Electromagnetic Wave
Propagation
The goal is to describe quantitatively the propagation of
electromagnetic waves in an unbounded medium
characterized, in general, by conductivity 𝜎 ,
permittivity 𝜀 = 𝜀𝑟 𝜀𝑜 and permeability 𝜇 = 𝜇𝑟 𝜇𝑜 .
The following media to be considered are:
1. Free space 𝜎 = 0, 𝜀 = 𝜀𝑜 , 𝜇 = 𝜇𝑜
2. Lossless dielectrics 𝜎 ≅ 0, 𝜀 = 𝜀𝑟 𝜀𝑜 , 𝜇 =
Recall the wave equations in a source-free medium are
given by,
2𝑬
𝜕𝑬
1
𝜕
𝛻 2 𝑬 = 𝜇𝜎
+ 2 2
𝜕𝑡 𝑢 𝜕𝑡
2𝑯
𝜕𝑯
1
𝜕
𝛻 2 𝑯 = 𝜇𝜎
+ 2 2
𝜕𝑡 𝑢 𝜕𝑡
where the speed of propagation is given by,
1
𝑢=
𝜇𝜖
Consider and EM wave propagating in 𝑧 direction in a
lossless medium. The corresponding wave equation for
the electric field is given,
𝜕2𝐸
1 𝜕2𝐸
= 2 2
2
𝜕𝑧
𝑢 𝜕𝑡
The general solution to the above partial differential
equation is,
𝐸 𝑧, 𝑡 = 𝐸 𝑧 ± 𝑢𝑡
For sinusoidal EM waves with frequency 𝑓 and
wavelength 𝜆,
𝑧
𝐸 𝑧, 𝑡 = 𝐸𝑜 𝑠𝑖𝑛2𝜋 𝑓𝑡 ±
𝜆
= 𝐸𝑜 𝑠𝑖𝑛 𝜔𝑡 ± 𝛽𝑧
Substitution into the wave equation yields,
𝑢 = 𝑓𝜆
Recall the time-harmonic Maxwell’s equations in a
source-free medium,
𝛻 2 𝑬𝑠 + 𝑘 2 𝑬𝑠 = 0
𝛻 2 𝑯𝑠 + 𝑘 2 𝑯𝑠 = 0
with
𝑘 2 = 𝜔2 𝜀𝑒𝑓𝑓
Define the effective propagation constant 𝛾𝑒𝑓𝑓 as,
𝛾𝑒𝑓𝑓 = 𝛼 + 𝑗𝛽 = 𝑗𝑘 =
𝑗𝜔𝜇 𝜎 + 𝑗𝜔𝜀
The time-harmonic Maxwell’s equations become,
 Es  
2
eff s
 Hs  
2
eff
2
2
E 0
Hs  0
Consider the solution of the Helmholtz equation for
the electric field phasor. Let,
Es  aˆxEsx  aˆyEsy  aˆzEsz
Then,
2
ˆ
ˆ
ˆ


  axEsx  ayEsy  azEsz    eff  aˆxEsx  aˆyEsy  aˆzEsz   0
2
Collecting terms of the same vector components,

aˆ   E

0
2
2
aˆx  2Esx   eff
Esx   aˆy 2Esy   eff
Esy 
z
2
sz

E
2
eff sz
Each component must be zero. Take the x-component,
 Esx  
2
 Esx  Esx  Esx
2
E 




effEsx  0
2
2
2
x
y
z
2
eff sx
2
2
2
Use the technique of separation of variables,
Esx  x, y,z   X  x  Y  y  Z z 
d2X
d2Y
d2Z
2
YZ 2  XZ 2  XY 2   effXYZ  0
dx
dy
dz
Dividing by XYZ ,
1 d2X 1 d2Y 1 d2Z
2


  eff  0
2
2
2
X dx
Y dy
Z dz
 y2
 x2
 z2
so that,
2
x
2
y
2
z
    
2
eff
 jωμ  σ  jωε
Take the differential equation for X,
1 dX
2
 x
2
X dx
2

dX
2
  xX
2
dx
2
Then,
X  x   Ae xx
where A is a complex amplitude.
Since the equations for Y and Z are similar, we get
Y  y   Be
Z z   Ce  zz
 y y
Therefore, the phasor for the x-component of the
electric field is,
Esx  X  x  Y  y  Z z   Ae

