Oblique incidence: Interface between dielectric mediaamanogawa.com/archive/docs/EM13.pdfOblique...

Electromagnetic Fields

© Amanogawa, 2006 – Digital Maestro Series 165

Oblique incidence: Interface between dielectric media Consider a planar interface between two dielectric media. A plane wave is incident at an angle from medium 1. The interface plane defines the boundary between the media. The plane of incidence contains the propagation vector and is both perpendicular to the interface plane and to the phase planes of the wave.

θi θrβ

Plane of incidence

Interface plane

Medium 1ε1 = εr1 εo µ1 = µr1 µo

Medium 2ε2 = εr2 εo µ2 = µr2 µo

Phase plane

θt



There are two elementary orientations (polarizations) for the electromagnetic fields: Perpendicular Polarization

The electric field is perpendicular to the plane of incidence and the magnetic field is parallel to the plane of incidence.

The fields are configured as in the Transverse Electric (TE) modes.

Parallel Polarization

The magnetic field is perpendicular to the plane of incidence and the electric field is parallel to the plane of incidence.

The fields are configured as in the Transverse Magnetic (TM) modes.

Any plane wave with general field orientation can be obtained by superposition of two waves with perpendicular and parallel polarization.



Perpendicular (TE) polarization

Medium 1 ε1 = εr1 εo µ1 = µr1 µo

z

x

y

iE

iβxβ

zβ

iθ

×

×

rθ×

rE

iH

rH rβ

× tHtE

tθ

tβ


Incident wave

Reflectedwave

Transmitted wave



The electric field phasors for the perpendicular polarization, with reference to the system of coordinates in the figure, are given by

The propagation vector components in medium 1 are expressed as

i

r

t

Ê

Ê

Ê

ix iz

rx rz

tx t z

j x j zyi y

j x j zyr y

j x j zyt y

E e i

E e i

E e i

β β

β β

β β

− ⋅ − ⋅

− ⋅ − ⋅

− ⋅ − ⋅

=

=

=

2 21 1 1

1 1

2 21

1 1

cos sin

cos sin

i ix iz

ix i iz i

r rx rz

rx r rz r

β β β β ω µ ε

β β θ β β θ

β β β β

β β θ β β θ

= + = =

= =

= + =

= − =



The propagation vector components in medium 2 are expressed as

The magnetic field components can be obtained as

2 22 2 2

2 2cos sint tx tz

tx t tz t

β β β β ω µ ε

β β θ β β θ

= + = =

= =

( )

( )

( )

ii

1 1

rr

1 1

tt

2 2

E ˆ ˆH sin cos

E ˆ ˆH sin cos

E ˆ ˆH sin cos

ix iz

rx rz

tx tz

j x j zyiii x i z

j x j zyrrr x r z

j x j zyttt x t z

Ei i e

Ei i e

Ei i e

β β

β β

β β

β θ θωµ η

β θ θωµ η

β θ θωµ η

− −

− −

− −

×= = − +

×= = − +

×= = − +



Assuming that the amplitude of the incident electric field is given, to completely specify the problem we need to find the amplitude of reflected and transmitted electric field. The boundary condition at the interface (x = 0) states that the tangential electric field must be continuous. Because of the perpendicular polarization, the tangential field is also the total field The relation above must be valid for any choice of “z” and we must have (phase conservation law)

The first equality indicates that the reflected angle is the same as the incident angle.

)0 i z rz tzj z j z j zyi yr ytE e E e ex Eβ β β− − −

+ ==

iz rz tzβ β β= =

1 1sin siniz rz i r i rβ β β θ β θ θ θ= ⇒ = ⇒ =



The second equality provides the transmitted angle

Since we have also the boundary condition for the electric field becomes

1 11

2 2sin sint i

µ εθ θ

µ ε−

⇒ =

Snell's Law

i z rz tzj z j z j ze e eβ β β− − −= =

yi yr ytE E E+ =

1 2sin siniz tz i tβ β β θ β θ= ⇒ =



The tangential magnetic field must also be continuous at the interface. This applies in our case to the z−components

