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