 ABC  e
 xx
e
 y y
e
  zz

 xx

Be
 Eoxe
 y y


Ce  zz
  xx y y  zz


Introduce the following vectors,
 effaˆp  aˆx x  aˆy y  aˆz z
where aˆp is a unit vector along the direction of
propagation of the electromagnetic wave,
R  aˆxx  aˆy y  aˆzz
is the position vector in Cartesian coordinates. Then,
 xx   y y   zz   effaˆp R
The phasor for the x-component of the electric field
becomes,
Esx  aˆxEoxe
  eff aˆp R
The y and z phasor components of the electric field
are obtained similarly as,
Esy  aˆyEoye
  effaˆp R
Esz  aˆzEoze
  effaˆp R
The total electric field phasor is,


Es  Esx  Esy  Esz  aˆxEox  aˆyEoy  aˆzEoz e
or,
Es  Esoe
  eff aˆp R
  effaˆp R
The instantaneous expression for the electric field is
obtained as,
 

E R, t  Re Ese
jt

 Re E e
 Re Esoe
so
 Esoe
  Re E e
so
 jt  j aˆ R 
p
 aˆp R
  aˆp R
e

  eff aˆp R

j t   aˆp R


e
jt


cos t   aˆp R

Note that (-) sign is the physically meaningful solution.
 
E R, t  Esoe
 aˆp R

cos t   aˆp R

For an attenuated EM wave travelling along +z
direction,
Now, from Maxwell’s equations,
free medium.

 Es   Esoe
  eff aˆp R
 E  e
so
Es  0 in a source  eff aˆp R





   xx y y zz
 Eso  aˆx
 aˆy
 aˆz  e
y
z 
 x
 x x  y y  zz 
 Es o aˆx x  aˆy y  aˆz z e


 Eso  effaˆpe
  effaˆp Esoe
  eff aˆp R
 aˆp R


  effaˆp Es  0
 aˆp Es  0


Now, from  Es
free medium.
 jωBs  jωμHs in a source-

1
  eff aˆp
Hs  
  Esoe
jωμ
R


1
  eff aˆp

 effaˆp  Esoe
jωμ



 eff
jωμ
 eff


aˆp  Esoe
jωμ
aˆp  Es
 effaˆp R

R

Introduce the impedance of the medium,
ηeff 
jωμ
 eff
jωμ

σ  jωε
Then,
1
Hs 
aˆp  Es
ηeff
Since the wave equation for the magnetic field intensity
is very similar to that of the electric field, we can write
the phasor for the magnetic field as,
Hs  Hsoe
  eff aˆp R
Given the phasor for the magnetic field component,
the electric field is derived as follows
1
Hs 
aˆp  Es
ηeff
1
 aˆp  Hs 
aˆp  aˆp  Es
ηeff
1 

aˆp aˆp E  Es aˆp aˆp 

ηeff 


Es  ηeffaˆp  Hs




Plane Electromagnetic Wave
E
aˆp
direction of
propagation
H
 eff  jωμ  σ  jωε 
ηeff
jωμ

σ  jωε
1
Hs 
aˆp  Es
ηeff
Es  ηeffaˆp  Hs
Propagation in Lossless Charge-Free Media
In this case,
 eff  jωμ  0  jωε  j2ω2με  jω με    j

 0
  ω με
The wave propagates without attenuation at the speed
given by,
ω
1
c
up  

β
με
μrεr
The impedance of a lossless medium is,
ηeff
jωμ
μrμo
μr
μ




ηo
0  jωε
ε
εrεo
εr
where ηo is the intrinsic impedance of free space,
μo
ηo 
 120π Ω
εo
Propagation in Lossy Dielectric Media
Energy loss due to finite conductivity or polarization
loss or both can be included by defining the complex
permittivity,
εc  ε  jε
'
"
From the time-harmonic Ampere’s law,


"
'

 Hs  σEs  jωεcEs   σ  ωε  jωε  Es
An effective conductivity can be defined as,
σ eff  σ  ωε
"
The general expressions for the attenuation and phase
constants can be obtained as follows,

