electromagnetic (e-m) theory of waves at a dielectric interface ii rr tt nini ntnt ûnûn x y. z b...

Electromagnetic (E-M) theory of waves at a dielectric interface

i r

t

ni

nt

ûn

x

y

.z

b

ik

rk

tk

iE

rE

tE

While it is possible to understand reflection and refraction from Fermat’s principle, we need to use E-M theory in order to understand quantitatively the relationship between the incident, reflected, and transmitted radiant flux densities:

We can accomplish this treatment by assuming incident monochromatic light waves which form plane waves with well defined k-vectors as shown in the diagram. The interface is shown with an origin and coordinates (x,y,z).

We will consider E-field polarizations which are (i) in the plane of incidence and (ii) perpendicular to the plane of incidence, as shown below.

),,(),,,(),,,( ttttrrrriiii kEIkEIkEI

Ii

Ir

It

E-field is perpendicular to the plane-of incidence

E-field is parallel to the plane of incidence

Maxwell’s Equations for time-dependent fields in matter

00

BSdB

t

DjHSd

t

DjrdH

t

BESd

t

BrdE

DdVSdD

HBEDEgj

HMMBH

EPPED

mo

eo

1

D – Displacement field

H – Magnetic Intensity

P – Polarization

M – Magnetization

- Magnetic permeability

- Permittivity

e - Dielectric Susceptibility

m - Magnetic Susc.

g – Conductivity

j – Current density

Summary of the boundary conditions for fields at an interface

Boundary

Side 1

Side 2

Maxwell’s equations in integral form allow for the derivation of the boundary conditions for the total fields on both sides of a boundary.

jHHu

uBB

EEu

uDD

n

n

n

n

12

12

12

12

ˆ

0ˆ

0ˆ

ˆ Normal component of D is discontinuous by the free surface charge density

Tangential components of E are continuous

Normal components of B are continuous

Tangential components of H are discontinuous by the free surface current density

For dielectrics, j = 0. Therefore, the components of E and H that are tangent to the interface must be continuous across it. Since we have Ei, Er, and Et the continuity of E components yield:

i r

t

ni

nt

ûn

x

y

.z

b

ik

rk

tk

iE

rE

tE

i 090

iii EE cossin

tnrin EuEEu

ˆˆ

iiinin EEuEu cossinˆˆ

Note that

tnrin EuEEu

ˆˆ

tttotn

rrrorniioin

tnrnin

trkEu

trkEutrkEu

EuEuEu

cosˆ

cosˆcosˆ

ˆˆˆ

Consider the expression on the interface (y = b) for all x, z and t. The above relationship must hold at all points and at any instant in time on the interface. Therefore

bytttbyrrrbyii trktrktrk |||

Since then we have tri

byttbyrrbyi rkrkrk |||

Thus, at the interface plane

cos.|ˆ

ˆ||0ˆ|

rconstrkandkkk

ukandkkurkk

byir

nirnrbyri

knu

r

r co

s

ttiitt

ii

ttii

itntbyti

rirriiirn

nncc

kk

kkurkkAlso

kkkku

sinsin,sin1

sin1

sinsin

0ˆ|,

sinsin0ˆ

which is again Snell’s Law

Case 1: E Plane of incidence

Continuity of the tangential components of E and H give

otoroi EEE

Cosines cancel

Using H = -1 B, the tangential components are

otoroitott

tioroi

i

i

tttt

iriii

riri

tttrrriii

tt

tr

i

ri

i

i

EEEandEn

EEn

Ev

EEv

vvand

vEBandvEBvEBSince

BBB

coscos)(

cos1

cos)(1

,

/,/,/

coscoscos

The last two equations give

coscos

cos2

coscos

coscos

t

ti

i

i

ii

i

oi

ot

t

ti

i

i

tt

ti

i

i

oi

or

nn

n

E

Eand

nn

nn

E

E

The symbol means E Plane of incidence. These are called the Fresnel equations; most often i t o.

Let r = amplitude reflection coefficient and t = amplitude transmission coefficient. Then, the Fresnel equations appear as

ttii

ii

oi

ot

ttii

ttii

oi

or

nn

n

E

Etand

nn

nn

E

Er

coscos

cos2

coscos

coscos

Note that t - r = 1

Case 2: E || Plane of Incidence

Continuity of the tangential components of E:

totrorioi EEE coscoscos

Continuity of the tangential components of -1 B:

it

tt

i

i

ii

i

oi

ot

tt

ti

i

i

ti

ii

t

t

oi

or

riri

ottt

orrr

oiii

nn

n

E

Etand

nn

nn

E

Er

and

Ev

Ev

Ev

coscos

cos2

coscos

coscos

;

111

||

||

||

||

If both media forming the interface are non-magnetic i t o then the amplitude coefficients become

itti

ii

itti

tiit

nn

ntand

nn

nnr

coscos

cos2

coscos

coscos||||

Using Snell’s law

the Fresnel Equations for dielectric media become

ttii nn sinsin

)cos()sin(

cossin2

)sin(

cossin2

,)tan(

)tan(

)sin(

)sin(

||

||

titi

it

ti

it

ti

ti

ti

ti

tandt

randr

Note that t - r = 1 holds for all i , whereas t|| + r|| = 1 is only true for normal incidence, i.e., i = 0.

