Jackson 7.4 Homework Problem Solution
Dr. Christopher S. Baird
University of Massachusetts Lowell
PROBLEM:
A plane-polarized electromagnetic wave of frequency ω in free space is incident normally on the flat
surface of a non-permeable medium of conductivity σ and dielectric constant ε.
(a) Calculate the amplitude and phase of the reflected wave relative to the incident wave for arbitrary σ
and ε.
(b) Discuss the limiting case of a very poor and a very good conductor, and show that for a good
conductor the reflection coefficient (ratio of reflected to incident intensity) is approximately
R≈1−2


c
where δ is the skin depth.
SOLUTION:
(a) If we set up the problem in the usual way with the incident wave E traveling in the positive z
direction, the transmitted wave E' traveling in the positive z direction, and the reflected wave E''
traveling in the negative z direction, we get from the boundary conditions:
E 0 '=E 0 E 0 ''
k ' E 0 '=k E 0−k E 0 ''
We can use both of these to solve for the reflected wave in terms of the incident wave:
E 0 '' =
E 0 ''=
k
k
E 0− E 0 ''−E 0
k'
k'
k −k '
E
k k ' 0
The wave number in free space is just k =

c
The wave number in the other material obeys: k '=   0 
For a conducting material we can use  =i

so that we have


k '=   0i

 
 0
Plugging these in above leads to:
E 0 ''
=
E0
1−
1




i
0
0 


i
0
0 
The problem asks us to explicitly expand this in terms of amplitude and phase. The amplitude of a
complex number is defined as ∣z∣=  z * z so that we have
E 0 ''
=
E0
∣ ∣

E 0 ''
=
E0
∣ ∣

1−
1






−i
0 0 


−i
0 0 

1−
1




i
0 0 


i
0 0 

2
2


1 2  2 2 −2 ℜ
i


 0 0 
0
0
1


2
2



2 ℜ
i
2
2 2
0
0 
 0 0 

−1
The phase of a complex number is Arg  z =tan −i
 
E 0 ''
Arg
=tan−1
E0
z−z *
zz *

 
2ℑ
1−



−i
0 0 
2
2

 20  20  2
(b) Discuss the limiting case of a very poor and a very good conductor, and show that for a good
conductor the reflection coefficient (ratio of reflected to incident intensity) is approximately
R≈1−2


c
where δ is the skin depth.
For a very poor conductor σ << ε0ω which means the same as σ/ε0ω << 1. This leads to:





−2
E 0 ''
0
0
=
E0


1 2
0
0
∣ ∣
1
∣ ∣ 

E 0 ''
=
E0

0

1
0
1−
Now recognize:
i
ℑ  x i y =ℑ  x 2 y 2 1 / 4 e 2
−1
tan  y/ x
1
ℑ  x i y= x 2 y 2 1 / 4 sin  tan−1  y / x 
2
If y << x then tan−1  y / x= y / x and sin
ℑ  x i y=
Arg
1
1
y / x= y / x so that
2
2
1
 x y / x leading to:
2
 
E 0 ''
=tan−1
E0
 

 
0  

−1
0
We must be careful because the arctan leads to some ambiguity of what quadrant the phase should be
in. We know from physical arguments that the phase difference should be π for a perfect dielectric, so
that our answer should reduce to π for σ→0. The answer must be in the third quadrant:
  

E 0 ''
0 
Arg
=

E0
−1  
0
For a very good conductor σ >> ε0ω





−2
E 0 ''
0 
0 
=
E0


1
2
0 
0 
∣ ∣
1
which leads to



0 
E 0 ''

after dropping small terms
=
E0

1 2 0

∣ ∣
1− 2
Use the binomial expansion  1x =11 /2 x... on top and bottom and drop all small terms

∣ ∣

∣ ∣ 
 
1
2 0

E 0 ''

2
=
which leads to
E0
0 
1
1  2

2
1−
E 0 ''
=
E0
∣ ∣
1

 
1 0 
− 2 0


2
1 
1− 0
2 
E 0 ''
 
=1− 2 0

E0
We can now calculate the reflection coefficient:
2
∣ ∣
 
E '' 2
 
 
 
R= 0 = 1− 2 0
=12 0 −2  2 0



E0

Drop the smallest term
R=1−2  2

0 

R=1−2
or


2
 where =
c
0 
The phase for a good conductor can also be found:
 
E 0 ''
Arg
=tan−1
E0
 
2


0 

−1
0 
Drop the 1 in the denominator because it is much smaller:
Arg
 
Arg
E 0 ''
=tan−1
E0
  
  
2
0 

E 0 ''
2 0 
=

E0