1

Prof. Ji Chen

Notes 18

Reflection and Transmission of Plane Waves

ECE 3317

Spring 2014

2

General Plane Wave

Consider a plane wave propagating at an arbitrary direction in space.

jkze

sin cos sin sin cosz x y z

Denote

so

ˆ ˆ ˆ ˆ

ˆ ˆˆ ˆ ˆ ˆ

ˆ ˆ ˆ ˆ ˆ ˆ

z xx yy zz z

x z x y z y z z z

x r x y r y z r z

x

y

z

r̂

z

ˆ ˆ ˆ ˆsin cos sin sin cosr x y z

, ,x y z

3

General Plane Wave (cont.)

Hence

x y zj k x k y k ze

sin cos

sin sin

cos

x

y

z

k k

k k

k k

x

y

z

r̂

z

Note: 2 2 2 2 2 2 2 2 2sin cos sin cosx y zk k k k k

2 2 2 2x y zk k k k (wavenumber equation)or

4

General Plane Wave (cont.)

We define the wavevector:

sin cos

sin sin

cos

x

y

z

k k

k k

k k

ˆ ˆ ˆx y zk x k y k z k

2 2 2 2 2 2x y z x y zk k k k k k k k

The k vector tells us which direction the wave is traveling in.

(This assumes that the wavevector is real.)

x

y

z

r̂

z

5

TM and TE Plane Waves

The electric and magnetic fields are both perpendicular to the direction of propagation.

There are two fundamental cases:

Transverse Magnetic (TMz ) Hz = 0 Transverse Electric (TEz) Ez = 0

x

TMz

y

z

E

H

S

x

TEz

y

z

EH

S

Note: The word “transverse” means “perpendicular to.”

ˆ ˆTM : E E H H

ˆ ˆTE : E E H H

z

z

6

Reflection and Transmission

As we will show, each type of plane wave (TEz and TMz) reflects differently from a material.

#1

x

z

qi qr

qt

#2

Incident Reflected

Transmitted

7

Boundary Conditions

Here we review the boundary conditions at an interface (from ECE 2317).

1 2

1 2

1 2

1 2

ˆ D D

ˆ E E 0

ˆ B B 0

ˆ H H J

s

s

n

n

n

n

n̂1 1,

2 2,

++++s Js

Note: The unit normal points towards region 1.

1 2

1 2

1 2

1 2

ˆ ˆD D

ˆ ˆE E

ˆ ˆB B

ˆ ˆH H

n n

n n

n n

n n

No sources on interface:

The tangential electric and magnetic fields are continuous. The normal components of the electric and magnetic flux densities are continuous.

8

Reflection at Interface

First we consider the (x, z) variation of the fields. (We will worry about the polarization later.)

Assume that the Poynting vector of the incident plane wave lies in the xz plane ( = 0). This is called the plane of incidence.

0E E xi zii jk x jk zi e - -

0E E xr zrr jk x jk zr e -

0E E xt ztt jk x jk zt e -

#1

x

z

i r

t#2

Incident Reflected

Transmitted

Note: The sign for the exponent term in the reflected wave is chosen to match the direction of the reflected wave.

9

Reflection at Interface (cont.)

Phase matching condition:

0E 0 E xii jk xix e -, 0E 0 E xrr jk x

rx e -, 0E 0 E xtt jk xtx e -,

xi xr xtk k k

This follows from the fact that the fields must match at the interface (z = 0).

#1

x

z

i r

t#2

Incident Reflected

Transmitted

x xrk k,

xtk

10

Law of Reflection

xi xrk k

sin cos

sin sin

cos

x

y

z

k k

k k

k k

1

1

sin

0

cos

xi i

yi

zi i

k k

k

k k

0 1

1

sin

0

cos

xr r

yr

zr r

k k

k

k k

1 1sin sini rk k i r

Similarly,

Law of reflection

#1

x

z

i r

t#2

Incident Reflected

Transmitted

11

Snell’s Law

xi xtk k

1

1

sin

0

cos

xi i

yi

zi i

k k

k

k k

2

2

sin

0

cos

xt t

yt

zt t

k k

k

k k

1 2sin sini tk k 1 2sin sini tn n

0 0 0/ /i i i i ri rin k k

We define the index of refraction:

Snell's law

Note: The wave is bent towards the normal when entering a more "dense" region.