2
eff
σ

 jωμ  σ  jωε   ω μ  ε  j 
ω

2
   j     
2
2
2
  j2
Equating real and imaginary parts and solving for  and ,
2


με
σ



 ω
1
1


2
ωε 



2


με
σ



 ω
1
1


2
ωε 



For low-loss dielectrics, σ ωε 1 . By using the binomial
n
approximation, 1  x  1  nx
2
2
1 σ 
 σ 
1
 1 

2  ωε 
 ωε 
This yield the expression for  and ,
σ μ

,   ω με
2 ε
In media with complex permittivity,
σ  σ eff and ε  ε
'
A measure of loss in dielectrics is given by the loss
tangent. Recall Ampere’s law,


 Hs   σ  ωε"  jωε'  Es
σ  ωε" σeff
tan 

'
ωε
ωε'
Loss Tangent in Materials
Propagation in Conductors
For good conductors, σ ωε
1. Then,
2

 σ
σ 

 1
1 


 ωε
ωε 



Substitution of this result into the expressions for
the attenuation and propagation constants yields,
με σ
ωμσ
  ω

2 ωε
2
Or in terms of the cyclic frequency,
     fμσ
Since σ ωε
by,
1 , the intrinsic impedance is approximated
ηeff
jωμ


σ  jωε
jωμ
σ
Now,
j  e
So that,
ηeff
i 2

1
2
e
i 4

1 j
2
ωμ
ωμ i45o
 i45o

e  2 e
1  j 
2σ
σ
σ
For a good conductor, the magnetic field lags the
electric field by 450.
The propagation speed and wavelength are reduced
significantly in a conductor,
ω
2ω
up  

μσ
,
 2

fμσ
These results are shown schematically below,
The distance from the surface of the conductor at
which the field amplitude has dropped to e-1 is called
the penetration depth or skin depth,

1


1
 fμσ
Because of the strong attenuation of the EM fields
inside a conductor at high frequencies, current resides
mainly on its surface. Let us introduce the concept of
sheet resistance.
1 L
1 L
R

σ wt σt w
For a square cross - section, L  W
Rsheet
1
R 
σt
 
Jx
w
L
For a semiinfinite conducting slab, an x-polarized
electric field decays as,
Ex  Exo e  z
From Ohm’s law, the corresponding current density is,
Jx  σExo e  z
The current flowing through a surface extending from
zero to infinity along the z-direction and of width w
along the y-direction is,
I


w
z 0 y 0

σExoe
 z
wσExo
 wσExo z 
dydz  
e  
 wσExo

 
0
For the distance L along the x-direction, the potential
drop is,
V  ExoL
The corresponding resistance drop is,
ExoL
V
1 L
R 

I wσExo σ w
By analogy with the sheet resistance, we define a skin
resistance as,
Rskin
1

σ
for a semiinfinite conducting slab. For a slab of finite
thickness, following the same analysis as above,
Rskin 

1
σ 1  e
t


Polarization of a Plane EM Wave
By convention, the direction of the polarization of a
plane electromagnetic wave is given by the direction of
the electric field component.
Consider a wave traveling along the +z-direction.
y
Ey
H
z

E
Ex
x
 is the direction of polarization
measured from the x - axis
To determine the direction of polarization, consider a
wave traveling along the +z-direction in a lossless
medium. In this case, the x- and y- components of the
electric field are given by,
E  z, t   aˆxExocos  ωt   z  φx 

 aˆyEoy cos ωt   z  φy

where φx and φy are the phase constants for the x and
y components of the wave, respectively.
Linear Polarization
The x and y components of the electric field are
either in-phase or out-of-phase by 1800. The
direction of the electric field is fixed for any
position and at any given time.
Then,
φx  φy or φy  1800
E  z, t   aˆxExocos  ωt   z  φx 
 aˆyEoy cos  ωt   z  φx 


E  z, t   aˆxEox  aˆyEoy cos  ωt   z  φx 
And the direction of polarization is given by,
E
 Eoy cos  ωt   z  φx  
-1  oy 
  Tan 
  Tan 

