华工电磁场与电磁波全英课Lecture 21

Field andWave Electromagnetics
R.L. Li
Field and Wave Electromagnetics
Lecture 21
7.2.4 Normal incidence at Multiple Dielectric Interfaces
Many practical applications
involve the reflection and
refraction of electromagnetic
waves from dielectric or
metallic surfaces that are
coated with another dielectric
materials to reduce reflections
and improve the coupling of
the wave energy.The
underlying principle in such
cases involves a multiple
dielectric interface problem as
shown in Figure7.2.9.
Figure 7.2.9Normal incidence at multiple dielectric interfaces.
The expressions for the total wave fields in the three media can be written as
follows.1111i 1r
1010ˆ+j z
j z
z
x
E e E e
E E E In medium 1:
1111i
1r
10101
1
ˆj z
j
z
z
y E
e
E e H H H In medium 2:
2222f
2r
2020ˆj
z
j
z
z
x
E e E e
E E E 2222f
2r
20202
1
ˆj
z
j
z
z
y
E e
E e H H
H In medium 3:
333f
30ˆj
z
z
xE
e E E 3
33f
303
1
ˆ.
j
z
z
y E e
H H
Using boundary conditions yields that
at z=0
1200z
z E E 10
10
20
20
E E E
E 120
z
z
H H 10
10
20
20
1
2
E E E
E at z=d
3222020302
3j
d
j
d j
d
E e
E e E
e
23z
d
z
d
E
E 23z
d
z
d
H H 322202030j
d
j
d
j
d
E e
E e
E e
2222132213210
102
2
1
3
2
2
1
3
2
.
j d j d
e E E
e
The solution for
is found to be
The condition of no reflection is
10
E 22
2
1
3
2
2
1
3
2
j d
e
10
E 222132
2132
j
d
j
d
e
e
231
2cos
d
22
31
2
sin
0.
d and
If , we require that 2
sin
d 13
2d
n
2
2
d n
which is a half-wave dielectric window.If , we require that 13
2
cos
d 2
21
2
d n 2
21
4
n d
and
2
31
This is a quarter-wave impedance transformer.
Example 7.2.6Coated glass surface.Determine the dielectric constant and minimum thickness of a film be deposited on the glass surface (r =2.3)such that no normally incident visible light of free-space wavelength 550nm is reflected.0
2
min
2
2550
112.
4
4
4 1.23
4
r d
d nm Solution
2
31
000
20
r r
2 1.5
r r
7.2.5 Normal incidence on a Lossy medium
Figure 7.2.10Normal incidence on a lossy medium.
The phasor fields for the incident,reflected,and transmitted waves
are
10ˆj z i i z xE
e E 101
ˆj z i i E z y
e
H 10ˆj z
r r z xE
e E 10
1ˆj z
r r E z y
e
H 0ˆz t t z xE
e E 0
2
ˆz
t t E z
y
e
H where
1
10
/
377
)
(j
j
)
/(
2
j j
For a good
conductor,
1
1
j
12(
)(1)
c
s s s
c
j
Z R jX
j where
is the skin depth.
1/2
()
f Figure 7.2.11Surface resistance concept.
The conductor presents an impedance Z s =c to the electromagnetic wave with equal inductive and resistive parts,defined as R s and jX s ,respectively.
The resistance part is simply the
resistance of a sheet of metal of l l and of
thickness .Actually the resistance is independent of the area (l l )of the square plate.Thus,the resistance between the two
shaded faces is given by
1
().
s
l R l
Since R s is independent of the linear dimension l ,it is called surface resistance,and c can be thought of as the surface impedance.
To find the reflection and transmission coefficients,we apply the boundary conditions:
Solving the equations,we find
i r t E E
E 0
1
.i i t s
E E E
Z
0101
,
j
r s i s E Z e
E
Z
00
1
2.
j
t s i s
E Z e E
Z For any reasonably good conductor,Z s is very small
compared
to
1
for free
space (i.e.,
377).For example,for copper at 1
MHz,
66,m 8
2.6
10
,
s R 178.999920.9999986,j e 645210.
j e For a good conductor,we can find
01
12(1)
,2
(1
)
r i
E j E j 14
.(1
)j where
is the wavelength in medium 1.
