*Note: The information below can be referenced to: Iskander, M., Electromagnetic
Fields and Waves, Waveland Press, Prospect Heights, IL, 1992, ISBN: 1-57766-115-X.
Edminister, J., Electromagnetics (Schaum’s Outline), McGraw-Hill, New York, NY,
1993, ISBN: 0-07-018993-5. Hecht, E., Optics (Schaum’s Outlines), McGraw-Hill, New
York, NY, ISBN: 0-07-027730-3. Serway, R., & Faughn, J., College Physics, ISBN: 003-022952-9.
Chapter 6
Oblique Incidence Plane Wave Reflection and Transmission
6.1 Plane Wave Propagation at Arbitrary Angle
Plane waves are not normally incident, so now we must consider the general problem of a
plane wave propagating along a specified axis that is arbitrarily relative to a rectangular
coordinate system. The most convenient way is in terms of the direction cosines of the
uniform plane wave, the equiphase surfaces are planes perpendicular to the direction of
propagation.
Definitions:
uniform planes – a free space plane wave at an infinite distance from the
generator, having constant amplitude electric and magnetic field vectors over the
equiphase surfaces.
equiphase surface – any surface in a wave over which the field vectors of a
particular instant have either 0° or 180° phase difference.
For a plane wave propagating along the +z axis
 ( z)   m e  j z a x
(6.1)
Equation (6.1) states that each z equal to a constant plane will represent an equiphase
surface with no spatial variation in the electric or magnetic fields. In other words,


0
x
y
 for a uniform plane wave
It will be necessary to replace z for a plane wave traveling in an arbitrary direction with
an expression when put equal to a constant (βz = constant), that will result in equiphase
surfaces.
The equation of an equiphase plane is given by
  r   n  r
Notes by: Debbie Prestridge
1
The radial vector (r) from the origin to any point on the plane, and β is the vector normal
to the plane is shown in Figure (6.1).
z
P
  n
M
r
z
y
O
x
x
y
W
As you can see from figure 6.1, the plane perpendicular to the vector β is seen from its
side appearing as a line P-W. The dot product nβ · r is the projection of the radial vector
r along the normal to the plane and will have the constant value OM for all points on the
plane. The equation β · r = constant is the characteristic property of a plane
perpendicular to the direction of propagation β.
The equiphase equation is
β · r = βxx + βyy + βzz
= β (cos θxx + cos θyy + cos θzz)
= constant
r=
x a x  y a y  z az
   x a x   y a y   z az
θx, θy, θz, are the angles the β vector makes with x, y, and z axes, respectively.
Notes by: Debbie Prestridge
2
Definition:
transverse electromagnetic wave (TEM) – electromagnetic wave having electric
field vectors and magnetic field vectors perpendicular to the direction of propagation.
H is perpendicular to E, and both E and H are perpendicular to the direction of


propagation β. The expressions for  and  are
   m e  j  r
(6.2)
 
n   

The unit vector nβ along β and η is the wave impedance in the propagation medium. See


Figure 6.2 for the illustration of orthogonal relations between  and  and the direction
of propagation.
Z
Plane of constant
phase
E
p=PxH
power density flow
H
nβ
r
X
Point on the
plane
Y
EXAMPLE 6.1
The vector amplitude of an electric field associated with a plane wave that propagates in
the negative z direction in free space is given by  m  2 a x  3 a y V
m
Notes by: Debbie Prestridge
3
Find the magnetic field strength.
Solution:
The direction of propagation nβ is –az. The vector amplitude of the magnetic field is then
a x a y az
n  
1
 1

 
given by


0 0 1  
3 a x  2 a y A m
m
 377



2 3 0

*note  

120π~377Ω (Appendix D – Table D.1)
EXAMPLE 6.2
The phasor electric field expression in a phase is given by


  a x   y a y  2  j5 a z e j 2.3( 0.6 x  0.8 y )
Find the following:
1.  y .
2. Vector magnetic field, assuming    and    .
3. Frequency and wavelength of this wave.
4. Equation of surface of constant phase.
Solution:
1. The general expression for a uniform plane wave propagating in an arbitrary
direction is given by
   m e  j  r
where the amplitude vector  m , in general, has components in the x, y, and z
directions. Comparing equation 6.3 with the general field equation for the plane
wave propagating in an arbitrary direction, we obtain
β · r = βxx + βyy + βzz
= β (cos θxx + cos θyy + cos θzz)
= 2.3(-0.6x + 0.8y + 0)
Notes by: Debbie Prestridge
4
Hence, a unit vector in the direction of propagation nβ is given by
nβ = -0.6ax + 0.8ay.
Because the electric field  must be perpendicular to the direction of propagation nβ, it
must satisfy the following relations:
nβ ·  = 0


Therefore, (-0.6ax + 0.8ay) · a x   y a y   2  j5 a z  0
Or
-0.6 + 0.8  y =
0
Hence,  y = 0.75. The electric field is given by


  a x   y a y  2  j5 a z e j 2.3( 0.6 x  0.8 y )