Solution of the system of boundary equations gives

1 1 2

1

2

cos cos cos

coscos

y i y r y ti i t

ty

z i z r z t

i y r y ti

E E E

E E

H H H

E

θ θ θη η η

η θη θ

− =

⇒ −

+ =

=

2 1

2 1

cos cos( )cos cos

yr i t

yi i t

EE

Eη θ η θη θ η θ⊥

−Γ = =

+Reflection coefficient

2

2 1

2 cos( )cos cos

yt i

yi i t

EE

Eη θτ

η θ η θ⊥ = =+

Transmission coefficient



For the magnetic field, we can define the reflection coefficient as

In terms of electric field, the magnetic field components are

The reflection coefficient for the magnetic field is then

1 2

2 1

cos cos( ) ( )cos cos

y r t i

y i i t

EH E

Eη θ η θη θ η θ⊥ ⊥

− −Γ = = −Γ =

+

1

1

cos cos

cos cos

yrzr i r i

yizi i i i

EH H

EH H

θ θη

θ θη

−= = −

= =

( ) zr rzi i

H HHH H⊥Γ = = −



The transmission coefficient is defined as

The magnetic field components are

The transmission coefficient for the magnetic field is then

( ) t

i

HHH

τ⊥ =

112 2 1

2 cos( ) ( )cos cos

t i

i i t

HH EH

η θητ τη η θ η θ⊥ ⊥= = =

+

2

1

y tt

y ii

EH

EH

η

η

=

=



Parallel (TM) polarization


z

x

y

iEiβ

xβ

zβ

iθ

×

×

rθ×

rE

iH

rH

rβ

× tH

tEtθ

tβ


Incident wave

Reflectedwave

Transmitted wave



The magnetic field phasors for the parallel polarization are given by

and the electric field components can be obtained as

i

r

t

ˆH

ˆH

ˆH

ix iz

rx rz

tx t z

j x j zyi y

j x j zyr y

j x j zyt y

H e i

H e i

H e i

β β

β β

β β

− ⋅ − ⋅

− ⋅ − ⋅

− ⋅ − ⋅

=

=

=

( )

( )

( )

ii 1

1

rr 1

1

tt 2

2

H ˆ Ê sin cos

H ˆ Ê sin cos

H ˆ Ê sin cos

ix iz

rx rz

tx tz

j x j ziyi i x i z

j x j zryr r x r z

j x j ztyt t x t z

H i i e

H i i e

H i i e

β β

β β

β β

β η θ θωεβ η θ θωε

β η θ θωε

− −

− −

− −

×= − = −

×= − = +

×= − = −



Also for parallel polarization one can verify that the same relationships between angles apply, as found earlier for the perpendicular polarization, including Snell’s law

We have again two boundary conditions at the interface. One condition is for continuity of the tangential magnetic field

yi yr ytH H H+ =

1 11

2 2sin sin

i r

t i

θ θ

µ εθ θ

µ ε−

=

=



A second condition is for continuity of the tangential electric field

From the equations provided by the boundary conditions we obtain the reflection and transmission coefficients for the magnetic field of a wave with parallel polarization as

1 1 2

2

1

cos cos cos

coscos

zi

i yi i yr t yt

tyi yr y

i

zr zt

t

H H H

H H H

E E E

η θ η θ η θ

η θη θ

− + = −

⇒ − =

+ =

1 2

1 2

1

1 2

cos cos( )cos cos

2 cos( )cos cos

yr i t

yi i t

yt i

yi i t

HH

H

HH

H

η θ η θη θ η θ

η θτη θ η θ

−Γ = =

+

= =+



The reflection coefficient for the electric field is defined as

The tangential components of the electric field can be expressed in terms of magnetic field as

The reflection coefficient for the electric field is

( ) zr rzi i

E EEE E

Γ = = −

2 1

1 2

cos cos( ) ( )cos cos

zr yr t i

zi yi i t

E HE H

E Hη θ η θη θ η θ

−Γ = = − = −Γ =

+

1

1

cos cos

cos coszr r i i yr

zi i i i yi

E E H

E E H

θ η θ

θ η θ

= =

= − = −



The transmission coefficient for the electric field is defined as

The electric field components are given by

The transmission coefficient for the electric field becomes

2 21 1

1 221 1 2 1 2

( ) ( )

2 cos 2 coscos cos cos cos

ytt

i yi

i i

i t i t

HEE HE H

η ητ τη η

η θ η θηη η θ η θ η θ η θ

= = =

= =+ +

( ) t

i

EE

Eτ =

2

1

t yt

i yi

E H

E H

η

η

=

=



Considerable simplifications are possible for the common case of nonmagnetic dielectric media with

First of all, Snell’s law becomes

or, equivalently

Snell’s law provides then a useful recipe to express the reflection and transmission coefficients only with angles, thus eliminating the explicit dependence on medium impedance.

