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Lecture 8: Reflection and
Transmission of Waves
2
Normal incidence – propagating
waves
So far, we have considered plane waves in an infinite homogeneous medium. A
natural question would arise: what happens if a plane wave hits some object?
Such object can be either dielectric or conductor.
Instructor:
Dr. Gleb V. Tcheslavski
To answer this question, we need to use boundary conditions. We study first
normal incidence on the boundary.
Contact:
gleb@ee.lamar.edu
We assume that a plane wave is
generated in the region z < 0 in a
lossless material with dielectric
constant 1 and that a second lossless
material is in the region z  0 with a
dielectric constant 2. The
permeabilities of both materials are 0.
Office Hours:
Room 2030
Class web site:
www.ee.lamar.edu/gleb/e
m/Index.htm
A portion of the wave is transmitted to
the medium 2, another portion is
reflected back to medium 1.
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Normal incidence – propagating
waves
The direction of wave’s propagation can be defined using a right hand rule. This
suggests that the polarization of the reflected field has to be altered after the
incident wave strikes the interface. We assume that the electric component is
unchanged, and the magnetic field will change its direction.
(8.3.1)
Therefore: Incident wave:
E y ,i ( z , t )  Ai e j t  k1z 
Reflected wave:
Transmitted wave:
E y ,r ( z , t )  Br e j t  k1z 
E y ,t ( z , t )  At e
j t  k2 z 
ELEN 3371 Electromagnetics
Fall 2008
Reflected wave:
Transmitted wave:
H x ,i ( z , t )  
H x,r ( z , t ) 
Ai j t  k1z 
e
Z c ,1
Br j t  k1z 
e
Z c ,1
H x ,t ( z , t )  
(8.3.3)
Since the materials are assumed as lossless, the waves will not attenuate (i.e.
equations or by using the characteristic impedances for two regions.
Incident wave:
(8.3.2)
Here, k1 and k2 are the wave numbers for the regions (media) 1 and 2
respectively, constants A and B indicate the terms propagating in the +z and –z
directions.
 = 0,  = k). The magnetic field intensities can be found from the Maxwell’s
4
Normal incidence – propagating
waves
At j t  k2 z 
e
Z c ,2
(8.4.1)
(8.4.2)
(8.4.3)
From the boundary conditions, the tangential components of the electric field must
be continuous and the tangential components of the magnetic field intensity must
differ by any surface current that is located at the interface. Usually, we assume
that there are no surface currents, which implies that the tangential components of
the magnetic field intensity are also continuous at the interface.
Also, since we chose the interface between two media to be at z = 0, the exponent
will be
j t  kz 
jt
e
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z 0
e
(8.4.4)
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Normal incidence – propagating
waves
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Normal incidence – propagating
waves
Therefore, at the boundary z = 0, we can write:
We specify the reflection and transmission coefficients as:
E y ,i ( z  0, t )  E y ,r ( z  0, t )  E y ,t ( z  0, t )
(8.5.1)
H x ,i ( z  0, t )  H x ,r ( z  0, t )  H x ,t ( z  0, t )
(8.5.2)
Reflection:

Br Z c ,2  Z c ,1

Ai Z c ,1  Z c ,2
(8.6.1)
Transmission:

2 Z c ,2
At

Ai Z c ,1  Z c ,2
(8.6.2)
Substituting the last results to (8.3.x) and (8.4.x), we obtain:
Ai  Br  At
(8.5.3)
Ai
A
B
 r  t
Z c ,1 Z c ,1 Z c ,2
(8.5.4)
Knowing the characteristic impedances of the materials allows us to determine
the propagation characteristics and amplitudes of both waves: transmitted to the
second medium and reflected back. If the characteristic impedances on both
sides are equal, all energy is transmitted into region 2 and none is reflected
back. This is called matching the media (lenses, glasses etc.).
Therefore, if one of three wave’s magnitudes is known, two other can be
computed.
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Normal incidence – propagating
waves
Since the characteristic impedance is:
Zc   
(8.7.1)
For two dielectric materials:
Reflection:
Transmission:

