕ

ΕΕΕ

coscos)(

coscoscos

000

000

Boundary condition 1, continuity of Etan:

Boundary condition 2, continuity of Btan/ :

tt

tri

i

i

ri

t

t

r

r

i

i

En

EEn

cnEvEB

BBB

000

000

)( ,//

Solving these two equations for E0r and E0t, using Cramer’s rule, we have

coscos

cos2

coscos

coscos

//0

0

//0

0

it

tt

i

i

ii

i

i

t

it

tt

i

i

ti

ii

t

t

i

r

nn

n

E

Enn

nn

E

E

For nonmagnetic dielectrics, the amplitude reflection (and transmission) coefficients:

coscos

cos2

coscos

coscos

//0

0//

//0

0//

itti

ii

i

t

itti

tiit

i

r

nn

n

E

Et

nn

nn

E

Er

Bi

Ei

kii r

t

ni

nu

Br

Er kr

Bt

Et

kt

ntx

y

23

Finally using Snell’s law

we have the Fresnel equations:

,sinsin

sinsini

t

t

ittii

nnnn

Note that the negative sign is from our definition of the positive E field direction for each beam.

)cos()sin(

cossin2

coscos

cos2

)sin(

cossin2

coscos

cos2

)tan(

)tan(

coscos

coscos

)sin(

)sin(

coscos

coscos

//

//

titi

it

itti

ii

ti

it

ttii

ii

ti

ti

itti

tiit

ti

ti

ttii

ttii

nn

nt

nn

nt

nn

nnr

nn

nnr

24

Read: Ch4: 6Homework: Ch4: 37,39,40,41Due: October 10

25

Sum and difference formulas:

Double-angle formulas:

Half-angle formulas:

Product to sum formulas:

Sum to product formulas:

Trigonometric Formulas

26

October 3 Reflectance and transmittance

4.6.3 Interpretation of the Fresnel equations

I. Amplitude coefficients

Externalreflection

Internalreflection

27

1) At normal incidence, . 00//

it

it

nn

nnrr

ii

2) When is the

polarization angle (Brewster angle), where r// = 0.

pi

titi n

n arctan ,90

3) For external reflection, at glancing incidence, .1

9090// ii

rr

4) For internal reflection, when is the critical angle,

where

ci

tit n

n arcsin ,90

.1// rr

E

r// = 0

Brewsterangle

i

)cos()sin(

cossin2

coscos

cos2

)sin(

cossin2

coscos

cos2

)tan(

)tan(

coscos

coscos

)sin(

)sin(

coscos

coscos

//

//

titi

it

itti

ii

ti

it

ttii

ii

ti

ti

itti

tiit

ti

ti

ttii

ttii

nn

nt

nn

nt

nn

nnr

nn

nnr

Externalreflection

Internalreflection

28

II. Phase shift:

mission.for trans )( argarg and

,reflectionfor argarg

0

0

0

0

tE

E

rE

E

i

t

i

r

Externalreflection

29

1 ,cos

cos

cos

cos

cos

cos

powerIncident

power dTransmitte)( nceTransmitta

cos

cos

powerIncident

power Reflected)( eReflectanc

2

1

2

1

2

2

0

0

2

2

0

0

200

20

TRtn

n

E

E

n

n

AI

AIT

rE

E

I

I

AI

AIR

EcnEvI

ii

tt

i

t

ii

tt

ii

tt

i

r

i

r

ii

rr

III. Reflectance and transmittance:

A

Acosi

)(

4 , incidence, normalAt

2//

2

it

it

it

it// nn

nnTTT

nn

nnRRR

Externalreflection

Internalreflection

30

Read: Ch4: 6Homework: Ch4: 45,49(Optional),56,58Due: October 10

31

October 6 Total internal reflection

4.7 Total internal reflectionTotal internal reflection: For internal reflection, when i ≥ c= arcsin(nt/ni), all the incoming energy is reflected back into the incident medium.

I) Reflection

)exp(||sincos

sincos

sincos

sincos

coscos

coscos

)exp(||sincos

sincos

sincos

sincos

coscos

coscos

arcsinarcsin

////222

222

222

222

//

22

22

22

22

irnin

nin

nn

nn

nn

nnr

irni

ni

n

n

nn

nnr

nn

n

tiiiti

tiiiti

itiiti

itiiti

itti

tiit

tiii

tiii

itii

itii

ttii

ttii

citii

tci

1) Reflectance: total reflectance.,1|||| ,1|||| 2////

2// rRrRrr

2) Phase:

iti

tii

i

tii

n

n

n

rr

cos

sin

2tan

cos

sin

2tan

then,)arg( ,)arg(Let

2

22//

22

////

32

// Relative phase difference between r┴ and r//:

i

tiii n

2

22

//

//

sin

sincos

2tan

2tan1

2tan

2tan

2tan

Maximum relative phase difference m:

ti

tim

ti

tii

tiii

ititi

i n

n

n

n

n

nn

d

d

2

1

2tan

1

2sin0

sinsin

sin)1(22tan 2

2

22

223

222

33

* (Reading) Producing a circular polarization:1) Let m =/2, we have nti= 0.414, or ni= 2.414 if nt= 1 (air). This means ni ≥ 2.414 is required. These materials are not easily obtainable.

2) Let us try two reflections. Let m =/4, we have ni= 1.497. For a glass of n =1.51, i = 48º37' and 54º37' will make =/4.

Fresnel’s rhomb:

54º37'Linearpolarization

Circularpolarization

34

II) Transmission: the evanescent wave

txn

kiy

nk

tyn

ikxn

ki

tyn

kxn

ki

tykxki

ti

ti

it

ti

itt

ti

it

ti

itt

ti

it

ti

itt

ttttt

ttt

sin

exp1sin

exp

1sinsin

exp

sin

1sin

exp

cossinexp

exp

2

2

0

2

2

0

2

2

0

0

0

Ε

Ε

Ε

Ε

rkΕE

t

i

kt

ktcost

ktsint

x

y

1) The wave propagates along the x axis.2) The amplitude decays rapidly in the y-direction within a few wavelengths.3) Energy circulates back and forth across the interface, but averaged in a zero net

energy flow through the boundary.

Properties of evanescent waves:

35

Frustrated total internal reflection (FTIR):When an evanescent wave extends into a medium with higher index of refraction, energy may flow across the boundary (similar to tunneling in quantum mechanics).

FTIRbeam splitter

d

36

Read: Ch4: 7Homework: Ch4: 59,60,61,69,76 (Optional)Due: October 17