Download Propagation of EM Waves in material media

Propagation of EM Waves in material media
S.M.Lea
2017
1 Wave propagation
As usual, we start with Maxwell’s equations with no free charges:
 ·
 =0
∇
 ·
 =0
∇

 ×
 = − 
∇




 ×
 =
∇
+ 

 =
 0 (·−) (or equivalently,
If we now assume that each field has the plane wave form 
we Fourier transform everything), the equations simplify:
 · 
 =0
(1)
 · 
 =0
(2)
 × 
 = 

(3)
 = − 
 + 
 × 
(4)

 = 
 
 = 

Then, for an LIH, conducting medium, we can eliminate 
and  =  
to get:
 =0
 · 
³
´
 × 
 = −
 − 
 = − 1 +   

(5)

For now let’s take  = 0 (non-conducting medium). Then we get the wave equation:
³
´
³
´
 ×  × 
 = − × 


 − 2 
=   · 

2 

=  2 
where we used equation (2). So the wave phase speed is

1
 = = √


1
(6)
and the refractive index is:

=
=

and
=
Then equation (3) becomes:
r

0 0



(7)
(8)
 = 

̂ × 
(9)


For most LIH materials in which  =  is a useful relation,  ' 0  so the refractive
index is primarily determined by the dielectric constant 0 
2 Reflection and transmission of waves at a boundary
Now let’s consider a wave incident on a flat boundary between two media with refractive
indices 1 and 2  We choose coordinates so that the boundary is the  − −plane. The
field in the incident wave is
³
´
 =
  exp  ·  − 

In general there will be a transmitted wave with
³
´
 =
  exp  ·  −   

and a reflected wave with
³
´
 =
  exp  ·  −   

The plane of incidence is the plane containing the normal to the boundary and the incident
ray, that is ̂ and  . We choose the axes so that the plane of incidence is the  − −plane.
The angle of incidence  is the angle between  and ̂ = ̂ Then
 ·  =  ( sin  +  cos )
The boundary conditions we have to satisfy are (Notes 1 equations 5, 7, 8 and 10 with 
  zero).
and 
 =  is continuous
̂ · 
(10)
 =  is continuous
̂ · 
(11)
 tan is continuous

(12)
 tan is continuous

(13)
 and
For most ordinary materials 0 ' 1 so we shall ignore the difference between 

0 
In order to satisfy the boundary conditions for all times  and at all  and  we must
have  ·  −  =   +   −  the same for each wave at the boundary  = 0. That is,
each wave has the same frequency  and  =  sin  is the same for each wave. From (8)
with fixed  we also have  =  and  = 21  (eqn 8). So the angle of incidence equals
the angle of reflection, and
2
 sin  =  sin  =
or
2
 sin 
1
1 sin  = 2 sin 
(14)
which is Snell’s law.
Notice that the physical argument that gives the laws of reflection and refraction
(boundary conditions must hold for all time and everywhere on the boundary) is independent
of the kind of wave and the specific form of the boundary conditions, and so these laws hold
for waves of all kinds (sound waves, seismic waves, surface water waves, etc.).
  and 
   However, two of the
We now have four equations to solve for the unknowns 
equations (12 and 13) are vector equations with two components, so we actually have six
 is perpendicular to
equations. Since we know the directions of the wave vectors  and 
 there are two components of 
  and 
  in the plane perpendicular to their respective 
so we have four unknowns. Our equations are not all independent. We can simplify by
decomposing the incident light into two specific linear polarizations.
2.1
Polarization perpendicular to the plane of incidence
 is perpendicular to the plane of incidence, (that is, with our chosen
In this polarization, 
 =  ̂). The vectors 
 and 
 in the waves are as shown in the diagram. The
coordinates, 
 is chosen so that 
 the Poynting vector, is in the correct direction for each
direction of 
wave.
In this case there are only two unknowns,  and   but the boundary condition (10) is
trivially satisfied, since  = 0 The remaining conditions cannot all be independent. Eqn.
3
(12) has only one non-zero component:
 +  = 
(15)
and similarly from (13) we have:
 +  = 
and from equation (9) we find  = −̂   so we may rewrite this equation in terms of
the electric field components.
1 ( −  ) cos  = 2  cos  = 2 ( +  ) cos 
(16)
where we used equation (15) in the last step. The final boundary condition is (11):
 + 
1 ( + ) sin 
= 
= 2  sin 
where we used  = ̂   With Snell’s law, this relation duplicates eqn (15), so we
have two equations for two unknowns. Rearranging eqn (16), we get
 (1 cos  − 2 cos  ) =  (1 cos  + 2 cos  )
Solving for   and using Snell’s law to eliminate  , we have
1 cos  − 2 cos 