1   2
1   2
  1  
2 1
1   2
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Normal incidence – propagating
waves (Example)
(8.7.2)
Example 7.1: Describe the expected reflectiontransmission characteristic of a time harmonic
EM wave normally incident at a layered
dielectric. Find the total reflection and
transmission coefficients of a single layer with
known thickness d and dielectric constants are
2 = r0; 1 = 3 = 0. Plot the frequency
dependence of these coefficients assuming d =
0.1 m and r = 4.
z
(8.7.3)
The amplitudes of the total transmitted and reflected fields are

Et ( z )   E ( z )n3
n 1

(8.8.1)
wave index
Er ( z )   E ( z ) n1
(8.8.2)
n 1
-z direction region number
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9
Normal incidence – propagating
waves (Example)
The wave incident to the boundary with
the second medium is partially
reflected back and partially transmitted
to medium 2. At the interface between
materials 2 and 3, the transmitted to
material 2 field is partially reflected
back and partially transmitted to the
material 3, etc.
Normal incidence – propagating
waves (Example)
The total reflected electric field is:
Ei
Ei 1
Ei 1  2 1
2
Er ( z )  Ei 1  1  2 1 e j 2 k2 d  1   2 1   2 1 e j 4 k2 d  1   2 1   2 1 e j 6 k2 d  ...


2
j 2 k2 d
2
2 j 2 k2 d
2 j 2 k2 d


 Ei 1  1  1  1e
1   1 e
(8.10.1)
   1 e   ... 

Ei 1

Ei 1  2
The total reflection coefficient is:

The phase of individual terms in
summations is different: an additional
phase difference of k2d appears after
each crossing the slab.
 1  12  e j 2 k2 d  1 1  e j 2 k2 d 
Er ( z )

 1 1 
Ei
1  12 e j 2 k2 d  1  12 e j 2 k2 d


(8.10.2)
Similarly, the total transmitted electric field is:
2
Et ( z )  Ei 1 1 e jk2 d  1   2 1  2 e j 3k2 d  1   2 1  2 e j 5 k2 d  ...


2
 Ei 1  12  e jk2 d 1   12 e j 2 k2 d    12 e j 2 k2 d   ... 


We assume next that the reflection coefficient from the first interface is 1 in the
+z direction and -1 in the –z direction. Since the medium 3 is identical to medium
1, 2 = -1, and -2 = 1. Similarly, the transmission coefficient through the first
interface will be 1+ = 1 + 1 in the +z direction and 1- = 1 - 1 in the -z direction.
ELEN 3371 Electromagnetics


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Fall 2008

(8.10.1)
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Normal incidence – propagating
waves
Normal incidence – propagating
waves (Example)
What if the medium where the wave is propagating is lossy, i.e.   0?
The total transmission coefficient is:
jk2 d
2
E ( z ) 1  1  e
 t

2 j 2 k2 d
Ei
1  1 e
The reflection coefficient of the first
boundary is:
1 r 1 2
1
1 

3
1 r 1 2
(8.11.1)
The magnetic field must be:
  H (r )  j E (r )   E (r )
The Helmholtz equation is:
 2 E (r )   2 E (r )  0
(8.11.2)
    j   j  1  j
(8.11.3)
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