2. The vector magnetic field  is given by
ax
ay
az
1
  n   

 0.6 0.8
0
 
377
1
0.75 2  j5
1
so that
0.8(2  j5)
 x 
  4.24  j10.6 103
377
0.6(2  j5)
 y 
  318
.  j 7.95 103
377
0.6  0.75  0.8
 z 
 3.31 103
377
The vector magnetic field is then given by
Notes by: Debbie Prestridge
5


  
 a 
 a 
 a e j 2.3( 0.6x 0.8 y)

x x
y y
z z
3. The wavelength λ is given by

2


2
 2.73 m
2.3
and the frequency
3  108
 011
. GHz
f 

2.73
c
4.
The equation of the surface of constant phase is
nβ · r = -0.6x + 0.8y = constant
The general expression of this equation in terms of the direction cosines is given
by
nβ · r = (cos θxx + cos θyy + cos θzz) = constant
Comparison between equation 6.4 and the general expression shows that the plane
given in equation 6.4 has no z dependence and, hence defines a plane parallel to
the z axis. In other words, equation 6.4 can be obtained by substituting θx = π/2 in
the general expression of the equiphase plane.
6.2 Reflection by Perfect Conductor – Arbitrary Angle of Incidence
By decomposing the general problem into two special cases we can simplify our analysis.
1. E field is polarized in the plane formed by the normal to the reflecting surface in the
direction βi of the incident wave.
2. E field is perpendicular to the plane of incidence.
Notes by: Debbie Prestridge
6
The plane formed by the normal to the reflecting surface and the direction of propagation
β is known as the plane of incidence. The general case can be considered as a
superposition of two cases –


E is parallel to the plane of incidence
E is perpendicular to the plane of incidence
6.2.1 E Field Parallel to Plane of Incidence
i
r
 y
βi
r
 y
θi
θr
X
Y
Perfect
Conductor
Z
The figure shows an incident wave polarized with the E field in the plane of incidence
and the power flow in the direction of i at angle  i with respect to the normal to the
surface of the perfect conductor.
The direction of propagation is given by the Poynting vector and the i , E, and H fields
need to be arranged so that i is in the same direction as  i   i at any time. The
magnetic field is out of the plane of the paper,    y a y for the direction of the electric
field shown. There is no transmitted field within the perfect conductor; however there
will be a reflected field with power flow at the angle  r with respect to the normal to the
interface. To maintain the power density flow  r   r will be in the same direction
 r as. The expression for the total electric field in free space is
   i   r   im e j i  r   rm e j r  r
Notes by: Debbie Prestridge
(6.5)
7

 i  r    cos i a z  sin  i a x   x a x  y a y  z a z

   x sin  i  z cos i 
(6.6)
 r  r    x sin  r  z cos r 
(6.7)
The total electric field has x and z components:
 j  r
 x  x, z   im cos i e j i r   rm cos r e r
 j  r
 z  x, z   im sin  i e j i r   rm sin r e r
 x
at z  0
j r
  im cos  i e j  i r   rm cos r e  r  0
 jxsin r
  im cos i e jx sin i   rm cos r e
0
(6.8)
Equation 6.8 shows the relationship between the incident and reflected amplitudes for a
perfect conductor the total tangential E field at the surface must be zero which satisfies
the boundary condition. To be zero at all values of x along the surface of the conducting
plane, the phase terms must be equal to each other –
i   r
(6.9)
Equation 6.9 is known as Snell’s law of reflection.
Definition:
Snell’s Law is a rule of Physics that applies to visible light passing from air (or
vacuum) to some medium with an index of refraction different from air.
Substitute equation 6.9 into equation 6.8 –
 im   rm
(6.10)
Therefore, the total electric field in free space is
 ( x , z )   x  x , z  a x   z  x , z  a z
Notes by: Debbie Prestridge
8


  mi cos i e jx sin i e j z cos i  e j z cos i a x


  im sin  i e jx sin i e j z cos i  e j z cos i a z
 2 j im cos i sin ( z cosi )e
 jx sin i
ax
(6.11)
 2 j im sin  i cos   z cos i e jx sin i a z
 2 im   j cos  i sin (  z cosi ) a x
 sin  i cos   z cos i a z e jx sin i
Take equation 6.11 and recover the time-domain form of the total electric field


r , t   Re   r  e j t
Observe the variation of the total field with the x variable indicating there is a traveling
wave in the x direction with a phase constant
x   sin  i
And in the z direction the field forms a standing wave.
The total magnetic field is
  x, z  
  x, z a  
 i  x, z a  
 r  x, z a

y
y
y
y
y
y
Use the relation  
n  

for each of the incident and reflected fields to employ the
expressions x and z components of the incident and reflected electric fields.
n
 i   i   i


Notes by: Debbie Prestridge
9

ax
sin  i
1

ay
0
az
cos  i
 im cos  i e j (sin i x  cos i z ) 0   im sin  ii e j (sin i x  cos i z
 i is the ay component
The solution of the determinant, the only nonzero component of 
given by

1
 i  a y  im cos2  i e j sin i x  cos i z    im sin2  i e j  sin i x  cos i z 


 im  j sin x cos z 
i
i ay

e

The reflected magnetic fields is given by
i
r  m  j sin i x cos i z 

 
e
ay


The total magnetic field  (x, z) is
i
 ( x, z)  a 2 m cos   z cos   e jx sin i

y
i

The average power flow parallel to the conducting surface is
 ave  x , z  

1
Re   
2
ax
1
 Re  x
2
0

ay
0
 
az
 z
y 0
The cross product yields two components:
Notes by: Debbie Prestridge
10