1 2 oµ µ µ= =

1 11 1 12 2 2

sin sin sin sint i iµ ε εθ θ θµ ε ε

− − = =

)2 21 1

sin (sin

i

t

n nn

θ εθ ε

= = = index of refraction



After some trigonometric manipulations, we obtain the following table of simplified coefficients for electric and magnetic field

( )( )( )( )

( )

( ) ( )

1

2

1

2

2sin cos( ) ( )sin

2sin cos( ) ( )

sin( ) ( )

sint

sin cos

an( ) ( )

tan

i t

i t

i t

t i

i t

i

t

t

i

i t i t

E H

E H

E

E H

H

ε θ θτ τθ θε

ε

θ θθ θ

θ θτ τ

θ θθ θ

θ θ θ θε

⊥ ⊥

⊥ ⊥

−Γ = −Γ = −

+

−Γ

= =+

= =+ −

= −Γ = −+



Power flow The time-average power flow normal to the interface must be continuous. We can express this as

We define the reflection and transmission coefficients for the time-average power as

2

2

21

2 2

coscos

r

i

t t

ii

ERE

ETE

θηη θ

= =

= =

reflected power

incident power

transmitted power

incident power

2 22

1 1 2

1 1 1cos cos cos2 2 2i tr

i i tE EEθ θ θη η η

− =

− =

incident p reflected power ower transmitted p ower



The following conversion formulas relate power and electric field coefficients

Note that reflection and transmission coefficients for the time-average power are always real positive quantities. The following power conservation condition is always verified

Since the power flow normal to the interface is considered, the results obtained above apply equally to perpendicular and parallel polarization.

2

2 12

( )

cos( )cos

t

i

R E

T E θητη θ

= Γ

=

1R T+ =



Non−magnetic perfect dielectric media

From Snell’s law

2 1Case ε ε>

Medium 1 ε1 = εr1 εo µ1 = µo

z

x

y

iβiθ

×

iθrβ

tθtβ


2 11 2sin sin ii ttε θ ε θ ε ε θ θ> ⇒ <=



Since θ t < θ i there is always a transmitted (refracted) beam.

The transmission coefficients are always positive

⇒ transmitted and incident wave are in phase at the boundary.

( )

( ) ( )

1

2

1

2

2sin cos( ) ( ) 0sin

2sin cos( ) ( ) 0sin cos

t i

i t

t i

i t i t

E H

E H

ε θ θτ τθ θε

ε θ θτ τθ θ θ θε

⊥ ⊥= = >+

= = >+ −

× ×

ytE

yiE ziH

ztH( ) yt

yi

EE

Eτ⊥ =



Perpendicular polarization

The reflection coefficient for the electric field is always negative

The reflection coefficient for the magnetic field is always positive

( )( )

( )( )

sin sin( ) ( )

sin sini t i t

i t i tE H

θ θ θ θθ θ θ θ⊥ ⊥− −

Γ = − Γ =+ +

and have always phase difference of 180yi yrE E °

and are always in phasezi zrH H

( ) yr

yi

EE

E⊥Γ = × yrEyiE ziH zrH( ) zr

zi

HH

H⊥Γ =



Parallel polarization ( )( )

( )( )

tan tan( ) ( )

tan tani t i t

i t i tE H

θ θ θ θθ θ θ θ− −

Γ = − Γ =+ +

( )

( )

When and have phase difference of 180 and are in phase

When an

90d are in phase

and have phase differenc

tan

e o

9

0

f 180

0 nta 0

zi zr

yi yr

zi z

i t i

r

yi y

t

t i

r

i tE EH H

E EH H

θ

θ

θ θ θ

θ θ θ

+ > °

⇒

°

⇒

+ < ° +

+ <

°

>

×yrHyiHziE zrE×

×yrHyiHziE zrE



( )90 taWhen ni t i tθ θ θ θ+ = ° +⇒ →∞ the reflection coefficients vanish (TOTAL TRANSMISSION)

For

90 and cos s

(Brewster angle)in

it

t

Bi

B

θ θθ θθ θ

⇒

⇒ =

= ° =+

From Snell’s law

( )( )

( )( )

tan tan( ) 0 ; ( ) 0

tan tani t i t

i t i tE H


Γ = − → Γ = →+ +

121

21

sin sin tansin co

ns

taB BB

t BB

θ θ εθ θθ ε

εεθ

−= = = ⇒ =



From Snell’s law

2 1Case ε ε<


z

x

y

iβiθ

×

iθrβ

tθtβMedium 2 ε2 = εr2 εo

µ2 = µo

2 11 2sin sin ii ttε θ ε θ ε ε θ θ< ⇒ >=



For angles of incidence such that

we have for perpendicular polarization

( )( )