(8.12.3)
If the electric field is linearly polarized in the x direction, wave equation reduces to:
 2 Ex ( z )
  2 Ex ( z )  0
x 2
The frequency dependence of the total
transmission and reflection coeffs:
ELEN 3371 Electromagnetics
(8.12.2)
where the complex propagation constant is:
The wave number in the slab:
k 2   r k  2k
(8.12.1)
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(8.12.4)
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Normal incidence – propagating
waves
The solution will be in form:
  z
Ex ( z )  E e
  z
E e
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Normal incidence – propagating
waves
(8.13.1)
If the material 2 is a good conductor, the characteristic impedance will be very
small and it would approach zero as the conductivity approaches infinity.
Therefore, it will be NO transmission of EM energy into the conductor, it all will
be reflected. In this case:
Traveling exponentially decaying waves
0
The first term – a wave propagating in +z direction:
  1
  z
E e
  z  j  z
E e
e
  z
E e
cos t   z 
(8.13.2)
The magnetic field will be:
H y ( z) 
E  e  z  E  e  z
Zc
(8.13.3)
Alternatively, loss can be incorporated by:
  0;   ' j "
(8.13.4)
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(8.14.2)
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Normal incidence – propagating
waves
Since the Poynting’s vector has no real part, NO average power is
delivered to the region 2 (perfect conductor)!
Fields exist only in the region z < 0 (medium 1 – dielectric):


H ( z)  Hi ( z)  H r ( z) 


E0
E
 e  jkc z  e  jkc z u y  2 0 coskc z
Zc
Zc
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Normal incidence – propagating
waves
At the interface, the amplitudes of both pulses add up to equal zero in order to
satisfy the requirement that the tangential component of the electric field must
be zero at a perfect conductor.
E ( z )  Ei ( z )  Er ( z )  E0  e  jkc z  e  jkc z u x   j 2 E0 sinkc zu x
(8.14.1)
Considering the most general solution
(7.12.3) of the wave equation and a
traveling pulse instead of a time-harmonic
wave, we may assume that a virtual pulse
with a negative amplitude was launched
at z = + in the –z direction
simultaneously with the real pulse. Both
pulses meet at z = 0 at the time t = t + 3t
after that moment, they keep propagating
but the virtual pulse becomes real and
the real one becomes virtual.
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Polarization change
(8.15.1)
The surface current on the conductor can be found from the boundary
conditions on the tangential magnetic field:
J s  u z  H ( z  0) 
(8.15.2)
2 E0
ux
Zc
(8.16.1)
Note that at z =0: E(z) = 0 and H(z) = 2E0/Zc uy.
This is where “no metal plates in a microwave oven” comes from!
The Poynting vector for the first region (z < 0) will be:
S0 ( z ) 
ELEN 3371 Electromagnetics
2E 2
1
E ( z )  H ( z )*  j 0 cos  k0 z sin  k0 z u z
2
Zc
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(8.15.3)
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17
Normal incidence – propagating
waves (Example)
Normal incidence – propagating
waves (Example)
During t, the car travels a distance z; therefore, the velocity of the car can be
estimated.
Example 7.2: Pulse radars can be used to determine the velocity of cars. Show
how such radars could work.
A repetitive EM pulse from the
radar is incident on the car.
Because of the high conductivity of
the car, the pulse is reflected back
to the radar where the total time of
travel t for the given pulse can be
estimated. The pulse repetition
time (the time between two
consecutive pulses) is t.
The difference
in arrival time
for two pulses
ti  ti 1 
The velocity is:
2 L 2( L  z ) 2z 2Vcar t



c
c
c
c
c  t i   t i 1 
vcar  
2 t
(8.18.1)
(8.18.2)
The actual distance between the car and the radar L is not important to determine
the car’s speed; although it can be computed as well.
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Oblique incidence – propagating
waves
When a plane EM wave incident at an oblique
angle on a dielectric interface, there are two cases
to be considered: incident electric field has
polarization parallel to the plane of incidence, and
incident electric field has polarization that is
perpendicular to the plane of incidence.
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Oblique incidence – propagating
waves
Incident wave:
Ei  E0  cos  i u x  sin i u z  e
Hi 
Reflected wave:
1. Parallel polarization:
Transmitted wave:
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E0
 jk x sin   z cost 
uye 2  t
Zc 2
(8.20.1)
(8.20.2)
E0
 jk x sin   z cos r 
u ye 1 r
Z c1
Et  E0  cos t u x  sin t u z  e
Ht 
Note: the angles are measured with respect to normal.
E0
 jk x sin   z cosi 
u y e 1 i
Z c1
Er  E0  cos  r u x  sin  r u z  e  jk1  x sinr  z cosr 
Hr 
The incident, reflected, and transmitted electric field
vectors lie in the plane of incidence: the x-z plane.
 jk1  x sin i  z cosi 
(8.20.3)
(8.20.4)
 jk2  x sin t  z cost 
(8.20.5)
(8.20.6)
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Oblique incidence – propagating
waves
From the boundary conditions: i.e., continuity of tangential electric and magnetic
fields at the interface z = 0, we derive:
cos i e jk1x sini   cos  r e  jk1x sinr   cos  t e jk2 x sint
e
 jk1 x sin i
Z c1