One in the x direction
One in the z direction
 ave 

1
Re   z  y a x   x  y a z
2

The expression of Pave will reduce to
 ave  x , z  
2  im



1
 a
Re   z 
y x
2
sin  i cos2  z cos  i  a x
Glancing Incident:


 i  90 ,  ave 
 
  i 2
2 m 



a x , the power flow is at maximum.
Normal Incident:
 i  0,  x , ave  0 (Power flow in the x direction is zero)
Average power flow perpendicular to the conducting surface is zero, because the average
Poynting Vector is zero in that direction


1
Re  x  y  0
2

Why? Because  x is multiplied by j, therefore  x and  y are out of phase by 90°.
P z , ave 
Therefore, a traveling-wave pattern occurs in the x direction, because the incident and
reflected waves travel in the same direction, the standing-wave pattern will be observed
in the z direction, because the incident and reflected waves travel in the opposite
directions.
The location of zeros (nodes) of the  x field can be found by letting sin   z cos i
At a distance z from the conducting plane given by
 = 0.
 z cos i  n
Or
Notes by: Debbie Prestridge
11
z=n

n  0, 1, 2 , ...
2 cos  i
The zeros will occur at distances larger than integer multiples of  2 . So, for normal
incidence, i  0, cos i 1 , and the positions of the zeros will are the same as those
discussed in chapter 5. For the oblique incidence, the locations of the standing-wave
nodes are  2 apart along the direction of propagation. The wavelength measured along
the z-axis is greater than the wavelength of the incident waves along the direction of
propagation. As shown in Figure 6.4 the relation between these wavelengths

is  z 
.
cos  i
Incident
wave fronts

cos  i
 z
θi
λ
Figure 6.4
The plane of the zero  x field occur at multiples of  2 along the direction of
propagation, and they are located at integer multiples of  z 2 along the z-axis which
appear separated by larger distances. Also note that the standing-wave pattern associated
with the  z component may appear as if there is no zero value of the electric field at
z = 0, but the  component is normal to the reflecting surface, therefore the boundary
z
condition is not in violation.
Notes by: Debbie Prestridge
12
6.2.2 Electric Field Normal to the Plane of Incidence
The entire electric field is (out of the paper) in the y direction and the magnetic field will
have both x and z components. See Figure 6.5.
The incident electric and magnetic fields are
 i   i e j i  r
m
i
 
n i   i


 im

  cos i a x  sin  i a z e j i  r
 iy
r
 ry
i
i
r
i
r
ax
Y
Perfect
Conductor
az
Figure 6.5


where i  r   sin i x  cosi z . Assume that the reflected field is also in the y
direction so the magnetic field must be perpendicular to both E and the Poynting Vector
P = E ^ H,
 r   rm e  j  r  r a y
 
r
n  r   r



 rm

Notes by: Debbie Prestridge
 cos r a x  sin  r a z e j  r r
13
Where  r  r    sin r x  cosr z . Determine the angle of reflection  r and the
amplitude of the reflected electric field  rm by using the boundary conditions at z = 0.
This also includes zero values of the tangential electrical field E and the normal
component of the magnetic field H.
 y  x, z   iy   ry  0
at z = 0
Therefore,
 y  x,0   im e j x sin i   rm e j x sin r  0
And
  x , 0  1  i sin  e jx sin i  1  r sin  e jx sin r  0

i
i
z
m
m


Note: These two conditions will provide the same results for the unknowns  r and rm ,
and be true for every value of x along z = 0 plane, so the phase factors must be equal.
 r  i
And
 rm   im
Negative sign indicates the opposite direction of the reflected electric field (i.e. into the
paper)
The total E field is

 y  x, z   im e j x sin i e j z cos  i  e j z cos i
 2 j i sin   z cos  e j x sin i
m

i


The total H field is
n


 
i 
 r   n i  a  i e j  i  r     i  a y  i e j  i  r 

y m
 

m


  
And the substitution of  rm   im has been made. The direction vectors of the incident
and reflective wave are
Notes by: Debbie Prestridge
14
n  i, r  sin  i a x  cos i a z
And
n  i, r  a y  sin  i a z  cos i a x
The components of the total magnetic field are
 2 im
 x  x, z 
cos i cos   z cos  i  e jx sin i

i
  x, z   2 j m sin  sin   z cos   e jx sin i

i
i
z

There is a standing-wave in the z direction because the reflected and incident waves
travel in the opposite direction along the z-axis. The fields traveling in the x direction
and having the only nonzero power flow in the direction parallel to the interface.
The concept can be illustrated by considering the average density flow associated with
the wave.
 ave  x , z  