 =
1 cos  + 2 cos 
r
³ ´2
1 cos  − 2 1 − 12 sin2 
r

=
³ ´
2

=
1 cos  + 2 1 − 12 sin2 
q
1 cos  − 22 − 21 sin2 
q

1 cos  + 22 − 21 sin2 
(17)
The reflected amplitude depends on the angle of incidence, and on the ratio of the two
refractive indices. From equation (17) we can conclude that  has the opposite sign from
 if 2  1  independent of the angle  If 2  1 
µ ¶2
1
sin2   1 − sin2  = cos2 
1−
2
s
µ ¶2
1
sin2   1 cos 
2 1 −
2
Thus the wave has a phase change of  if 2  1  as we learn in elementary optics.
Finally from equation (15), we find the transmitted amplitude:
21 cos 
q
 = 
1 cos  + 22 − 21 sin2 
The time-averaged power transmitted is given by (waveguide notes eqn 30)
¸
∙
´
³
  = 1 Re 
 × (  ̂ × 
 ∗ ) = ̂  ||2
 ×
 ∗ = Re 

2
20
20
4
(18)
Thus the sum of power transmitted normally across the boundary plus power reflected
normally is
 · ̂  = 1 | |2 cos  + 2 | |2 cos 
 · ̂  +  
 
20
20
⎧ ⎡
q
⎤2
⎪
2
2
2
| |2 ⎨ ⎣ 1 cos  − 2 − 1 sin  ⎦
q
1
=
cos 
20 ⎪
⎩
1 cos  + 22 − 21 sin2 
⎫
⎡
⎤2 q
2 ⎪
2
2
2 − 1 sin  ⎬
21 cos 
⎦
q
+2 ⎣
⎪
2
⎭
1 cos  + 22 − 21 sin2 
q
2
2
2
cos

+
2
cos

22 − 21 sin2  + 22 − 21 sin2 

1
1
| |
=
1 cos 
¶2
µ
q
20
1 cos  + 22 − 21 sin2 
2
=
2.2
| |
 · ̂ 
1 cos  =  
20
as expected!
Polarization parallel to the plane of incidence
 and 
 fields in this polarization look like this:
The 
and boundary condition (11) is trivially satisfied. Boundary condition (13) with
1 = 2 = 0 together with eqn (9) gives
 +  = 
1 ( +  ) = 2 
5
(19)
Then (12) gives:
1
( −  ) cos  =  cos  =
( +  )
2
s
1−
The final boundary condition (10) is
µ
1
2
¶2
sin2 
(20)
1 ( +  ) sin  = 2  sin 
21
and, since 1 ∝
(eqn 7), this duplicates eqn (19).
Eqn (20) gives  :
⎛
⎛
⎞
⎞
s
s
µ ¶2
µ ¶2




1
1
1
1
1−
sin2 ⎠ =  ⎝cos  +
1−
sin2 ⎠
 ⎝cos  −
2
2
2
2
So

= 
cos  −
1
2
r
³ ´2
1 − 12 sin2 
r
³ ´
2
1 − 12 sin2 
q
22 cos  − 1 22 − 21 sin2 
q
= 
22 cos  + 1 22 − 21 sin2 
cos  +
and then from (19):

=
1
2
222 cos 
1
q

2 2 cos  +  2 − 2 sin2 
1
2
2
1
21 2 cos 
q
= 
22 cos  + 1 22 − 21 sin2 
2.3
(21)
(22)
Polarization by reflection
Since the reflected amplitudes (21) and (17) are not the same in the two different
polarizations, the reflected light is always partially polarized when the incident light is
unpolarized. Equation (21) shows that the reflected amplitude in the polarization parallel to
the plane of incidence is zero if:
q
22 cos  − 1 22 − 21 sin2  = 0
or, squaring and writing 1 = cos2  + sin2  we have
¢
¤
£ ¡
42 cos2  = 21 22 cos2  + sin2  − 21 sin2 
¡
¡
¢
¢
22 cos2  22 − 21 = 21 sin2  22 − 21
Thus either 1 = 2 (no boundary), or
tan  =
6
2
1
(23)
This is Brewster’s angle.
Can this also happen for the other polarization? We would need:
q
1 cos  − 22 − 21 sin2  = 0
21 cos2 
= 22 − 21 sin2 
or 2 = 1  So the reflected field amplitude for this polarization is not zero unless there is
no boundary. Thus when unpolarized light is incident at Brewster’s angle, the reflected light
is 100% polarized perpendicular to the plane of incidence. At other angles of incidence, the
reflected light is partially polarized.
Why does this happen? At Brewster’s angle,
1
sin 1
sin  =
sin  =
= cos 
2
tan 
Thus