1
1
2/
The current density J in the conductor is
The total current per unit width of the conductor is
2
0ˆ(/).z
t i x E e A
m J E 0
(/).
z
i s
i E J E e
dz
A
m Since the magnetic field of the transmitted wave is and
ˆ,z
t t c
E z y
e
H substituting
,we have
c
s
Z 100
t H
H 0
1ˆ0
i s
E y
Z
H 010.
i s y E Z H Substituting in current density leads to
1
111
1
(0)
(
)(1)
(0)(0).
(1)s y s
y y Z H j J H H j 1ˆ(0).s
n
J H or
The time-average power loss per unit area in the xy plane may be evaluated from the complex Poynting vector at the surface.We
have
2
*20*
*0*
11*411Re 21Re 2
1~~Re 21~~Re 21i s
i s i i *y x av
E Z E Z E E H E
S H E 1()(1)s Z
j All of this power must be dissipated in the conductor due to ohmic losses,which can be evaluated by integrating over a volume with unit area in the xy
plane:
2
00
/
22010100
*
4
1
2
)~)(~(21~~21i z i
x x V loss
E dz
e
E dxdydz E E dv
P *J E Since
we can
express the power in
a more
useful form
as
0/
,i s E J s s s loss
R J J P 2
2
~2
1~4
1which underscores the term surface resistance for R s .
1
1j
7.3.1 Plane Wave Propagation in General Directions
0,,x y z jk x jk y jk z
x y z
e
E E The general form for a plane wave propagating in an arbitrary direction is
It can be easily proved that
2222
2
x
y
z
k
k
k
k
Define a wave number vector as
ˆˆˆˆx y z k
xk yk zk
kk
ˆk
where is a unit vector in the direction of propagation. Thus
0jk r
r
e
E E ˆˆˆ.r xx
yy
zz
where
7.3 Oblique Incidence of Plane Waves
Therefore constant represents planes of constant phase.k r 00
jk r
jke
E 0
E
Since , we have 00
jk r
e
E 00
jk r
jk r
e
e
E E 0
k E which implies that the electric field is transverse to the direction of propagation.
or
ˆ0
k
E Figure 7.3.1Plane waves propagating in general directions
k
r
The magnetic field associated
is
011ˆ
1ˆ
jk r
jk r r
r
j j
k r
k e
E H E E E ˆ0.k
r H Obviously,Therefore a uniform plane wave in an arbitrary direction is a transverse Electromagnetic Waves (TEM)ˆk
*2
01
11ˆRe Re 2
2
1ˆˆRe 2ˆ2
av
S k k k
E k E H E E E E E E The time-average Poynting vector is
Example 7.3.1A uniform plane in free space is expressed as
Solution 68ˆ,10(/)
j x z
x z
y
e V
m E ˆ
k
,,E x z
t ,,H x z t .av S Find: (a) f
; (b) ; (c) ; (d) ; (e); (f)2
22
68100
10(/)
k k rad m (a)8
9
00
1
1310
3
10
102
2
2
2
k
f
Hz
2/0.628k
m
(b)
(c)ˆˆ34
ˆ
ˆˆ55
x y xk
yk
k
x z
k ˆk
68ˆˆ,,Re 1010cos 68(/)
j x z
j
t
E x z t
y e e y
t x z V m 9111
ˆ
ˆˆ,,,,cos 31068(/)
20
15
H x z t
k E x z t
z x t x z
A m 2
0211ˆ
ˆˆ(/)
2
43
av
E S k x y W m (d)(e)(f)
r
i
t
t
H ~i H ~
r
H
~1
1,2
2
,
Region 2
Region 1
i
E ~
z x
r E ~t
E ~7.3.2 Oblique Incidence of Plane Waves at a Dielectric Interface
Parallel Polarization Figure 7.3.2Oblique incidence of a
plane wave with parallel polarization.
For parallel polarization,the electric field lies in the xz plane so that the incident fields can be written as
1(sin
cos )
0ˆˆ(cos sin )i
i jk x z i i
i E x z
e E 1(sin cos )
1
ˆi
i jk x z i
E ye
H ,
1
11
k .