1

Re   
2
ax
1
 Re 0
2
S

2  im

flow is in the x direction.
S
 2  im
W

 2  im


ay
az
T
0
0
W


sin  i sin2  z cos  i a x  This indicates that the power
cos i cos   z cos i  e  j  x sini , T  2  im sin   z cos i  e  j  x sin i ,
sin i sin   z cosi  e  j  x sini
EXAMPLES:
Notes by: Debbie Prestridge
15
Find the peak value of an induced surface current when a plane wave is incident at am
angle on a large plane, perfectly conducting sheet. The surface of the sheet is located at z
= 0 and
x
z  V

 i  10 cos1010 t  

 ay m

2
2
Solution
From the equation of the incident electric field, the propagation vector is given by


2
ax 

2
az


that is,  i  45
Because the electric field is along the y direction – that is, perpendicular to the plane of
incidence, the equations given in the section above will be used.
The sheet current J (in ampere per meter) is determined by the total tangential
magnetic field at the surface. From the boundary condition,
  sin 45 a x  cos 45 a z ,
J  n  
where the normal n to the surface for the geometry of Figure 6.5 is n = -az. The magnetic
field in this case has two components:
 2 im
 x 
cos i cos  z cos i  e jx sin i

 2 j im
 z 
sin  i sin   z cos  i  e jx sin i

The surface is then current is then

 im
J at z  0   a z    a y 2
cos  i e jx sin i

And the peak value of the surface current at z = 0 is given by
Notes by: Debbie Prestridge
16
2 im
2(10) cos 45
J

cos i 
 3.75x 102 A m
peakvalue

377
EXAMPLE:
The electric field associated with a plane wave propagating in an arbitrary direction is
given by
  (7.83 a x  4 a y  4.5 a z) e j 7(0.5x 0.87 z )
If this incident on a perfectly conducting plane oriented perpendicular to the z axis, find
the following:
1. Reflected electric field.
2. Total electric field in region in front of the perfect conductor.
3. Total magnetic field.
Solution
Because a vector in the direction of propagation and a unit vector normal to the reflecting
surface are contained in the x-z plane, we consider the x-z plane to be the plane of
incidence as shown in Figure 6.6. The given electric field may, therefore, be decomposed
into two components. The parallel polarization case in which the electric field is
perpendicular to the plane of incidence  || and the perpendicular polarization case in
which the electric field is perpendicular to the plane of incidence   . From the given
equation of the electric field,
 ||  (7.83 a x  4.5 a z ) e j 7( 0.5x  0.87 z )
Comparing this with the equation of the electric field in the parallel polarization case,
where the incident electric field is given by
 i   im (cos i a x  sin  i ) e j (sin i x cos i z)
Notes by: Debbie Prestridge
17
βi
X
Y
Z
Figure 6.6
Observe that:
cos  i  0.5 

 That is, i  30
sin  i  0.87 
The magnitude of the incident electric field  im is therefore = 7.83/0.87 = 9 or 4.5/0.5 =
9. Hence, the electric field associated with the parallel polarization case can be expressed
in the form
 i||  9(0.87 a x  0.5 a z ) e j ( 0.5x  0.87 z )
Based on the analysis of section 6.2.1, we have  r  30 , and the amplitude of the
reflected electric field  r   i  9 . Hence
||
||


 r  9 (cos 30 a x  sin 30 a z ) e j 7(sin30 x cos30 z )
We treat the perpendicular polarization case where
 i  4 a y e j 7( 0.5x  0.87 z )
Based on the analysis of section 6.2.2, it can be shown that
 r  4 a y e j 7( 0.5x  0.87 z )
The total reflected electric field is then
 r  ( 7.83 a x  4 a y  4.5 a z ) e j 7( 0.5x 0.87 z )
Notes by: Debbie Prestridge
18
Parts 2 and 3 can easily be obtained by the following the analysis of section 6.2.
For example, the magnetic field associated with the electric field in the parallel
polarization case is given by
9
 i||  e j 7(0.5x 0.87 z ) a y

The reflected magnetic field intensity for this polarization is
9
 ||r  e j 7( 0.5x 0.87 z ) a y

For the perpendicular polarization case, the magnetic field has two components,
4
 i  (  cos  i a x  sin  i a z ) e j 7( 0.5x  0.87 z )

 4

4
   cos  i a x  sin  i a z e j 7( 0.5x  0.87 z )
 


Because, for the case,  r   i
 r   4 (cos 30 a  sin 30 a ) e j 7( 0.5x  0.87 z )

x
z


The total reflected magnetic field is then
 r   1 ( 4 cos 30 a  9 a  4 sin 30 a ) e j 7( 0.5x  0.87 z )

x
y
z

6.3 Reflection and Refraction at Plane Interface between Two Media:
Oblique Incidence
Figure 6.7 shows two media with electrical properties  1 and  1 in medium 1, and  2 and
 2 in medium 2. Here a plane wave incident angle  i on a boundary between the two
media will be partially transmitted into and partially reflected at the dielectric surface.
The transmitted wave is reflected into the second medium, so its direction of propagation
is different from the incidence wave. The figure also shows two rays for each the
Notes by: Debbie Prestridge
19
incident, reflected, and transmitted waves. A ray is a line drawn normal to the equiphase
surfaces, and the line is along the direction of propagation.
2
Incident
rays
Reflected
rays
1
r
1
i
C
2
E
 1 , 1
 2 , 2
A
B
t
Reflected
rays
Figure 6.7
The incident ray 2 travels the distance CB, while on the contrary the reflected ray 1
travels the distance AE. For both AC and BE to be the incident and reflected wave fronts
or planes of equiphase, the incident wave should take the same time to cover the distance
AE. The reason being that the incident and reflected wave rays are located in the same
medium, therefore their velocities will be equal,
CB AE