e
 jk1 x sin  r
Z c1

e
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Oblique incidence – propagating
waves
(8.21.1)
 jk2 x sin t
For parallel polarization, a special angle of incidence exists, known as the
Brewster’s angle, or polarizing angle, i = B, for which the reflection
coefficient is zero:  = 0.
It happens when
Z c 2 cos t  Z c1 cos i
(8.21.2)
Zc2
To satisfy these conditions, the Snell’s laws of reflection and refraction must hold:
i   r ;k1 sin i  k2 sin t
Therefore:
(8.21.3)

Z c 2 cos t  Z c1 cos i
Z c 2 cos t  Z c1 cos i
(8.21.4)

2 Z c 2 cos i
Z c 2 cos t  Z c1 cos i
(8.21.5)
(8.22.2)

1 1
2
If the incidence angle equals to Brewster’s angle, the reflected field will be
polarized perpendicularly to the plane of incidence.
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sin  B 
These simplifications lead to:
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(8.22.1)
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Oblique incidence – propagating
waves
1. Perpendicular polarization:
The incident, reflected, and transmitted electric field vectors are perpendicular to
the plane of incidence: the x-z plane.
 jk1  x sin i  z cosi 
Incident wave:
Ei  E0u y e
Hi 
Reflected wave:
E0
  cos i ux  sin i uz  e jk1  x sini  z cosi 
Z c1
Er  E0u y e  jk1  x sinr  z cosr 
Et  E0u y e
Ht 
ELEN 3371 Electromagnetics
From the continuity of tangential electric and magnetic fields at the interface z =
0, we again derive:
e  jk1x sin i  e jk1x sin r  e  jk2 x sin t
(8.23.1)
E0
Hr 
cos  r ux  sin  r uz  e jk1  x sinr  z cosr 
Z c1
Transmitted wave:
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Oblique incidence – propagating
waves
 jk2  x sin t  z cost 
E0
  cos t ux  sin t uz  e jk2  x sint  z cost 
Zc 2
Fall 2008
(8.23.2)

 jk1 x sin i
cos  i e
Z c1

 jk1 x sin  r
 cos  r e
Z c1

(8.24.1)
 jk2 x sin t
 cos  t e
Zc 2
(8.24.2)
As before, the Snell’s laws of reflection and refraction must hold:
(8.23.3)
(8.23.4)
i   r ;k1 sin i  k2 sin t
(8.24.3)
These simplifications lead to very similar expressions:
(8.23.5)
(8.23.6)
ELEN 3371 Electromagnetics

Z c 2 cos i  Z c1 cos t
Z c 2 cos i  Z c1 cos t
(8.24.4)

