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Fields and Waves I
Lecture 24
Plane Waves at Oblique Incidence
K. A. Connor
Electrical, Computer, and Systems Engineering Department
Rensselaer Polytechnic Institute, Troy, NY
These Slides Were Prepared by Prof. Kenneth A. Connor Using
Original Materials Written Mostly by the Following:
 Kenneth A. Connor – ECSE Department, Rensselaer Polytechnic
Institute, Troy, NY
 J. Darryl Michael – GE Global Research Center, Niskayuna, NY
 Thomas P. Crowley – National Institute of Standards and
Technology, Boulder, CO
 Sheppard J. Salon – ECSE Department, Rensselaer Polytechnic
Institute, Troy, NY
 Lale Ergene – ITU Informatics Institute, Istanbul, Turkey
 Jeffrey Braunstein – Chung-Ang University, Seoul, Korea
Materials from other sources are referenced where they are used.
Those listed as Ulaby are figures from Ulaby’s textbook.
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Overview

EM Waves in Lossless Media
• Wave Equation & General Solution
• Energy and Power

EM Waves in Lossy Media
• Skin Depth
• Approximate wave parameters
 Low Loss Dielectrics
 Good Conductors
• Power and Power Deposition

Wave Polarization
• Linear, circular & elliptical

Reflection and Transmission at Normal Incidence
• Dielectric-Conductor Interface
• Dielectric-Dielectric Interface
• Multiple Boundaries

Plane Waves at Oblique Incidence
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Example 1 – Arbitrary Propagation Angle

The direction of E and  of a electromagnetic wave with = 500nm
are shown below. The wave is traveling through air. The electric field
has a magnitude of 30 V/m. What are the E and H phasors?
x
E

30°
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y axis is out of the page
z
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Example 1
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Example 1
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Arbitrary Propagation Angle
In phasor form we have had
 j z

 j z
E x ( z)  E m e
H y ( z) 
We can generalize this with

e
Em e


 jk r

k  k x a x  k y a y  k z a z
r  xa x  ya y  za z
 
k  r  a z  xa x  ya y  za z  z

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
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Arbitrary Propagation Angle
For propagation in more than the z direction, let us consider just
adding x propagation, since that is all we will need to do oblique
incidence.

k  k x a x  0a y  k z a z  k x a x  k z a z
 
   jkr
  jk x x  jk z z
E (r )  E m e
 a E E m e
  jk x x  jk z z
 
Em e
H (r )  a H

where we have left unspecified the unit vectors for E & H
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Oblique Incidence – Parallel Polarization
For the first choice,
we can assume that
the electric field in
directed in the plane
of incidence. This is
called parallel
polarization since E is
parallel to this plane.
Note that H is only
tangent to the
boundary while E has
both normal and
tangential
components.
Ulaby
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Oblique Incidence – Perpendicular Polarization
For the second choice,
we can assume that
the electric field in
directed out of the
plane of incidence.
This is called
perpendicular
polarization since E is
perpendicular to this
plane. Note that E is
only tangential while
H has both
components.
Ulaby
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Oblique Incidence
With the two possible polarizations, we have two sets of boundary
conditions. Thus, they will behave differently.
Note also that the combination of the two polarizations gives us all
possible vector components for E and H.
Now we must apply the boundary conditions to determine how the
incident, reflected and transmitted waves relate to one another.
VERY IMPORTANT POINT: Because of the x-directed
propagation, the phase of the E and H fields vary along the
boundary. Thus, our first task is to match the phase and then we will
match the amplitudes. The matching of the phase will allow us to
derive one of the most fundamental laws of optics.
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Oblique Incidence – Matching the Phase of the Electric
and Magnetic Fields at a Boundary
The incident electric field:
 

  jk1 r
1
m1

E

E
e
   jk sin x  jk cos z
 Em1 e 1 i 1 i
The reflected electric field:
     jk  r    jk sin x  jk cos z
E1  Em1 e 1  Em1 e 1 r 1 r
To match the phase of the terms at z = 0:
 jk1 sin i x   jk1 sin r x
i  r  1
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Oblique Incidence – Matching the Phase of the Electric
and Magnetic Fields at a Boundary
Thus, we have that the angle of incidence equals the angle of
reflection, a result that all of us have seen before. Now, we need to
see what happens to the transmitted angle.
Ulaby
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Oblique Incidence – Matching the Phase of the Electric
and Magnetic Fields at a Boundary
Consider now all three waves – incident, reflected and transmitted:
 