/11
1where The reflected and transmitted fields can be written as
)
cos
sin
(||01)sin ˆcos ˆ(~r
r
z x jk r
r
r e
z
x E E )
cos sin
(1||
1ˆ~r
r
z x jk r
e y
E H
In the above,and
are the reflection and transmission coefficients for
parallel polarization.
We can obtain two complex equations for these unknowns by applying the boundary conditions at the interface at z=0.We then obtain
)
cos sin
(||02)sin ˆcos ˆ(~t t
z x jk t t t e z x
E E )
cos sin (2
||02ˆ~t t z x jk t
e y
E H ,
2
2
2
k .
/
2
2
2where ||
||121sin
sin
sin
||||cos cos cos i
t
r
jk x jk x jk x i r r e
e
e 121
||||sin
sin
sin
1
1
1
1
i
t
r
jk x jk x jk x e
e
e
These equations apply for all x.Thus we have
1
12sin
sin
sin
i
r
t
k k
k which results in the well-known Snell’s law of reflection and refraction:i
r 12sin sin
i
t
k k Dutch
physicist
Now we can solve for the reflection and transmission coefficients as
For this polarization,a special angle of
incidence,
b
,called the Brewster
angle,exists where ,which leads to
||
21
||
21cos cos ,cos
cos
t i t
i
2||
212cos .
cos cos
i t i
2
1
cos
cos
t b
2
22
12
cos cos t
b
2
2
2
12
sin
sin
t
b
k k 12
b
21sin
sin
t
b
k k Snell’s law:
2
2
2
2
112
2
1
(1sin
)
sin
b
b
k k )
(
2
1Total Transmission
r
i
t
t
H ~
i
H ~r H ~1
1,
2
2,
Region 2
Region 1
i
E ~z
x
r
E ~t
E ~
Perpendicular Polarization
Figure 7.3.3Oblique incidence of a plane
wave with perpendicular polarization.
For perpendicular polarization,the electric field is normal to the xz plane.The incident fields are given by
1(sin
cos )
0ˆi
i jk x z i E ye E 1(sin
cos )
1
ˆˆ(cos sin ).
i
i jk x z i
i
i E x
z
e H The reflected and transmitted fields can be expressed as )
cos sin (01ˆ~r r z x jk r e
E y E 1(sin
cos
)
1
ˆˆ(cos sin ).
r
r
jk x z r
r
r
E x
z
e
H )
cos sin (02ˆ~t t
z x jk t
e E y
E 1(sin
cos )
2
ˆˆ(cos sin ).
t
t jk x z t
t
t E x
z
e H
Applying the boundary conditions at the interface at z=0,we have
121sin
sin
sin
i
t
r
jk x jk x jk x e
e
e
121sin
sin
sin
1
1
2
1
cos cos cos i
t
r
jk x jk x jk x i r t e
e
e
By the same phase matching argument that was used in the parallel case,we obtain Snell’s laws
112sin sin sin i r t k k k identical to the parallel case.
We can
solve for
the reflection and transmission coefficients of the perpendicular case as
2
1
21cos cos ,cos
cos
i t i
t
2212cos .
cos cos
i i t
Figure 7.3.3Reflection coefficients for parallel and perpendicular polarizations of an obliquely incident plane wave.
For the perpendicular polarization no Brewster angle exist where
,as we can see by examining
2
1
cos
cos
i
t
0and using Snell’s law to give
2222222221
22
11
(
)()sin .
i k k
k
But this leads to a contradiction,since the term in parentheses on the right-hand side is identically zero for dielectric media.Thus,no Brewster angle exists for perpendicular polarization,for dielectric media.
7.3.3 Total Reflection
Snell’s law can be rewritten
as
sin
sin .
t
i Now consider the case
where 1
>2.
As i increases,the refraction
angle t will increase,but at a faster rate
than i increases.The incident angle
than i
for
which t
=90is called the critical
angle,c ,thus
2
1
/
c At this
angle and beyond,the incident wave will totally reflected,as the transmitted wave is not propagating into region
2.Figure 7.
3.4Total reflection of an oblique incidence.