 Eoxcos  ωt   z   φx 
 Eox 
-1
i.e., the direction of polarization is constant.
x
E  z, t 
H  z, t 
y
z
Elliptical or Circular Polarization
The direction of the electric field rotates as a
function of position and at any given time.
This is achieved if the x- and y- components of the
electric field are moving out-of-phase. Without loss
of generality, let
φx  0 and φy  φ
Then,
E z,t   aˆxExocos  ωt   z   aˆyEoycos  ωt   z  φ
Consider the curve traced out by the tip of the electric
field vector at z = 0.
E z,t   aˆxExocos  ωt   aˆyEoycos  ωt  φ
The magnitude of the electric field at any time is given
by,
2
2
2
E  Exo cos  ωt   Eoy cos  ωt  φ 
2
which traces out an ellipse.
2
For simplicity, consider the special case
2
2
φ  π

E  Exo cos  ωt   Eyo cos ωt  
2
2
2
2
2
2
 Exo cos  ωt   Eyo sin  ωt 
2
And the direction of polarization is given by,
 Eyo sin  ωt  
  Tan 

 Exocos  ωt  
E

-1  yo
 Tan 
tan  ωt  
 Exo

-1
i.e., the direction of polarization rotates.
2
2

For Eox  Eoy , the polarization is elliptical. For Eox  Eoy ,
the polarization is circular.
For circular polarization,
  ωt
y
y

E
E

z
x
right- circular polarization
z
x
left- circular polarization
In terms of phasors, the electric field of an
electromagnetic wave propagating along the +z direction is
written as
Es  aˆxExoe
jφx
 aˆyEyoe
For a left-hand circularly polarized wave,

Es  Exo aˆx  jaˆy
jφy
φy  φx  π

For a right-hand circularly polarized wave, φy

Es  Exo aˆx  jaˆy
2

 φx  π
2
Poynting’s Theorem
Recall the two curl equations in Maxwell’s equations,
B
E  
t
D
H  J 
t
Use the following vector identity,



 
 E  H  H  E  E  H

 B 
 E H  H  
E

 t 





D 
 J 

t 

 


 


 μH
 B 
1  μH H
 1 2
H
H

  μH 

 t 
t
2
t
t  2



 εE

ε
E
E
 D 
1
 1 2
E
E

  εE 

 t 

t
2

t

t
2




For EM wave propagation, we ignore the presence of
sources so that
J  σE
and,
E J  σE E  σE
2
Then,
1  μH
 E H  
2 t


2
  1  εE   σE
2
2
t
2
Integrating both sides of the equation within a finite
volume V enclosed by a surface area S in the medium
in which the EM wave propagates,
V
S
 1  μH2  1  εE2 

2


V dV  E  H   V dV  2 t  2 t  σE 




Using Stoke’s theorem to convert the volume integral
of the divergence to a surface integral,

S


 
E  H dS     dV
t  V
Rate of flow of
EM energy out of
the area enclosing
the volume
1
1
2
2 
2

μH

εE

σE
 2    V   dV
 2 

Rate of decrease of the total
energy stored in the EM field
Rate of EM power
dissipated in the
volume
In more compact form,

  P dS 
S
t
  w
V
e

 wm  dV   pσ dV
V
P  E H
1 2
we  εE
2
1
2
wm  μH
2

Poynting's vector

electric energy density

magnetic energy density
pσ  σE2

Ohmic power density
Poynting’s Theorem for TimeHarmonic EM Fields
Recall the two curl equations in time-harmonic Maxwell’s
equations for a source-free conducting medium,
 Es  jωBs
 Hs   Es  jωDs
To derive Poynting’s theorem for this case, we need to
involve the complex conjugates of phasor quantities.
Thus,

 Es  H
*
s
  H  E   E  H 
*
s
s
s
*
s

 Es  H
*
s

H
*
s
 jωBs  Es  E
*
s

   Es E  jω H Bs  Es D
*
s


 jωD
*
s
*
s
*
s

   Es E  jω μHs H  εEs E
*
s
*
s
*
s
so that,


   Es  H dv    Es E dv
V
*
s
*
s
V


 jω μHs Hs*  εEs Es* dv
V

Using Stoke’s theorem to convert the volume integral
of the divergence to a surface integral,




  Es  Hs* ds    Es Es*dv  jω μHs Hs*  εEs Es* dv
S
V
real part
Ohmic power loss
V
imaginary part
reactive power loss
Note that the integrands on the right sides of the
equation are real quantities since they are products of
a complex number and its conjugates.
It is also of interest to calculate the time-average
value of the product of two quantities. Let’s illustrate
this with the Poynting’s vector.