#1

x

z

i r

t#2

Incident Reflected

Transmitted

1,2i

12

Snell’s Law (cont.)

The bending of light (or EM waves in general) is called refraction.

Incident Transmitted

Reflected

http://en.wikipedia.org/wiki/Refraction

Acrylic block

Normal

13

1 2sin sini tn n

1sin 1.7689 sini t

o45i Given:

Note that in going from a less dense to a more dense medium, the wavevector is bent towards the normal.

Note: If the wave is incident from the water region at an incident angle of 32.1o, the wave will exit into the air region at an angle of 45o.

Example

o32.1t

Note: At microwave frequencies and below, the relative permittivity of pure water is about 81. At optical frequencies it is about 1.7689.

Air

1.7689r

#1

x

z

i r

t#2

Incident Reflected

Transmitted

Water

2 1.7689r

Find the transmitted angle.

14

Critical Angle

The wave is incident from a more dense region onto a less dense region.

1 2sin sini tn n

1 2n n

1

2

sin sint i

n

n

o 1

2

sin 90 sin c

n

n

At the critical angle:

1 2

1

sinc

n

n

2 2sinxi xr xt tk k k k k

#1x

z

qi qr

qt #2

Incident Reflected

Transmitted

qi < c

#1x

z

qi = c

qr

qt #2

IncidentReflected

Transmitted

qi

qt = 90o

qt = 90o

15

Example

1 2

1

1

sin

1sin

1.7689

c

n

n

o48.75c

#1x

z

Water

qr

#2

IncidentReflected

Transmitted

qc

1 1.7689r

Air

Find the critical angle.

16

Critical Angle (cont.)

At the critical angle:

2xi xr xtk k k k

2 22

2 22 2

0

zt xtk k k

k k

There is no vertical variation of the field in the less-dense (transmitted) region.

#1x

z

qi = c

qr

qt #2

IncidentReflected

Transmitted

qi

qt = 90o

17

Critical Angle (cont.)

Beyond the critical angle:

2xik k

2 22

2 22

2 22

2 2 21 2

2 2 20 1 2

sin

sin

zt xt

xi

xi

i

i

zt

k k k

k k

j k k

j k k

jk n n

j

There is an exponential decay of the field in the vertical direction in the less-dense region.

2 2 21 0 1 2ˆ ˆˆ ˆsin sint xt zt ik x k z k x k z jk n n

, xt zt xt ztj k x k z j k x zt x z e e e 2 2 2

0 1 2sinzt ik n n

(complex)

#1x

z

qi > c

qr

#2

Incident Reflectedqi

18

Critical Angle (cont.)

Beyond the critical angle:

The power flows completely horizontally. (No power crosses the boundary and enters into the less dense region.)

#1x

z

qi > c

qr

#2

Incident Reflectedqi

Re S

This must be true from conservation of energy, since the field decays exponentially in the lossless region 2.

19

Critical Angle (cont.)

Example: "fish-eye" effect

Water

Air

c

o48.75c

The critical angle explains the “fish eye” effect that you can observe in a swimming pool.

1.7689r

A fish can see everything above the water by only looking no further than 49o from the vertical.

Artificial “metamaterials” that have been designed that have exotic permittivity and/or permeability performance.

20

Negative index metamaterial array configuration, which was constructed of copper split-ring resonators and wires mounted on interlocking sheets of fiberglass circuit board. The total array consists of 3 by 20×20 unit cells with overall dimensions of 10×100×100 mm.

http://en.wikipedia.org/wiki/Mhttp://en.wikipedia.org/wiki/Metamaterialetamaterial

0

0r

r

(over a certain bandwidth of operation)

Exotic Materials

21

Cloaking of objects is one area of research in metamaterials.

The Duke cloaking device masks an object at one microwave frequency.

Image courtesy Dr. David R. Smith.

Exotic Materials

22

TEz Reflection

Note that the electric field vector is in the y direction.

(The wave is polarized perpendicular to the plane of incidence.)

#1

x

z

qi qr

qt

#2

TE

TET

Ei

Hi

23

TEz Reflection (cont.)