V1 V 2
OR
AB sin  i  AB sin  r
With this being the case then it follows that
i   r
What is the relationship between the angles of incidence i and refraction  r ?
It takes the incident ray the equal amount of time to cover distance CB as it takes the
refracted ray to cover distance AD –
Notes by: Debbie Prestridge
20
CB AD

V1 V 2
And the magnitude of the velocity V1 in medium 1 is:
V1 
1
1  1
And in medium 2:
V2 
1
2   2
Also,
CB  AB sin  i
AD  AB sin  i
Therefore,
CB sin  i V 1



AD sin  t V 2
2   2
1   1
For most dielectrics  2  1  
Therefore,
sin  i

 2
sin  t
1
(6.12)
1  2  
Equation 6.12 is known as Snell’s Law of Refraction.
Notes by: Debbie Prestridge
21
6.3.1 Parallel Polarization Case – E is in Plane of Incidence
||r
i
r
||
 i||
i
 |r|
i
Region 1
 1 , 1
Y
(Out of
paper)
Region 2
 2 , 2
r
X
t
 ||t
 ||t
t
Z
Figure 6.9
The unknown amplitudes of the reflected and transmitted electric fields ||r and ||t can be
determined by simply applying the boundary conditions at the dielectric interface. The
electric fields ||r and ||t will now be used in the analysis to emphasize the case of parallel
r
and  tm .
polarization, instead of using the electric fields  m
The tangential component of H should be continuous across the boundary. Therefore,
 t e jir a
 r e jir a  
 i e jir a  

y
y
y
||
||
||
There is no need to carry the ay vector, because the magnetic fields only have one
component in the y direction. Recall that this relation is valid at z = 0,
 r e j i (sin r ix )   t e j  i (sin t x )
 i|| e ji (sin i x )  
||
||
Notes by: Debbie Prestridge
(6.13)
22
1 & 2 1 are the magnitudes of  in regions 1 & 2, respectively. In order for this to be
valid at any value of x at any point on the interface, and knowing  i   r :
1 sin i  2 sin  t
Or

sin  i  2 V 2 V 1



sin  t  1  V 2
V1
* This is the same relation that was determined earlier from Snell’s Law. Substitute
sin  i V 1

into equation 6.13 to obtain
sin  t V 2
 i   r   t

||
||
||
At z = 0
(6.14)
E and H are related by  , so equation 6.14 can be rewritten as
 i||   ||r 
1 t

 2 ||
(6.15)
Tangential components of E must be continuous across the boundary, therefore
 i|| cos  i   ||r cos r   ||t cos  t
At z = 0
(6.16)
*Remember the exponential terms cancel out z = 0, (Snell’s Law).
Equations 6.15 & 6.16 are solved by –
 cos i  2 cos t
 ||r   i|| 1
`1 cos i  2 cos t
And
 ||t   i||
22 cos  i
`1 cos  i  2 cos  t
(6.17)
*Making use of the fact that  i   r . Define the reflection coefficient  || and the
transmission  || :
Notes by: Debbie Prestridge
23
 || 
cos  t 
 ||r
2
cos  i
1
2 cos t  1 cos  i



 i|| 2 cos  t  1 cos  i
cos  t  2 cos  i

1  2  
1
And
 || 
 ||t
2  2 cos  t  1 cos  t 

2 cos  t  1 cos  t
 i||
2 cos  i
2
cos  i
1
cos  i 
1 2  
The total electric field in region 1 is
 ||tot   i||   ||r   im (cos i a x  sin  i a z) e j i  r +  rm ( cos r a x  sin  r a z) e j  rr

 j z cos i
j zcos i
 cos i  im e jx sin i (e
  || e
)ax



 sin  i  im e  jx sin  i  e  j  z cosi   || e j  z cosi a z


 


Traveling  wave
S tan ding plus
part
travelingwaves
(6.18)
Substituted  i  r , r  r from expressions derived earlier, and  rm  im   || .
Equation 6.18 states that there is a traveling-wave field in the x direction, and a traveling
and standing wave field in the z direction. The difference is that  ||  1 , but
that  ||    rm  im . By rearranging the second term in ax component of the total field –
1   e jz cos i  2    z cos  

||

||

i


This expression indicates that a wave of amplitude 1   || is propagating in the z
 
direction and another wave of amplitude 2  || has the characteristics of a standing wave
along the z axis. The characteristic of the wave along the z axis is a combination of a
traveling and standing wave. If  ||  1 the amplitude of the traveling wave will be zero,
and the wave characteristic along the z axis will be a totally standing wave. If   0 , the
||
Notes by: Debbie Prestridge
24
amplitude of the standing wave will be zero and the wave characteristic in the z direction
would be a totally traveling wave.
The magnetic field in region 1 is
 i e j  ir a +  r e j  rr a
 ||tot   i||   ||r  
y
y
m
m
i
 i
 j z cos i 
m
 r e j x cos i ) a y
= m e j xsin i (e