( )( )

( )( )

( )( )

sin sin( ) 0

sin sin

sin sin( )

and are always in phase

and have always phase differenc

0sin sin

e of 180

i t t i

i t i t

i t t i

i t i

yi yr

zi zr

t

E

H

E

H

E

H

θ θ θ θθ θ θ θ

θ θ θ θθ θ θ θ

⊥

⊥

− −Γ = − = >

+ +

− −Γ = = − <

+ +

°

sin 1tθ <

×yrEyiE ziH zrH

×



For parallel polarization ( )( )

( )( )

( )( )

( )( )

tan tan( )

tan tan

tan tan( )

tan tan

i t t i

i t i t

i t t i

i t i t

E

H

θ θ θ θ

θ θ θ θ

θ θ θ θ

θ θ θ θ

− −Γ = − =

+ +

− −Γ = = −

+ +

( )

( )

When and are in phase and have phase difference of 180

When an

90d have phase difference of 180

and a

tan

re

9

0

in ph

0 0

e

an

as

t

zi zr

yi yr

zi z

i

i

r

yi yr

t

t i t

i t

E EH H

E EH H

θ θ θ

θ

θ

θ θ θ+ <

+ > ° +

⇒°

⇒

+

°

°

>

<

×yrHyiHziE zrE×

×yrHyiHziE zrE



( )90 taWhen ni t i tθ θ θ θ+ = ° +⇒ →∞ Also in this case the reflection coefficients vanish and we have TOTAL TRANSMISSION

Total transmission occurs again, for parallel polarization only, at the Brewster angle

( )( )

( )( )

tan tan( ) 0 ; ( ) 0

tan tant i t i

i t i tE H


Γ = → Γ = − →+ +

1 21

tani Bεθ θε

−= =



When

we have a limit condition for TOTAL REFLECTION, valid for both polarizations. This particular angle of incidence is called

2 21 1

sin sin sin 01 9i t t tε εθ θ θε ε

θ⇒ = ⇒ == = °

1 21

sini cεθ θε

−= = critical angle

90tθ = °

cθ



For angles of incidence beyond the critical angle The negative sign is selected, in order to obtain the proper wave vector in medium 2, as shown later.

imaginary

12

2 21 22 1

sin sin 1

cos 1 sin sin

cos

, " "

i t i

t

t

c

t ij

εθ θε

ε εθ θ θε ε

θ

θ

θ

=

±

>

⇓

⇒ = >

⇒

= −

−

− − = −

choose



The reflection and transmission coefficients become complex 2

2 1

12

22 1

2 2 1

1 2 12

222 1

122

2

22 2 1

1

1

1

cos sin( ) ( )

cos sin

cos sin( ) ( )

cos sin

2cos( ) ( )cos sin

2 / cos( ) ( )cos sin

i

i

i i

i i

i i

i i

i

i

i i

jE H

j

jE

E Hj

E

H

Hj

j

θ θ ε ε

θ θ ε εε θ θ ε εεε θ θ ε εε

ε θτ τε θ θ ε ε

ε ε ε θτ τ εε θ θ ε ε

ε

⊥

⊥ ⊥

⊥+ −

Γ = −Γ =− −

− − −Γ = −Γ =

= =

= =

−

− −

−

−

−



If we consider the coefficients for time-average power flow, we have, for both polarizations This means that incident and reflected waves carry the same time-average power, and no power is transmitted to medium 2. But this does not mean that the field disappears in medium 2. The instantaneous power that enters medium 2 is eventually reflected back to medium 1. The electric field phasor of the transmitted wave has the form

2 2cos sintE t x t z t tj x j z j x j z

t tE e e E e eβ β β θ β θ− ⋅ − ⋅ − ⋅ − ⋅= =

( ) ( )*

011R E E

T R= Γ ⋅Γ =

= − =



The wave vector components are

The field in medium 2 corresponds to a surface wave, moving along the z−direction and exponentially decaying (evanescent) along the x−direction

21 22 2

2 1

2 22 21 1

1 1

12 2 1

2

cos sin

sin sin

sin sin sin

t o i

o i i t

t o i i i z

j

j j j

ε εβ θ ω µ ε θε ε

ε εω µ ε θ β θ αε ε

εβ θ ω µ ε θ β θ βε

= − −

= − − = − − = −

= = =

( )tE i i zt tzj z j zj j x

t txE e e E eeβ βα α− ⋅ − ⋅− − − ⋅⋅= =



The surface wave moves parallel to the surface, with a phase velocity equal to the apparent phase velocity along z of the incident wave in medium 1

For the surface wave, planes of constant amplitude are parallel and planes of constant phase are normal to the interface. These planes do not coincide, therefore the surface wave is a nontransverse wave.