jωt
jωt




 Re Hse
P  E  H Re Es e
 

Now,
Es  Es e

jθEs
Hs  Hs e


jθHs

jθEs jωt
jθHs jωt




P  Re Es e e
 Re Hs e e

 





j θEs  ωt   
j θHs  ωt  

P  Re Es e
 Re Hs e

 





  Es cos θEs  ωt    Hs cos θHs  ωt 

 





 Es  Hs cos θEs  ωt cos θHs  ωt 


Use the following trigonometric identity,
cos  α ±β  cos  α  cos β  sin  α  sin β 
we get,
1
cos  α  cos β   cos  α  β   cos  α  β  
2
So, if we do the following substitution
α  θEs  ωt
β  θHs  ωt

 
1
cos  θ  θ
2

cos θEs  ωt cos θHs  ωt 
Es
Hs



 2ωt  cos θEs  θHs 

The expression for the Poynting’s vector becomes,
1
P  Es  Hs cos θEs  θHs  2ωt  cos θEs  θHs 


2


The time-average of this quantity is,
P
t
1 T
  P dt
T 0
where T is the period of the oscillation.


P
t
1

Es  Hs
2T




 T cos θ  θ  2ωt dt  T cos θ  θ dt 
Es
Hs
Es
Hs
0
 0

T
1

Es  Hs  cos θEs  θHs dt
0
2T

1
 Es  Hs cos θEs  θHs
2



1
j θEs  θHs  

 Es  Hs Re e


2
1
jθEs
 jθHs


 Re Es e  Hs e


2
Or, finally
P
t
1
*

 E  H  Re Es  Hs 
2
This result can be generalized to the calculation of
the time-average of the product of two quantities, i.e.
the time-average is obtained as one-half of the real
part of the product of one quantity times the complex
conjugate of the other.
Another form for the time-averaged value of
the Poynting’s vector can be derived as follows.
P
t
1

1
1

*
* 
 Re Es  Hs   Re Es   aˆp  Es  
2
2 
η

1
1
1
*
*
 Re  Es  Es aˆp  Es  aˆp Es 
2 η
η


let, η  η e



jη
P
t

Es
2
2η
e
2 aˆp R
cosη aˆp
Boundary Conditions
medium 1
ân2
E1 , D1 , B1 , H1
medium 2
E2 , D2 , B2, H2
Tangential components
E1t  E2t


aˆn2  H1  H2  JS
Normal components
B1n  B2n


aˆn2 D1  D2  ρS
EM Wave Propagation at Interfaces
1. Write down the phasor representation for the
electric and magnetic field.
2. Determine the total fields for each medium.
3. Apply the boundary conditions at the interface.
Normal Incidence
medium 1
x
Esi
ε1 , μ1 ,  1
H
i
s
aˆi  aˆz
aˆr  aˆz
Incident
Hst
Esr
x
ˆ
ε2 , μ2 ,  2
aˆt  aˆz
z
y
Hsr
Reflected
Esi  aˆxEoi e  1az R
 aˆxEoi e 1z e  j1z
1
i
Hs  aˆz  Esi
η1
Eoi 1z  j1z
 aˆy e e
η1
medium 2
Est
Esr  aˆxEor e 1  z 
 aˆxEor e1z e j1z
1
r
Hs   aˆz   Esr
η1
Eor 1z j1z
 aˆy e e
η1
-  aˆ
R
Transmitted
ˆ
Est  aˆxEot e  2az R
 aˆxEot e -2z e -j2z
1
t
Hs  aˆz  Est
η2
Eot 2z  j2z
 aˆy
e e
η2
The total fields in medium 1 is a sum of the incident and the reflected
electromagnetic waves,
Es1  Esi  Esr  aˆxEoi e 1z e  j1z  aˆxEor e1z e j1z
i
r
E
E
Hs1  Hsi  Hsr  aˆy o e 1z e  j1z  aˆy o e1z e j1z
η1
η1
In medium 2, the total field consists only of the transmitted wave,
Es2  Est  aˆxEot e 2z e  j2z
Eot 2z  j2z
H  H  aˆy
e e
η2
2
s
t
s
Apply the boundary conditions at z  0 ,
E1t  E2t