1
m

 im  j x sin  j z cos
 e j z cos i ) a
i 
i (e

e
y
||
1
The transmitted fields in medium 2 are
 i||   tm  cos  t a x  sin  t a z  e jt  r
=  ||  im  cos t a x  sin  t a z e jt  r
And

 i
t 
 t a e j t  r  ||  m e j t  rt a

y
||
m y
2
Where t  r  2  x sin t  z cost  and  tm  im  || .
Definition:
Brewster Angle – (from Brewster’s Law), the polarizing angle of which (when
light is incident) the reflected and refracted index is equal to the tangent of the polarizing
angle. In other words, the angle of incidence of which there is no reflection.
From the reflection coefficient expression-
 || 
2 cos  t  1 cos  i
2 cos t  1 cos  i
It can be seen that there is an angle of incidence at  | |  0 . This angle can be obtained
when
1 cos i  2 cos t
Notes by: Debbie Prestridge
25
Or
cos  i 
2
cos  t
1
(6.19)
The angle of incidence  i , at which  ||  0 , is known as the Brewster angle. The
expression for this angle in terms of the dielectric properties of media 1 & 2, considering
Snell’s Law for the special case 1   2   is
sin  i V 1  2

sin  t V 2  1
 1 2  
This condition is important, because it is usually satisfied by the materials often used in
optical applications.
Equation 6.19 will take the form –
cos i 
1
cos t
2
(6.20)
Square both sides of equation 6.20 and use Snell’s Law for the special case of
1   2   for the following result:
cos2  i


1
 cos2  t  1 1  sin2 t
2
2

 1 1  sin 2  i
2



The last substitution was based on Snell’s Law of refraction. Therefore,

1  sin2 
Notes by: Debbie Prestridge
 1  12 2
 sin  i
i 
 2  22
26

1  1  sin2  i
2

 1 

 12 

 22 
And
sin2  i 
2
(6.21)
 2  1
The Brewster angle of incidence is
sin  i 
2
(6.22)
 2  1
A specific value of θi can be obtained from equation 6.21 1  cos2  i 
Or
cos2  i  1 
cos  i 
2
 2  1
2
 2  1

1
 2  1

1
 2  1
(6.23)
From equations 6.22 & 6.23 –
tan  i 
2
1
This specific angle of incidence  i is called the Brewster angle   .
   tan1
Notes by: Debbie Prestridge
2
1
27
6.3.2 Perpendicular Polarization case – E Normal to Plane of Incidence
As shown in figure 6.10 is a perpendicular polarized wave incident at angle  i a
dielectric medium 2. Snell’s Law states that a reflected wave will be at the same
angle  r   i , and the transmitted wave in medium 2 at angle  t can be calculated using
this law. The amplitude of the reflected and transmitted waves can be determined by
applying the continuity of the tangential components of E & H at the boundary.
This is given by –
 i cos  
 t cos
 r cos  = 

i
i
t




i
i
 i
1,  1
r
r

i
 r
r
 2 , 2
X
t
 t
βt
Z
Since E & H are related by  ,
 i
 r
 t
cos  i 
cos  i 
cos  t
1
1
2
 i   r   t
At z = 0
(6.24)
(6.25)
*Note: The exponential factors were canceled after substituting z = 0 and using Snell’s
Laws in the above two equations.
Notes by: Debbie Prestridge
28
r
 cos i  1 cos t
    = 2

2 cos i  1 cos t
 i

And for nonmagnetic materials, 1   2   ,
cos t 
2
cos t
1
cos t 
2
cos t
1
  
at z = 0,
22 cos  i
 t
   i =
2 cos  i  1 cos  t
 
For nonmagnetic material,
2 cos i
  
cos i
2
cos t
1
6.4 Comparison between Reflection Coefficients  || and   for Parallel
and Perpendicular Polarizations
The significant differences between the two will be illustrated in the following example:
EXAMPLE
1. Define what is meant by the Brewster angle.


2. Calculate the polarization angle (Brewster angle) for an air water  r  81
interface at which plane waves pass from the following:
(a) Air into water.
(b) Water into air.
Notes by: Debbie Prestridge
29
SOLUTION
1. Brewster angle is defined as the angle of incidence at which there will be no
reflected wave. It occurs when the incident wave is polarized such that the E field
is parallel to the plane of incidence.
2. (a) Air into water:
 r1  1 and  r 2  81
The Brewster angle is then given by
   tan1
2
= 6.34°
1
Therefore,
   tan1 81 = 83.7°
(b) Water into air:
r1  81 and r 2  1
Hence,
1
= 6.34°
   tan1
81
To relate the Brewster angles in both cases, let us calculate the angle of
refraction.
sin  i

 2
sin  t
1
Therefore, in case a,
sin  
 81
sin  t
Therefore,
sin  t 
Notes by: Debbie Prestridge
sin 83.7
 011
.
9
30
Or  t  6.34  , which is the same as the Brewster angle for case b. Also, the angle of
refraction in case b is given by Snell’s Law as:
1
sin  