Consider limiting cases for nearly normal incidence: i 0.

In which case, we have:

ti

ti

iirr

sin

sin00|| since 1sintan xxx

ittti

tittiti n

n cossincossincossincossinsin

Also, using the following identity with Snell’s law

Therefore, the amplitude reflection coefficient can be written as:

iitt

iitt

iti

t

iti

t

nn

nn

nnnn

ri

coscos

coscos

coscos

coscos

0

ti

i

it

it

nn

nttand

nn

nnrr

iiii

20||00||0

In the limiting case for normal incidence i=t = 0, we have:

Note that these equalities occur for near normal incidence as a consequence of the fact that the plane of incidence is no longer specified when i t 0.

Consider a specific example of an air-glass interface:

ni = 1

nt = 1.5

ttii nn sinsin

We will consider a particular angle called the Brewster’s angle: p + t = 90 .tan

i

tp n

n

3.56

1

5.1tan 1

p

At the polarization angle p, only the component of light polarized normal to the incident plane and therefore parallel to the surface will be reflected.

i

t

External reflection nt > ni

Internal reflection ni > nt

External Reflection (nt > ni) Internal Reflection (nt < ni)

n

i

tcttii n

nnn

tci sinsinsin 90

8.415.1/1sin 1c

7.33

5.1

1tan

'

'

p

i

tp n

n

3.56

1

5.1tan 1

p

1

5.1

i

tti n

nnConsider

5.1

1

i

tti n

nnConsider

Concept of Phase Shifts () in E-M waves : trkEE o

cos

0)sin(

)sin(

ti

ti

oi

or

E

Er

Since when nt > ni and t < i

as in the Air Glass interface,

we expect a reversal of sign in the electric field for the Ecase when

.1 forei

We need to define phase shift for two cases:

A. When two fields E or B are to the plane of incidence, they are said to be (i) in-phase (=0) if the two E or B fields are parallel and (ii) out-of-phase ( = ) if the fields are anti-parallel.

B. When two fields E or B are parallel to the plane of incidence, the fields are (i) in-phase if the y-components of the field are parallel and (ii) out-of-phase if the y-components of the field are anti-parallel.

Examples of Phase shifts for two particular cases:

,,

,

0,,

,

phaseofoutBB

EE

phaseinEE

BB

ri

ri

ti

ti

)a( )b(

0,,

,,

phaseinallBBB

EEE

tri

tri

Glass (n = 1.5) Air (n = 1)

Air (n = 1) Glass (n = 1.5)

Analogy between a wave on a string and an E-M wave traversing the air-glass interface.

= 0 = 0

= = 0

Compare with the case of .0 ior forE

Examples of phase-shifts using our air-glass interface:

In order to understand these phase shifts, it’s important to understand the definition of .

Reflected E-field orientations at various angles for the case of External Reflection (ni < nt). It is worth checking and comparing with the various plots for the phase shift on the previous slides.

Reflected E-field orientations at various angles for the case of Internal Reflection (ni > nt). It is worth checking and comparing with the various plots for the phase shift on the previous slides.

Reflectance and Transmittance

Remember that the power/area crossing a surface in vacuum (whose normal is parallel to the Poynting vector) is given by .2 BEcS o

The radiant flux density or irradiance (W/m2) is

TToo

TEcEvE

cSI 222

2

/1vPhase velocity

From the geometry and total area A of the beam at the interface, the power (P) for the (i) incident, (ii) reflected and (ii) transmitted beams are:

ttt

rrr

iii

AIPiii

AIPii

AIPi

cos)(

cos)(

cos)(

Define Reflectance and Transmittance:

2

2

2

2

2

2

cos

cos

cos

cos

cos

cos

2/

2/

cos

cos

tn

n

E

E

n

n

AI

AI

P

PT

RrE

E

Ev

Ev

I

I

AI

AI

P

PR

ii

tt

oi

ot

ii

tt

ii

tt

i

t

oi

or

oiii

orrr

i

r

ii

rr

i

r

Note that

2

2222

22

,

1

2

11

2

1

2

1

2

1

2

1

o

oo

oo

oo

oo

oo

EnITherefore

Ec

nEc

c

vEEvEI

Conservation of Energy at the interface yields:

riritottrorrioii

ttrrii

nnEnEnEn

AIAIAI

,222 ;coscoscos

coscoscos

Therefore, TRE

E

n

n

E

E

oi

ot

ii

tt

oi

or

22

cos

cos1

We can write this expression in the form of componets and:||

2||||

22||||

2

cos

cos

cos

cos, t

n

nTandt

n

nTrRandrR

ii

tt

ii

tt

11, |||| TRandTRTherefore

1

5.1

i

tti n

nnConsider

We must use the previously calculated values for ttrr ,,, ||||

1

42

4

,0

2

22

2||

2

||

it

itiitt

it

it

it

iti

nn

nnnnnnTR

nn

nnTTTand

nn

nnRRRWhen

It’s possible to verify for the special case of normal incidence:

Consider the case of Total Internal Reflection (TIR):

ttii nn sinsin

5.1

1

i

tti n

nnConsider

t

i

nt = 1

ni = 1.5

8.41

5.1

1sin,90

c

i

tit n

nWhen

c