12 1sin

pp pz p

i

vv v vθ

= = >

i cθ θ>

x

Et

Constant amplitude planes

Constant phase planes



If you consider a beam incident on the interface, it is found that the power is totally reflected but after penetrating for some distance into medium 2. The reflected beam emerges displaced by a distance D (called Goos-Hänchen shift, discovered in 1947) From experiments, the displacement is found to be

2 22 2

11

2 20.52 0.52sin t

i

D π πε εαεβ θ

ε

≈ =−

i cθ θ>D



Examples:

?Bθ =

Medium 1

o

o

air µ µ

ε ε

=

=

{80 microwaves

Medium

1.8 optic

2

al

o

o

o

water µ µ

εε

ε

=

=

1

1

tan 80 83.6 microwaves

tan 1.8 53.3 opticalB

B

θ

θ

−

−

= ≅ °

= ≅ °

1 21

tanBεθε

−=



At the Brewster angle

Verification with Snell’s law

906.38 microwaves36.7 optical

i t B t

t

θ θ θ θ

θ

+ = + = °

°⇒ ≅ °

1 2

1

2

sin sin

6.38 microwavessinsin36.7 optical

B t

Bt

ε θ ε θ

θθε

−

=

°= ≅ °



?Bθ =

Medium 2

o

o

air µ µ

ε ε

=

=

{80 microwaves

Medium

1.8 optic

1

al

o

o

o

water µ µ

εε

ε

=

=

1 21

tanBεθε

−=

1

1

1tan 6.38 microwaves801tan 36.7 optical

1.8

B

B

θ

θ

−

−

= ≅ °

= ≅ °



At the Brewster angle

Verification with Snell’s law

9083.6 microwaves53.3 optical

i t B t

t

θ θ θ θ

θ

+ = + = °

°⇒ ≅ °

1 2

1 2

sin sin

83.6 microwavessinsin53.3 optical

B t

Bt

o

ε θ ε θ

θ εθ

ε−

=

°= ≅ °



The total reflection angle does not exist since

?cθ =

Medium 1

o

o

air µ µ

ε ε

=

=

{80 microwaves

Medium

1.8 optic

2

al

o

o

o

water µ µ

εε

ε

=

=

1 21

sincεθε

−=

2 1ε ε>



?cθ =

Medium 2

o

o

air µ µ

ε ε

=

=

{80 microwaves

Medium

1.8 optic

1

al

o

o

o

water µ µ

εε

ε

=

=

1 21

sincεθε

−=

1

1

1sin 6.4193 microwaves801sin 48.19 optical

1.8

c

c

θ

θ

−

−

= ≅ °

= ≅ °



Consider a perpendicularly polarized wave. Find the Brewster angle and the critical angle:

60iθ = °

Medium 2

o

o

air µ µ

ε ε

=

=

Medium 1

4

o

o

µ µ

ε ε

=

=

1 1

1 1

1 1tan tan 26.5654 2

1 1sin sin 304 2

B

c

θ

θ

− −

− −

= = ≈ °

= = = °



Find the components of the incident propagation vector and of the x-component of the transmitted propagation vector in terms of

o o oβ ω µ ε=

choose " "

1

1

2 2

2 2

2cos 4 cos602

3sin 4 sin 60 2 32

3

3 2

oix i o o o

iz i o o o o

tx t tz t o tz iz o

tx o o o tj j

ββ β θ ω µ ε β

β β θ ω µ ε β β

β β β β β β β β

β β β β α

−

= = ° = =

= = ° = =

= − = = =

= ± − = − = −



In the second medium, find the distance at which the field strength is 1/e of that at the interface

What is the value of the magnitude of the reflection coefficient at the interface?

The reflection coefficient is a complex quantity when the incident angle exceeds the critical angle. Because of total reflection we know that it must be

since the time-average power reflection coefficient is

1 12t o

dα β

= =

( ) 1E⊥Γ =

2( ) 1R E⊥= Γ =

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