81
sin  t
81 
Therefore,
sin  t 
sin 6.34
 0.99
1
81

Or  t  83.7 , which is the Brewster angle for case a.
6.5 Total Reflection at Critical Angle of Incidence
In the previous section it was shown that for common dielectrics, the phenomenon of
total transmission exists only where the electric field is parallel to the plane of incidence
known as parallel polarization.
There is a second phenomenon existing for both polarizations:
 Total reflection occurring at the interface between two dielectric media
 A wave passing from a medium with a larger dielectric constant to a medium with
smaller value of ε
Snell’s Law of refraction shows –
sin  i

 2
sin  t
1
or
sin  i 
sin  t
2
1
(6.26)
Therefore, if  1   2 , and  t   i then a wave incident at an angle  i will pass into
medium 2 at a larger angle t .
Definition:
 c , (critical angle of incidence) is the value of  i that makes t = π/2, see Figure
6.13.
Substitute t = π/2 in equation 6.26 to get –
sin  c 
Notes by: Debbie Prestridge
2

, or  c  sin1 2
1
1
31
c
i
1
2
t   2
t
1   2
Figure 6.13 illustrates the fact that t  i , if 1   2 . The critical angle  c is defined as
the value of  i at which t = π/2.
Envision a beam of light impinging on an interface between two transparent media
where ni  nt . At normal incidence (  i = 0) most of the incoming light is transmitted
into the less dense medium. As  i increases, more and more light is reflected back into
the dense medium, while t increases. When t = 90°,  i is defined to be  c and the
transmittance becomes zero. For  i >  c all of the light is totally internally reflected,
remaining in the incident medium.
EXAMPLES:

Use Snell’s Law to derive an expression for θc. Compute the value of θc for a
water-air interface ( n w =1.33).
Rewrite
As
ni sin  i = nti sin t
sin t = nti sin t
Where nti < 1. Requiring that t = 90° for  i =  c leads to
sin  c = nti
Notes by: Debbie Prestridge
32
At water-air interface
 c sin 1 (
1
) = sin 1 0.752 = 48.8°
1.33

Imagine yourself lying on the floor of a pool filled with water, looking straight
upwards. How larger a plane angle doe the field of view beyond the pool
apparently subtend?
Rays striking the air-water interface from above at glancing incidence will
enter the water at a transmission angle equal to  c . The plane angle subtended at the
observer is therefore 2  c . Here,
sin  c =
1
1.33
Whence  c = 48.8° and 2  c = 97.6°.