Eoi  Eor  Eot
Since there are no surface currents generated at the interface
between two dielectric media,


aˆn2  H1  H2  0 
H1t  H2t
1 i
1 t
r
Eo  Eo   Eo

η1
η2
Solving for the reflected and transmitted wave amplitudes in terms of
the incident wave,
η2  η1 i
E 
Eo
η2  η1
r
o
2η2 i
E 
Eo
η2  η1
t
o
Define the reflection coefficient ,
E η2  η1


E η2  η1
r
o
i
o
Define the transmission coefficient  ,
Eot
2η2
 i 
Eo η2  η1
It is easily shown that,
1  
For lossless media, the superposition of the incident and reflected
waves in medium 1 leads to the formation of standing waves.
Es1  Esi  Esr  aˆxEoi e  j1z  aˆxEor e j1z

 aˆ E  e
 aˆxEoi e  j1z  e j1z
i
x o
 j1z

  e j e j1z

E1  z, t   aˆxEoi  cos  ωt  1z    cos  ωt  1z     
The maxima occurs when  = 0 and the amplitude is,
1
Emax
 Eoi 1   
The minima occurs when  =  and the amplitude is,
1
Emin
 Eoi 1   
The standing wave ratio for the total wave in medium is defined as,
1
1 
Emax
SWR  1 
Emin 1  
Oblique Incidence
x
aˆi  aˆz
Esi
θi
y
H
i
s
Perpendicular polarization
(TE or s – polarized)
- the electric field is
perpendicular to the plane
of incidence
x
z
θi
Esi
Hsi
aˆi  aˆz
y
Parallel polarization
(TM or p – polarized)
- the electric field is
parallel to the plane of
incidence
Note: The plane of incidence is the plane containing the vector
corresponding to the direction of propagation and the normal
to the boundary.
z
TE- or s-polarized
medium 1
ε1 , μ1
medium 2
ε2 , μ2
Incident
aˆpi  aˆx sinθi  aˆzcosθi
E  aˆ E e
i
s
i
y o
 j1aˆpi R
 j xsinθ zcosθi 
 aˆyEoi e 1  i
i
1
E
Hsi  aˆpi  Esi  o  aˆxcosθi  aˆz sinθi  e  j1  xsinθi zcosθi 
η1
η1
Reflected
aˆpr  aˆx sinθr  aˆzcosθr
E  aˆ E e
r
s
r
y o
 j1aˆpr R
 aˆyEor e
 j1  xsinθr zcosθr 
r
1 r
E
r
H  aˆp  Es  o  aˆxcosθr  aˆz sinθr  e  j1  xsinθr zcosθr 
η1
η1
r
s
Transmitted
aˆpt  aˆx sinθt  aˆzcosθt
E  aˆ E e
t
s
t
y o
 j2aˆpt R
 j xsinθ zcosθt 
 aˆyEot e 2  t
t
1
E
Hst  aˆpt  Est  o  aˆxcosθt  aˆz sinθt  e  j2  xsinθt zcosθt 
η2
η2
Apply the boundary conditions at z=0,

E1t  E2t
Eoi e
 j1  xsinθi 
 Eore
 j1  xsinθr 
 Eote
 j2  xsinθt 
Impose the phase matching condition,
1  xsinθi   1  xsinθr   2  xsinθt 
 1  xsinθi   1  xsinθr 
This leads to Snell’s law of reflection,
θi  θr
ie, the angle of incidence is equal to the angle of reflection.

1  xsinθi   2  xsinθt 
This leads to Snell’s law of refraction,
1 sinθt

2 sinθi
Total Internal Reflection
For 1  2, there is a critical incident angle in which the
incident EM wave is totally internally reflected.
sinθi 
n2
sinθt
n1
The critical angle θi  θc corresponds to θt  π
θcritical
2
.
 2 
 Sin  
 1 
-1
Forθi  θcritical
, the incident wave is totally reflected.
Going back to the other boundary conditions at z=0,

H1t  H2t
Eoi
Eor
Eot
 j1  xsinθr 
 j2  xsinθt 
 j1  xsinθi 
 cosθe

cosθ
e


cosθ
e
i
r
t
η1
η1
η2
After imposing the phase matching condition,
Eoi  Eor  Eot
Eot
1 i
r
Eo  Eo  cosθi  cosθt