Defining the refractive index,n,as
r
n
we can express the critical angle as
21sin
/.
c
n n
Applications
Dielectric slab waveguides
and surface waves.
Optical waveguides.
Figure7.3.5Applications of total reflection.
Summary and Review of Chapter 7
Uniform Plane Waves
0122
2
2
t
E
v E p 0
122
2
2
t
H v H
p
For time-harmonic waves,
~~2
2
E E k 0
~~2
2
H
H k .
/
1p
v .
k
where where
Where
for free space.
00(,)(,)
,cos
cos
.
x x x E z t E z t E z t
E t kz
E t
kz ˆ.x z
xE
z
E 00jkz
jkz
x E z
E e
E e
00
1ˆjkz
jkz
z
y
E e
E
e
H 1
ˆ
.z
z
z H E 00
120
377
where
for free space.
00(,)(,)
,cos
cos
.
x x x E z t E z t E z t
E t kz
E t
kz ˆ.x z
xE
z
E 00jkz
jkz
x E z
E e
E e
00
1ˆjkz
jkz
z
y
E e
E
e
H 1
ˆ
.z
z
z H E 00
120
377
Polarization of Plane Waves
01
02
ˆˆ,cos cos .
x y E z t
xE
t kz
yE
t kz
Linear polarization, if
1
2
0.
or
Circular polarization (CP), if
1
2
000/2.
x y and E E E
Right-Handed CP:
Left-Handed CP:
12
,
2
12
.
2
Uniform Plane Waves in a Lossy Medium
00(,)
(,)
,cos
cos
.
x x z
z
x E z t E z t E z t
E e
t
z
E e
t
z 0000()
()
x x z
z
z
j z
z
j z
x E z E z E z
E e
E e
E e
e
E e
e
.
)
(
j j
j
where
01
ˆˆ()z
j z
y c
z
yH
z y E e
e
H
,cos()
z
y c
E H z t
e
t
z
1(1/2)tan (/
)
j j
c c
e
j
Low-Loss Dielectric
2
2
11
8
1/2
1
1
2
c
c
j
j
Good Conductors
2
45
j c
c
j
e
Skin Depth and Skin Effect
12
1f
Normal Incidence on a Perfect Conductor
111
ˆ,2cos()cos()
i E H z t
y z
t 101ˆ,2sin()sin()i E z t x E z
t 0
1
2ˆ(/)
i s E x
A m J
=Normal Incidence on a Lossless Dielectric 0
210
2
1
r i E E
020
2
1
2
t i E E Reflection coefficient
Transmission coefficient
11210ˆ1
j z
j z
i z
xE
e e
E 112
11
ˆ()
1
.
j z
j z
i E z y
e
e
H 22
1
1
ˆ(1)
2i av
E S
z max min
11
E S
E
Normal Incidence on a Lossy Medium
Conditions for no reflection
13
22d
n If , 132214n
d and 23
1If , 1
(
).s
R s s loss R J P
2~2
11ˆ(0).s n
J H
Normal Incidence on a Lossy Medium
Conditions for no reflection
13
22d
n If , 132214n
d and 23
1If , 1
(
).s
R s s loss R J P
2~2
11ˆ(0).s n
J H
Plane Waves Propagating in General Directions Oblique Incidence of Plane Waves
0jk r r
e E E
1ˆr
k r H E 2
ˆ2av E S k Parallel Polarization 21||21cos cos ,cos cos t i t
i 2||
212cos .cos cos i t
i
121
sin b 2121cos cos ,cos cos i
t i
t 2212cos .cos cos i
i
t i r 12sin sin
i t k k Perpendicular Polarization Snell’s Law
Brewster angle
Critical angle 21
sin /c (Total reflection)
(Total transmission)
Homework
P.8-22, P.8-28 and P.8-44
References &
Acknowledgements
1.M.J.Rhee’s Lectures on Electromagnetic Theory,2005.
2.W.H.Hayt,J.A.Buck,Engineering Electromagnetics,7th Ed.,
McGraw-Hill,2006.
3.U.S.Inan,A.S.Inan,Engineering Electromagnetics,Addison
Wesley Longman,2000.
4. D.Cheng,Field and Wave Electromagnetics,Second Edition,
Addison Wesley,1992.。