 = − 
2
The angle between the reflected and tramsmitted waves is




 = −  + −  = −  +  =
2
2
2
2
Thus electrons accelerated by the elelctric field in medium 2 would need to radiate along
the direction of the acceleration in order to create the reflected wave. This is impossible.
(wavemks notes, see eg eqn 38.)
3 Waves in a Conducting medium
3.1
Propagation
If the medium is a conductor, (non-zero  in (5)), the dispersion relation (6) becomes:
³
 ´
2 =  2  1 + 
(24)

and thus the solution for  is complex.
where
 = ||  = || (cos  +  sin ) =  + 
2
2
=   1 +
||
tan 2 =
Then
r



³  ´2

· = ̂· − ̂·
The imaginary part  of  indicates spatial attenuation of the wave,
∙
³  ´2 ¸14
√
Im  =  =   1 +
sin 

7
(25)
(26)
while the real part  of  gives the wave phase speed
ph = 
If  is small ( ¿ 1), we may expand the functions (25) and (26) to first order:
µ
¶
√
1 ³  ´2
√
|| =   1 +
'  
4 

 =
2
and we get back the expected results as  → 0 The imaginary part of  is small because 
is small.
r

 
√
Im () '  
=
2
2 
and the wave phase speed is almost unchanged:
1
ph ' √

But if  is large ( À 1),  → 4, and
r

√
√
|| '  
= 

r
1

√
= Re ()
Im () '  √ =
2
2
The wave is damped within a short distance
r
2
(27)
 ∼ 1 =

The distance  is called the skin depth.
In a conducting medium the relations between the fields are (eqn 3):
³
´
 
 = ||  ̂ × 


so  is also the phase shift between the fields. If  is small ( ¿  low conductivity) then
 is almost the same as the amplitude of 
 times √ and the phase shift
the amplitude of 
¯ ¯
¯¯
is also small. But if  is large ( '  high conductivity  À 1) then ¯
¯ is much larger
¯ ¯
√ ¯¯
than  ¯ ¯  and the phase shift is almost 4 . Also the phase speed is given by
r
√
√
 0 0
 0 0
20

=
= p  =
¿1

Re ()

2
 and 
 are
To summarize, in a good conductor the wave fields are primarily magnetic, 
out of phase by 4 and the wave phase speed is very slow.
3.2
Reflection and refraction at a boundary with a conducting medium
How does this affect the reflection and refraction of a wave? Consider a wave with wave
8
number  incident on a conducting medium. Inside the conductor the fields behave like
s
Ã
!
2
2
exp (  sin  +   cos  − ) = exp  sin  +   1 − 2 sin  − 

where  is the incident wave number. For a good conductor, with
r
1

'=
=
2

we have
r
r
µ
¶−12

2
21  √ 1
√
 =  0 1
'
= 2


2 2
So  is small when 2 is large. Then the coefficient of  is
q
q
q
 cos  =  2 − 2 sin2  =  2 −  2 − 2 sin2  + 2 =  −2 sin2  + 2
or
 cos 
= − sin 
r
1−
2 2
2
sin2 
¶2 #14
2
−2
= − sin  1 +
2  2 sin2 
√
√
2 −2
2 −2
' −
=−
( ¿ 1)




where
and thus  ' 2 and −2
are proportional to
"
µ
(28)
2
tan  = 2 2 2
√   sin 
' (1 − )  2 Thus the field components in the conductor
h 
i
exp − (1 − )

This shows that the fields in the conducting medium propagate only a distance of order  in
the −direction (normal to the boundary).
For polarization perpendicular to the plane of incidence (only  non-zero), equation
(15) still holds, but the second boundary condition (equation 13) becomes (using 28):
√

2 −2
( −  ) cos  =  cos  =  



(1 + )
' 

Combining with equation (15), we have:
µ
¶
µ
¶
+1
1 +1
' 
2 =  1 +
 cos 
 cos 
or
 cos 
=   cos  (1 − )
 = 2
(1 + )
which is very small (| | ¿  ). Thus almost all the wave energy is reflected. A similar
result holds for the other polarization. (See also Jackson problems 7.4 and 7.5.)
9