  jk1 r
  jk1 sin 1x  jk1 cos1z
1
m1
m1
 

  jk1 r
  jk1 sin 1x  jk1 cos1z
1
m1
m1


  jk2 r
  jk2 sin  t x  jk2 cos t z
2
m2
m2

E

E

E

E

E

E

E
e
e

E e

E e
e
e
Matching the phases at z = 0:
 k1 sin 1 x   k 2 sin t x
k i sin i  k1 sin 1  k 2 sin t  k 2 sin 2
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Oblique Incidence – Matching the Phase of the Electric
and Magnetic Fields at a Boundary
This is Snell’s Law:
k1 sin 1  k 2 sin 2
To put it in its more normal form:
k   o     o  o  r 

c
r 

c
n
n1 sin 1  n2 sin 2
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Snell’s Law:
There are many useful wave representations:
Ulaby
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Snell’s Law:
Using the wave front representation, we can see that Snell’s Law is
required to match the wave variations on the two sides of the
boundary.
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Applying the Boundary Conditions for Both Polarizations
Gives the Reflection and Transmission Coefficients
Ulaby
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Applying the Boundary Conditions for Both Polarizations
Gives the Reflection and Transmission Coefficients
Perpendicular Polarization:
 
 
E m1
E m1


E m2
E m1


2 cos1  1 cos2

2 cos1  1 cos2
22 cos1

2 cos1  1 cos2
1     
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Applying the Boundary Conditions for Both Polarizations
Gives the Reflection and Transmission Coefficients
Parallel Polarization:
2 cos2  1 cos1
|| 
2 cos2  1 cos1
22 cos1
 || 
2 cos2  1 cos1
cos1
1  ||   ||
cos2
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Example 2 – Oblique Incidence

. za y
A plane wave described by Ei  100 cost  x  173

is incident
 upon a dielectric material with r = 4.
a. Write E i in phasor
form.

b. What are i and i ?
c. What are  t and  t ?
d. What are the reflection and transmission coefficients?
e. Write the total electric field phasors in both regions.
r =1
x
y
E
r =4
z
i
i
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Example 2
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Example 2
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Example 2
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Example 2
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Critical Angle
For total reflection:
n1 sin 1  n2 sin 2
n2 k 2
sin c 


n1 k1
2
1
Ulaby
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Example 3 – Snell’s Law and Critical Angle
For visible light, the index of refraction for water is n = 1.33. If we
put a light source 1 meter under water and observe it from above
the surface of the water, what is the largest i for which light will be
transmitted?
How large will the circle of illumination be?
Ai r
i
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r
Wa ter
1m
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Example 3
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Example 4 -- Polarization
Consider the same material properties and incident angle as
Example 2, but assume the opposite polarization.
a. What are the reflection and transmission coefficients?
Which polarization has a lower reflection coefficient (magnitude)?
b. Now allow i to vary. At what value of i is the wave completely
transmitted? (i.e. What's the Brewster angle?)
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Example 4
Example 2
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Reflection as a Function of Angle
Note that the
reflection varies
with angle.
Perpendicular
reflects more that
Parallel. There is
also an angle for
which there is no
reflection for
parallel
polarization.
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Ulaby
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Brewster’s Angle for Parallel Polarization
sin  B 
B  tan
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1

1 1
1
2
2
1
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Optical Fibers
Light is guided down the fiber.
Ulaby
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Optical Fibers
Cladding is added to eliminate surface problems since part of the
wave actually propagates outside the core. Also note that the
pulses spread and decay due to a variety of losses.
Ulaby
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Rensselaer & Other Info Sources
 Prof. E. F. Schubert
http://www.ecse.rpi.edu/~schubert/Light-EmittingDiodes-dot-org/chap22/chap22.htm
 Prof. D. J. Wagner
http://www.rpi.edu/dept/phys/ScIT/
 Prof. F. Ulaby (From his CD)
 http://www.amanogawa.com/index.html
 Movies of Waves from Prof. H. C. Han at Iowa State
http://www.ee.iastate.edu/%7Ehsiu/em_movies.html
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From Prof. Schubert’s Notes
This is
why
the sky
is blue.
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Power and Energy
Note that power density (the Poynting Vector) is not necessarily
conserved across the boundary. However, the total power is.
Because of the
boundary conditions,
the Poynting Vector is
conserved for
perpendicular but not
for parallel polarization.
All formulas are
summarized in Table 8-2
of Ulaby.
Ulaby
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