Determine the critical angle for a water ( n w =1.33) –glass ( n g =1.50) interface.
We have
sin  c = nti
Or
Notes by: Debbie Prestridge
 c = sin 1
1.33
= sin 1 0.887 = 62.5°
1.50
33
6.6 Electromagnetic Spectrum
Figure 6.16 Electromagnetic spectrum from radio waves to X and γ
Wavelength
1μm
1 mm
1m
Radio
Microwave
3 GHz
1nm
10-6 m
10-3 m
Infrared
3 x 1012 Hz
rays.
10-9 m
V
i
s
b
l
e
U
l
t
r
a
v
i
o
l
e
t
3 x105 Hz
γ rays
X rays
3 x1018 Hz
Frequency
6.7 Application to Optics
The figure above shows the spectrum of electromagnetic radiation extending from the
long- wavelength radio waves to X rays and gamma rays the shortest wavelength.
Topics to be discussed will include control of polarization of incident waves, role of
Brewster windows in light amplification, and use of the concept of angle of total
reflection in optical fibers.
Notes by: Debbie Prestridge
34
6.7.1 Polarization by Reflection
Definition
Unpolarized light – light in which the wave orientation is random around the axis
of the beam.
Unpolarized light has both polarization cases
 Parallel polarization, where the electric field is the plane of incidence
 Perpendicular polarization where the electric field is perpendicular to the plane of
incidence
In certain cases, there may be a need to separate the two polarizations. One method that
can be used is the Brewster angle of incidence, also called the polarization angle, to
separate the two orthogonal polarizations.
Example
Consider an Unpolarized light that is incident at the Brewster angle on a piece of glass
with index of refraction n   r  15
. . The polarization with a electric field parallel to the
plane of incidence will be entirely transmitted and the other polarization with a electric
field perpendicular to the plane of incidence will be partially reflected and partially
transmitted. Why is the electric field parallel to the plane of incident totally transmitted?
*Because it is incident at the Brewster angle.
The second interface which is glass to air as illustrated in example 6.7 has an angle of
incidence also known as the Brewster angle for light incident from the glass side to free
space. So, again the polarization with E parallel to the plane of incident will be entirely
transmitted, and E perpendicular will be partially reflected and partially transmitted.
In Figure 6.17:
 Reflected wave is entirely polarized, E perpendicular to the plane of incidence
 Transmitted wave possess both polarizations
 Larger amplitude is the E parallel to plane of incidence – entirely transmitted
throughout the interfaces
 More glass elements and the transmitted light could be essentially completely
polarized, E parallel to the plane of incidence
Notes by: Debbie Prestridge
35
6.7.2 Brewster Windows or Brewster Cuts in LASER
In a normal situation there are more electrons in the ground state (level 1) than in the
excited states (level 2 & 3). In other words, there are more electrons in level 1 ready to
absorb photons that there are electrons in level 2 & 3 to emit photons. A net emission of
photons could be the result if this situation could be inverted. Such a condition is called
population inversion. This in fact is the fundamental principle involved in the operation
of a laser. Figure 6.8 illustrates this principle.
Definition:
Laser (Light Amplification by Stimulated Emission) – A device that produces
coherent radiation in the visible-light range, between 7500 and 3900 angstroms
Summarized steps leading to LASER action in three-level ruby laser material:
1. The laser material is in the shape of a long rod that is subjected to radiation from
an extremely intense light source that causes interatomic transition from energy
levels 1 to 3. (Figure 6.18b)
2. If the nonradiative transition between level 3 and level 2 is fast enough, then
electrons in level 3 will transfer to level instead of returning to level 1.
3. As a result of direct transition the population of electrons in level 2 will increase
from level 1. This is during the radiation from the light source, as well as the
transfer from level 3. (Figure 6.18c)
4. If the pumping action is large and fast enough the electron population at level 2
can be made larger than level 1. Radiation of light quanta at frequency f21 occurs
when the electrons can make the transition from level 2 to level 1.
5. By placing mirrors at the end of the laser and forcing the radiation to be reflected
back and forth maintaining the high-photon density, stimulated emission will
increase resulting in a large photon density build up or in other words an
avalanche of photons.
6. An intense light beam will result emerging from the end of the laser rod.
Notes by: Debbie Prestridge
36
Partially
polarized light;
mostly E
parallel to plane
of incidence
Polarized
light with E
perpendicular
to plane of
incidence
 1
2
n
n
n
n
Figure 6.17 Light polarizations by multiple reflections.
Figure 6.18 is a schematic diagram illustrating the sequence of events.
The role of the Brewster angle:
Known Factors
 The output of many lasers is linearly polarized
 The ratio of the light polarized in one direction exceeds the light polarized in the
orthogonal direction by 1000:1
As in most cases, a high degree of linear polarization will be the result of a Brewster
surface within the laser. A Brewster surface is usually used in the construction of a laser.
The light must be transmitted out of the medium of the laser to avoid minimal loss.
Notes by: Debbie Prestridge
37
(a) Thermal equilibrium
(b) Absorption of pump radiation
(c)Nonradiative transfer
to upper level
(d) Coherent radiative transition
and emitting laser
Figure 6.18 Sequence of events occurring in laser action.
Figure 6.19 is a schematic illustrating the use of Brewster windows in a gas discharge
laser. The Brewster angle makes sure that light in one polarization direction is
transmitted out of the medium of the laser to the reflecting mirrors and back into the
medium of the laser with no loss. Where the light is polarized perpendicular to the plane
of incidence a large loss at the Brewster surface will take place due to the reflection out
of the medium of the laser. The preferred polarization case (linear polarization) will lase
(emit coherent light) that will account for the high degree of polarization taking place at
the output.
Figure 6.19 Schematic illustrating the use of Brewster windows in a gas discharge LASER
Output beam
Gas discharge
tube (plasma)
Brewster windows
External mirror
(totally reflecting)
External mirror
(partially reflecting)
The device in Figure 6.19 exhibits stimulated emission of radiation. For and example
lets say the mixture of gases are helium and neon. These gases are confined to the glass
tube sealed at both ends by mirrors. An oscillator is connected to the tube to that causes
Notes by: Debbie Prestridge
38
electrons to sweep through the tube, colliding with atoms of gas and raising them to
exited states. Some neon atoms are excited to a higher state during this process that will
also result in a collision with excited helium atoms. Stimulated emission occurs as the
neon atoms make a transition to a lower state and neighboring excited atoms are
stimulated to emit at the same frequency and phase. This will result in a production of
coherent light.
6.7.3 Fiber Optics
Fiber optics deals with the transmission of light through small filamentary fibers called
dielectric waveguides. This is based on the phenomenon of total internal reflection
occurring at the point where the light is obliquely incident on an interface between two
media with different refractive indexes at an angle greater than the critical angle. Light is
incident at an angle θi as shown in Figure 6.20 and is required to determine the range of
values of the index of refraction n so the internal reflections will occur for any value of
θi .
Snell’s Law of refraction is the relationship between θi and θt as the wave enter the fiber
is
sin  i
2
1   
(6.27)

n
sin  t
1
If  2 is suppose to be larger than  c , then
sin  2 = cos t ≥ sin  c
(6.28)
Refraction from fiber to air sin  c = 1/n, therefore, from equation 6.27 & 6.28 –
2
1
 1
1    sin2  i 
 n
n
sin  2  cos t  1  sin2  t 
(6.29)
Solve for n,
n 2  1 sin 2 i
For equation 6.30 to be true then
requiring
n2 ≥ 2
or
(6.30)
= π/2, all incident light will be passed by the fiber
n≥
2
Most types of glass have n ≈ 1.5; therefore, we have a valid equation.
Notes by: Debbie Prestridge
39
Reflected point
Smallest critical angle
θi
θt θ2
n≥
2
Figure 6.20 Schematic illustrating the principle of light propagation in optical fibers.
Notes by: Debbie Prestridge
40