- -0ˆE E xi zijk x jk zi ey

Incident Wave

-0E ˆ E xr zr

TEjk x jk zr e y

Reflected Wave

- -0

ˆE E xt ztjk x jk ztTET ey

Transmitted Wave

T

T

E

ET

where

Reflection Coefficient

Transmission Coefficient

ˆ ˆ

ˆ

ˆ ˆ

ˆi x

t xt

r xr zr

z

z

i i

t

k x k

k x k z k

z

k x k z k

k

Incident Wave Vector

Reflected Wave Vector

Transmitted Wave V

ector

Note: kzr is positive since we have already explicitly accounted for the sign in the reflected wave.

1 coszr rk k

#1

x

z

i r

t#2

TE

TET

Ei

Hi

24

0 00 ˆ Eˆ =ˆ E E rx rz ti tzx xizjk x jk jk x jk zTE

jk x zT

z jkE Te ee y yy

Boundary condition at z = 0:

Recall that the tangential component of the electric field must be continuous at an interface.

00 0= ˆˆ EEˆ E rx ti xx jk xTE

jkT

jk x xEe T ee yyy

0 0 =ˆ E ˆ ˆETE TET yy y

1 =TE TET

TEz Reflection (cont.)

E E Ei r ty y y

#1

x

z

i r

t#2

TE

TET

Ei

Hi

25

We now look at the magnetic fields.

( )0

( )0

1

ˆE E

Eˆ ˆH ( )

xi zi

xi zi

j k x k zi

j k x k zizi xi

y e

x k z k e

E Hj

1H E

j

( )0

( )0

1

ˆE E

Eˆ ˆH ( )

xr zr

xr zr

j k x k zrTE

j k x k zr TEzr xr

y e

x k z k e

( )0

( )0

2

ˆE E

Eˆ ˆH ( )

xt zt

xt zt

j k x k ztTE

j k x k zt TEzt xt

y T e

Tx k z k e

TEz Reflection (cont.)

#1

x

z

i r

t#2

TE

TET

Ei

Hi

26

1tan 2 tanH H H H Hi r tx x x

Recall that the tangential component of the magnetic field must be continuous at an interface (no surface currents).

1 1 2

zt TE Ei zrz Tk k Tk

Hence we have:

TEz Reflection (cont.)

#1

x

z

i r

t#2

TE

TET

Ei

Hi

27

Enforcing both boundary conditions, we have:

211

zr zt Tzi ETE k Tk k

1 =TE TETThe solution is:

2 1

2 2

1

zt ziTE

zt zi

TE TE

k k

k k

T

TEz Reflection (cont.)

#1

x

z

i r

t#2

TE

TET

Ei

Hi

28

Transmission Line Analogy

2 1

2 1

1

TE TE

TE TE TE

TE TE

Z Z

Z Z

T

2

2 TE

ztZ

k

11 TE

zi

Zk

1 TEZ

2 TEZ

TE

TET

Incident

TEz Reflection (cont.)

E V

-H I

y

x

#1

x

z

i r

t#2

TE

TET

Ei

Hi

29

TMz Reflection

Note that the electric field vector is in the xz plane.

(The wave is polarized parallel to the plane of incidence.)

#1

x

z

qi qr

qt

#2

TM

TMT

Ei

Hi

Word of caution: The notation used for the reflection coefficient in the TMz case is different from what is in the Shen & Kong book. (We use reflection coefficient to represent the reflection of the electric field, not the magnetic field.)

30

TMz Reflection (cont.)

- -10ˆH H xi zijk x jk zi

zi

ek

y

Incident Wave

1 -0H ˆ H xr zr

TMzr

jk x jk zr

ke

y

Reflected Wave

- -0

1ˆH H xt ztjk x jk ztTM

zt

T ek

y

Transmitted Wave

T

T

M

MT

where

Reflection Coefficient

Transmission Coefficient

ˆ ˆ

ˆ

ˆ ˆ

ˆi x

t xt

r xr zr

z

z

i i

t

k x k

k x k z k

z

k x k z k

k

Incident Wave Vector

Reflected Wave Vector

Transmitted Wave V

ector

#1

x

z

i r

t#2

TM

TMT

Ei

Hi

31

TMz Reflection (cont.)