2 Z c 2 cos i
Z c 2 cos i  Z c1 cos t
(8.24.5)
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Oblique incidence – propagating
waves
Oblique incidence – propagating
waves
Total internal reflection and surface waves
A magnitude of the reflection
coefficient for the parallel and
perpendicular polarization:
Assume that the uniform plane wave incident on an interface between two
perfect dielectrics with k1 > k2, for instance, water to air. From the Snell’s law:
2
k 
cos t  1   1  sin 2 i
 k2 
(8.26.1)
For a particular angle of incidence, the quantity under the square root
becomes zero. This angle is called critical angle:
 c  sin 1
k2
k1
(8.26.2)
At the incidence angles exceeding the critical angle, the phenomenon of
total internal reflection occurs.
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Oblique incidence – propagating
waves
Let us denote:
cos t   j ,wheni   c
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Fabry-Perot resonator – standing
waves
Let us consider again the electric component of a linearly polarized EM field
normally incident on a perfect conductor with a reflection coefficient  = -1.
(8.27.1)
In both cases of parallel and perpendicular polarization, the
reflection coefficient will be:
  j

  j




E y ( z , t )  Re B e j (t  kz )  e j (t  kz )   Re  Be jt e jkz  e  jkz 
 Re  j 2 Be
(8.27.2)
jt
sin kz
E y ( z , t )  A sin tsin kz
Here, both  and  are real. As a consequence, the magnitude of the reflection
coefficient || = 1, and all incident power is reflected off the interface.
(8.28.1)
(8.28.2)
Where A = 2B.
As a result, for instance for the perpendicular polarization, the transmitted
electric field is:
Et  TE0u y e  z e  jk2 x sin i
(8.27.3)
The tangential electric field Ey(z,t) = 0 at the interface z = 0. In this case, the
signal consisting of two oppositely propagating waves appears to be
stationary in space and oscillating in time. This is a standing wave.
If the incidence angle exceeds the critical angle, the field in region 2 propagates
in the x direction but rapidly exponentially decays in the z direction – away from
the interface. This is a surface wave.
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Fabry-Perot resonator – standing
waves
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Fabry-Perot resonator – standing
waves (Example)
Example 7.3: An EM wave propagating in a vacuum in the region z < 0 is normally
incident upon a perfect conductor at z = 0. The frequency of the wave is 3 GHz,
the amplitude of incident electric field is 10 V/m, and it is polarized in the uy
direction. Find the phasor and the instantaneous expressions for the incident and
the reflected field components.
The standing wave results
from the constructive and
destructive interference of
the two counter
propagating waves.
k
Observe that the
separation distance
between two successive
null points (nodes) equals
to the separation distance
between two successive
maxima (antinodes) and
equals to one half of the
wavelength.
The incident field
in the phasor form:
2 f 2  3 109

 20 m 1
c
3 108
Ei ( z )  Ae  jkz u y  10e j 20 z u y V m
Hi ( z) 
Or:
u z  Ei ( z )
10  j 20 z
e
u x  A m

120
Zc
(8.30.3)
(8.30.4)
10
cos  6 109  20 z  u x  A m
H i ( z , t )  Re  H i ( z )e   
120
(8.30.5)
ELEN 3371 Electromagnetics
Fall 2008
(8.30.2)
Ei ( z , t )  Re  Ei ( z )e jt   10 cos  6 109  20 z  u y V m
jt
ELEN 3371 Electromagnetics
(8.30.1)
Fall 2008
31
Fabry-Perot resonator – standing
waves (Example)
The reflected field in the phasor form:
Er ( z )    Ei ( z )  10e j 20 z u y V m
Hi ( z) 
u z  Er ( z )
10  j 20 z
e
u x  A m

Zc
120
32
Fabry-Perot resonator – standing
waves
(8.31.1)
(8.31.2)
Examination of a standing wave suggests that it should be possible to insert
another conductor (a conductive wall) at any of the nodes where the tangential
electric field is zero without changing a structure of electric field!
The applicable boundary condition, therefore, is that the tangential electric field
must be zero at a conductive surface.
Or in the instantaneous form:
Er ( z , t )  Re  Er ( z )e jt   10 cos  6 109  20 z  u y V m
(8.31.3)
H r ( z , t )  Re  H r ( z )e jt   
(8.31.4)
10
cos  6 109  20 z  u x  A m
120
Let us assume that the plates were
inserted instantaneously and the EM
energy was “trapped” between the
plates.
Constructive and destructive interference will lead to appearance of
a standing wave.
ELEN 3371 Electromagnetics
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Fabry-Perot resonator – standing
waves
For the 1D Helmholtz equation
d 2 Ey ( z)
dz 2
 k Ey ( z)  0
2
(8.33.1)
and a time-harmonic signal, the solution will be in a form:
E y ( z )  A sin kz  B cos kz
(8.33.2)
The integration constants A and B can be found from the boundary condition
that the tangential electric field must be zero at a metal wall. Therefore, B = 0,
k
n
(8.33.3)
L
where n is an integer (resonator mode) and L is the distance between the metal
walls. If the maximum magnitude of electric field is Ey0, the electric field is
 n z 
E y ( z )  E y 0 sin 