η1
η2
Solving for the reflection and transmission coefficient yields,
TE
Eor η2cosθi  η1cosθt
 i 
Eo η2cosθi  η1cosθt
 TE
E
2η2cosθi


E η1cosθt  η2cosθi
t
o
i
o
The following relation also holds true for TE-polarized wave.
 TE  1  TE
The Brewster’s angle is the angle of incidence with zero reflectance.
TE  0
 η2cosθBTE  η1cosθt  0
or
η2 cos θBTE  η1 cos θt
2
2
2
2
2
2
 2 
 1 
2
2

 1  sin θBTE  
 1  sin θt 
 2 
 1 


From Snell’s law,
11sin θB  2 2sin θt
2
2
TE
Solving for the Brewster’s angle yields,
sinθBTE
με

1   1 2
μ2ε1 


2
μ
1   1 
 μ2 
For non-magnetic media, there is no Brewster’s angle for TE or
s polarized wave.
TM- or p-polarized
medium 1
ε1 , μ1
medium 2
ε2 , μ2
Incident Fields
Esi  Eoi e
 j1  xsinθi zcosθi 
cosθiaˆx  sinθiaˆz 
i
E
Hsi  o e  j1  xsinθi zcosθi  aˆy
η1
Reflected Fields
Esr  Eore
 j1  xsinθr zcosθr 
cosθraˆx  sinθraˆz 
r
E
 j xsinθ zcosθr 
Hsr   o e 1  r
aˆy
η1
Transmitted Fields
Est  Eot e
 j2  xsinθt zcosθt 
cosθtaˆx  sinθtaˆz 
t
E
 j xsinθ zcosθt 
Hst  o e 2  t
aˆy
η1
For TM or p polarized wave,
TM
E η2cosθt  η1cosθi


E η2cosθt  η1cosθi
 TM
E
2η2cosθi


E η1cosθi  η2cosθt
r
o
i
o
t
o
i
o
Also,
 TM
cosi
 1  TM 
cost
The Brewster’s angle is the angle of incidence with zero reflectance.
TM  0
 η2cosθt  η1cosθBTM  0
or
η12cos2θBTM  η22cos2θt


η12 1  sin2θBTM  η22 1  sin2θt 
From Snell’s law,
12sin2θB
TM
2

 22sin2θt  sin2θt  12 sin2θB
2
TM
The Brewster’s angle is,
sinθBTM 
22 22  12 
22 12  12 22
For lossless non-magnetic media,
sinθBTM 
1
ε
1   r1 
 εr2 
or θBTM
 εr2 
 Tan 

ε
r1


-1
For non-magnetic media, the transmitted wave is TM or p polarized
at Brewster’s angle.
TE- or s-polarized
medium 1
ε1 , μ1
medium 2
ε2 , μ2
Incident
E  aˆ E e
i
s
i
y o
 j1aˆpi R
 aˆyEoi e
 j1  xsinθi zcosθi 
i
1 i
E
i
H  aˆp  Es  o  aˆxcosθi  aˆz sinθi  e  j1  xsinθi zcosθi 
η1
η1
i
s
Reflected
E  aˆ E e
r
s
r
y o
 j1aˆpr R
 aˆyEor e
 j1  xsinθr zcosθr 
r
1 r
E
r
H  aˆp  Es  o  aˆxcosθr  aˆz sinθr  e  j1  xsinθr zcosθr 
η1
η1
r
s
Transmitted
E  aˆ E e
t
s
t
y o
 j2aˆpt R
 j xsinθ zcosθt 
 aˆyEot e 2  t
t
1
E
Hst  aˆpt  Est  o  aˆxcosθt  aˆz sinθt  e  j2  xsinθt zcosθt 
η2
η2
Total Internal Reflection
For 1  2, there is a critical incident angle in which the
incident EM wave is totally internally reflected.
sini 
2
sint
1
The critical angle θi  θc corresponds to θt  π
critical
2
.
 2 
 sin  
 1 
-1
Fori  critical, the incident wave is totally reflected.
For i  critical,
sint 
1
sini  1
2
2


where    1 sin   1
 2

2
cost   1   sint    j
For the transmitted wave,
aˆp  aˆx sin t  aˆzcos t  aˆx
1
sini  jaˆz
2
 j  x   
 j xsin zcost 
Est  aˆyEote 2  t
 aˆyEote 2 1
Hst 
=
1
2
aˆp  Est 
2 sint z  j  
 aˆyEote  j1xsini e