We now look at the electric fields.

H Ej

1E H

j

- -10

( )0

ˆH H

ˆ ˆE ( / )H

xi zi

xi zi

jk x jk zi

zi

j k x k zixi zi

ek

x z k k e

-

y 1

( )0

-0

ˆ ˆE ( / ) H

H ˆ H

xr zr

xr zr

j k x k zrxr zr TM

TMzr

jk x jk zr

x z k k e

ky e

- -20

( )0

ˆH H

ˆ ˆE ( / ) H

xt zt

xt zt

jk x jk ztTM

zt

j k x k ztxt zt TM

y T ek

x z k k T e

-

Note that TM is the reflection coefficient for the tangential electric field.

#1

x

z

i r

t#2

TM

TMT

Ei

Hi

32

TMz Reflection (cont.)

Enforcing both boundary conditions, we have

1 21TTM

zrM

ztzi

Tkkk

1 =TM TMT

The solution is:

2 1

2 1

1

zt zi

TMzt zi

TM TM

k k

k k

T

H H Hi r ty y y

E E Ei r tx x x

Boundary conditions:

#1

x

z

i r

t#2

TM

TMT

Ei

Hi

33

Transmission Line Analogy

2 1

2 1

1

TM TM

TM TM TM

TM TM

Z Z

Z Z

T

2

2

TM ztkZ

11

TM zikZ

TMz Reflection (cont.)

E V

H Ix

y

#1

x

z

i r

t#2

TM

TMT

Ei

Hi

1 TMZ

2 TMZ

TM

TMT

Incident

34

Summary of Transmission Line Modeling Equations

TE ii

zi

Zk

TMz Reflection (cont.)

TM zii

i

kZ

22 2 21 1 1 1 1sin cosz zi x i ik k k k k k k

1, 2i

2 22 2 2 22 2 2 1 2 2 2sin sin cosz zt x i t tk k k k k k k k k

1 Z

2 Z

T

Incident

#1

x

z

i r

t#2

Incident Reflected

Transmitted

35

Power Reflection

2

2

100

100 1

%

%

power reflected

power transmitted100

ReS%

ReS zr

zi

power reflected

TM TE or

#1

x

z

i r

t#2

TM

TMT

Ei

Hi

#1

x

z

i r

t#2

TE

TET

Ei

Hi

36

Power Reflection Beyond Critical Angle

2100% power reflected

#1x

z

qi > c

qr

#2

Incident Reflectedqi

2 1

2 1

Z Z

Z Z

22

=TM ztkZ

22 =TE

zt

Zk

zt ztk j2 =Z imaginary

1

TM TE or

All of the incident power is reflected.

37

Find:

% power reflected and transmitted for a TEz wave

% power reflected and transmitted for a TMz wave

30i

Example

#1x

z

qi qr

qt #2

T

1

1

1

1r

r

2

2

2

1r

r

Given:

20.70t 1 2sin sini tn n Snell’s law:

o1sin 30 2 sin t

38

2 1

2 1

TM TM

TM TM TM

Z Z

Z Z

2 222

2 2 2

2 0

2 2

o

]

coscos

cos cos

377.7303cos 20.70

2249.8 [

TM tzt t

t t

r

k kZ

First look at the TMz case:

1 0

]

cos

327.1 [

TMiZ

0.1339TM

2 1

2 1

1

TM TM

TM TM TM

TM TM

Z Z

Z Z

T

0.8661TMT % 1.79

% 98.21

power reflected

power transmitted

Example (cont.)

39

2 1

2 1

TE TE

TE TE TE

Z Z

Z Z

2 22

2 2

2 0

2 2

cos

1sec

cos

285.5 [ ]

TE

zt t

tt r

Zk

1 0 sec

436.2 [ ]

TEiZ

0.2088

Next, look at the TEz part:

2 1

2 1

1

TE TE

TE TE TE

TE TE

Z Z

Z Z

T

Example (cont.)

0.7912TET % 4.36

% 95.64

power reflected

power transmitted

40

Find:

% power reflected and transmitted for a TEz wave

% power reflected and transmitted for a TMz wave

30i

Example

Given:

#1x

z

qi qr

qt #2

T

1

1

1

1r

r

Sea water2

2

2

81

1

4 [S/m]

r

r

1 GHzf

41

30i

Example (cont.)