 L 
ELEN 3371 Electromagnetics
34
Fabry-Perot resonator – standing
waves
The structure consisting of a parallel plate cavity is called a Fabry-Perot
resonator. If the frequency of a wave “matches” the dimensions of the
resonator (a resonant frequency) – the length of cavity equals an integer
number of half-wavelengths – a standing wave will be formed. All other
frequency components will be canceled out by a destructive
interference.
The Q-factor (2 ratio of stored energy to
the power dissipated per cycle) of this
resonator may be very high (approaches
a million).
Fabry-Perot resonators are widely used
in EM and optics: a He-Ne laser is
basically a Fabry-Perot resonator.
(8.33.4)
ELEN 3371 Electromagnetics
Fall 2008
Fall 2008
35
Fabry-Perot resonator – standing
waves
Recall that the wave number is a function of
frequency and the velocity of light between
plates.
k
r
c

n
L
The resonant frequency for a free space (or air) can be computed or measured
for known resonator’s dimensions. The relative frequency difference is


1
 1 r2  1
r1
r1
r
(8.35.1)
Therefore, we can find the resonant
frequency as
r 
n
L 0
(8.35.2)
ELEN 3371 Electromagnetics
n
L
 1
1


  


r 0 0
 0 0
Fall 2008
(8.36.1)
Therefore, if we can measure this frequency difference, we can estimate the
permittivity of unknown material placed in the resonator and, thus, identify the
material.
Considering two resonators of the same length L but one of the filled with air
(left) and the other filled with a dielectric , we can find that they will resonate at
two different frequencies. The frequency difference will be
  r 1  r 2 
36
Fabry-Perot resonator – standing
waves




(8.35.3)
Example 7.4: An empty Fabry-Perot resonator has a resonant
frequency of 35 GHz. Determine the thickness of a sheet of
paper that is inserted later between the plates if the resonant
frequency changes to 34.99 GHz. The separation between
plates is 50 cm. Assume that the integer n specifying the
mode doesn’t change and that there is no reflection from the
paper.

3
paper
ELEN 3371 Electromagnetics
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Fabry-Perot resonator – standing
waves (Example)
Example 7.5: A helium-neon laser emits light at a wavelength of 6328 Å in air.
Calculate the frequency of oscillation of the laser, the period of oscillation, and
the wave number. 1 Å (Åmgstrom) = 10-10 m.
The relative dielectric constant separating the plates with paper inserted can
be approximated as
r 
L  L
L
L
  paper
 1    paper  1
L
L
L

  paperadded
1
 vacuum
 1
 1
vacuum
r
1  
Therefore:
1
paper
L

  1
L 2 L paper
 1
L
35  34.99
L

(3  1)
35
2  0.5
38
Fabry-Perot resonator – standing
waves (Example)
(8.37.1)
The frequency:
f 
(8.37.2)
c

The period:
T 
(8.37.3)

2.998 108
 4.738 1014 Hz473.8THz
6.328 107
1
1

 2.11 1015 s 2.11 fs
f 4.738 1014
The wave number:
k
2


2
 9.93 106 m 1
6.328 107
Finally:
L  1.4 104 m
ELEN 3371 Electromagnetics
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??QUESTIONS??
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