1  1
ˆ
ˆ
ˆ t  j1xsini e
a
sin


j
a

 x
i
z   ayEo e
2  2

1  1
sini
 aˆz
2  2

jaˆx  Eote  j1xsini e

z
z
z
For z0, we take the upper sign. The time-averaged Poynting’s
vector for the transmitted wave is then obtained as,
Pave

1
1
t
t
 Re Es  Hs   Re  aˆyEote  j1xsini e
2
2 


z

 1  1
   aˆz
sini


2
 2
 t  j1xsini  z 
ˆ
jax  Eo e
e 


t 2
 Et 2


E

1

 o
1
o sini 2 z
2 z 
1
 Re 
e
sini  jaˆz    aˆx
e
 aˆx
2  2
22 2
 2
 

For i  critical, the transmitted wave is non-propagating.




Going back to the other boundary conditions at z=0,

H1t  H2t
Eoi
Eor
Eot
 j1  xsinθr 
 j2  xsinθt 
 j1  xsinθi 
 cosθe

cosθ
e


cosθ
e
i
r
t
η1
η1
η2
After imposing the phase matching condition,
Eoi  Eor  Eot
Eot
1 i
r
Eo  Eo  cosθi  cosθt

η1
η2
Solving for the reflection and transmission coefficient yields,
TE
Eor η2cosθi  η1cosθt
 i 
Eo η2cosθi  η1cosθt
 TE
E
2η2cosθi


E η1cosθt  η2cosθi
t
o
i
o
The following relation also holds true for TE-polarized wave.
 TE  1  TE
The Brewster’s angle is the angle of incidence with zero reflectance.
TE  0
 η2cosθBTE  η1cosθt  0
or
η2 cos θBTE  η1 cos θt
2
2
2
2
2
2
 2 
 1 
2
2

 1  sin θBTE  
 1  sin θt 
 2 
 1 


From Snell’s law,
11sin θB  2 2sin θt
2
2
TE
Solving for the Brewster’s angle yields,
sinθBTE
με

1   1 2
μ2ε1 


2
μ
1   1 
 μ2 
For non-magnetic media, there is no Brewster’s angle for TE or
s polarized wave.
TM- or p-polarized
medium 1
ε1 , μ1
medium 2
ε2 , μ2
Incident Fields
Esi  Eoi e
 j1  xsinθi zcosθi 
cosθiaˆx  sinθiaˆz 
i
E
Hsi  o e  j1  xsinθi zcosθi  aˆy
η1
Reflected Fields
Esr  Eore
 j1  xsinθr zcosθr 
cosθraˆx  sinθraˆz 
r
E
 j xsinθ zcosθr 
Hsr   o e 1  r
aˆy
η1
Transmitted Fields
Est  Eot e
 j2  xsinθt zcosθt 
cosθtaˆx  sinθtaˆz 
t
E
 j xsinθ zcosθt 
Hst  o e 2  t
aˆy
η1
For TM or p polarized wave,
TM
E η2cosθt  η1cosθi


E η2cosθt  η1cosθi
 TM
E
2η2cosθi


E η1cosθi  η2cosθt
r
o
i
o
t
o
i
o
Also,
 TM
cosi
 1  TM 
cost
The Brewster’s angle is the angle of incidence with zero reflectance.
TM  0
 η2cosθt  η1cosθBTM  0
or
η12cos2θBTM  η22cos2θt


η12 1  sin2θBTM  η22 1  sin2θt 
From Snell’s law,
12sin2θB
TM
2

 22sin2θt  sin2θt  12 sin2θB
2
TM
The Brewster’s angle is,
sinθBTM 
22 22  12 
22 12  12 22
For lossless non-magnetic media,
sinθBTM 
1
ε
1   r1 
 εr2 
or θBTM
 εr2 
 Tan 

ε
r1


-1
For non-magnetic media, the transmitted wave is TM or p polarized
at Brewster’s angle.