Given:#1

x

z

qi qr

qt #2

T

1

1

1

1r

r

Sea water2

2

2

81

1

4 [S/m]

r

r

22 2c j

1 GHzf

We avoid using Snell's law since it will give us a complex angle in region 2!

2 2 0 0 2 0 2/ /c rcn k k k

1 11 2

2

sin sin sin sini t t i

nn n

n

22 2

0rc r j

2 81 71.902rc j

42

30i

Example (cont.)

Given:

#1x

z

qi qr

qt #2

T

1

1

1

1r

r

Sea water2

2

2

81

1

4 [S/m]

r

r

2 2 2 2 2 2 22 2 2 2 1 sinzt z xt xi ik k k k k k k k

1 GHzf

Recommendation: Work with the wavenumber equation directly.

complex

43

2 1

2 1

TM TM

TM TM TM

Z Z

Z Z

2 2 22 1

22 2

2 2 22 2 1 1

2

20 2 0 1

2

22 1

02

22

02

sin

sin

sin

sin

sin

iTM zt

c c

c i

c

c i

c

rc r i

rc

rc i

rc

k kkZ

First look at the TMz case:

1 0

]

cos

326.3 [

TMiZ

0.8099 0.0644TM j

2 1

2 1

1

TM TM

TM TM TM

TM TM

Z Z

Z Z

T

% 66.0

% 34.0

power reflected

power transmitted

Example (cont.)

2 33.82 12.82 [ ]TMZ j

44

2 1

2 1

TE TE

TE TE TE

Z Z

Z Z

2 22 2 2 2

2 1

2

2 2 22 2 1 1

0

20 2 0 1

0 22 1

0 22

sin

sin

sin

1

sin

1

sin

TE

zt i

c i

c i

rc r i

rc i

Zk k k

1 0 sec

435.0 [ ]

TEiZ

0.8542 0.0510j

Next, look at the TEz part:

2 1

2 1

1

TE TE

TE TE TE

TE TE

Z Z

Z Z

T

Example (cont.)

% 73.2

% 26.8

power reflected

power transmitted

2 33.86 12.89TEZ j

45

Consider TMz polarization

Brewster Angle

2 1

2 1

TM TM

TM TM TM

Z Z

Z Z

Set 0TM 2 1 TM TMZ Z

Assume lossless regions

#1

x

z

i r

t#2

TM

TMT

Ei

Hi

46

Brewster Angle (cont.)

2 1 TM TMZ Z 22

2 2 TM z ztk kZ

11

1 1 TM z zik kZ

Hence we have

1 2

zi ztk k

2 2 22 11

1 2

sincos iik kk

2 2 2 2 21 2 1

2 21 2

(1 sin ) sini ik k k

47

Brewster Angle (cont.)

2 2 2 2 21 2 1

2 21 2

(1 sin ) sini ik k k

2 22 1

21 2

1 sin sini i

Assume m1 = m2:

2 2 2 2 22 2 1 2 1sin sini i

2

2 2 2 1 2 2 1 2 2 1 22 2 2 22 1 2 1 2 1 2 1 1 2

( ) ( )sin i

2 21 2 1

2 21 2

1 sin sini i

48

Brewster Angle (cont.)

Hence

2

1 2

sin i

q i

1 2

1

2

1 2

1

tani

Geometrical angle picture:

49

Brewster Angle (cont.)

1 2

1

tani b

For non-magnetic media, only the TMz polarization has a Brewster angle.

A Brewster angle exists for any material contrast ratio (it doesn’t matter which side is denser).

This special angle is called the Brewster angle b.

50

#1

x

z

i r

t#2

TM

TMT

Ei

Hi

Brewster Angle (cont.)

Example

1 1 12 2

1 1

tan tan tan 1.7689ri b

r

o53.06b

Air

Water

1.7689r

51

Brewster Angle (cont.)

Polaroid Sunglasses

The reflections from the puddle of water (the “glare”) are reduced.

TMz+TEz

Sunlight

Puddle of water

TEz

Polarizing filter (blocks TEz)

Eye