6 slides

Dr. Rakhesh Singh Kshetrimayum
6. Plane waves reflection from media
interface
Dr. Rakhesh Singh Kshetrimayum
3/25/20141 Electromagnetic FieldTheory by R. S. Kshetrimayum

6.1 Introduction
Plane waves reflection
from a media interface
Normal
incidence
TE
Oblique
incidencePerfect
conductor
3/25/2014Electromagnetic FieldTheory by R. S. Kshetrimayum2
TE
Lossless medium
Good
conductor
Fig. 6.1 Plane waves reflection from media interface
Lossy conducting
medium
TM
Brewster angle
Total internal reflection
Effect on
polarization

6.1 Introduction
Till now, we have studied plane waves in various medium
Let us try to explore how plane waves will behave at a media
interface
In practical scenarios of wireless and mobile
communications,
radio wave will reflect from
walls &
other obstacles on its path
When a radio wave reflects from a surface,
the strength of the reflected waves is less than that of the
incident wave
3/25/20143 Electromagnetic FieldTheory by R. S. Kshetrimayum

6.1 Introduction
The ratio of the two (reflected wave/incident wave) is
known as the‘reflection coefficient’ of the surface
This ratio depends on the
conductivity (σ),
permittivity (ε) and
permittivity (ε) and
permeability (µ)
of the material that forms the reflective surface
as well as material properties of the
air
from which the radio wave is incident

6.1 Introduction
Some part of the wave will be transmitted through the
material
How much of the incident wave has been transmitted
through the material is
also dependent on the material parameters mentioned above
also dependent on the material parameters mentioned above
It is given by another ratio known as‘transmission
coefficient’
It is the ratio of the transmitted wave divided by the incident
wave

6.1 Introduction
In plane wave reflection from media interface,
what we will be doing is
basically writing down the
electric and
magnetic field expressions
magnetic field expressions
in all the regions of interest and
apply the boundary conditions
to get the values of the
transmission and
reflection coefficients

6.1 Introduction
One of the possible applications of such an exercise is in
wireless communication,
where we have to find the
transmission and
reflection coefficients of
multipaths
multipaths
Electromagnetic waves are often
reflected or
scattered or
diffracted
at one or more obstacles before arriving at the receiver

6.2 Plane wave reflection from media
interface at normal incidence
For smooth surfaces,
EM waves are reflected;
For rough surfaces,
EM waves are scattered;
For edges of surfaces,
For edges of surfaces,
EM waves are diffracted
Fig. 6.2 depicts multipath for a mobile receiver in the car from
direct,
reflected and
diffracted waves
of the signal sent from the base station

6.1 Introduction
Fig. 6.2 Illustration of reflected and diffracted wave in mobile
communications (this chapter’s study will be very helpful in
understanding wireless and mobile communications)

We will consider the case of normal incidence,
when the incident wave propagation vector is along the normal
to the interface between two media
6.2.1 Lossy conducting medium
We will assume plane waves with electric field vector
We will assume plane waves with electric field vector
oriented along the x-axis and
propagating along the positive z-axis without loss of
generality
For z<0 (we will refer this region as region I and it is
assumed to be a lossy medium)

Let us do a more generalized analysis by assuming this region
I as a lossy medium
It could be a lossless medium like free space
The incident electric and magnetic fields can be expressed as
1
ˆ z
i oE xE e γ−
=
r
1
1
1
ˆ z
i oH y E e γ
η
−
=
r

where η1 is the medium 1 wave impedance and
Eo is the arbitrary amplitude of the incident electric field
The expression for intrinsic wave impedance and propagation
constant
( )
1 1 1 1
1 1 1 1 1 1
1 1 1 11 1 1
; 1
j j j j
j j
jj j
ωµ ωµ ωµ σ
η γ α β ω µ ε
γ ωε σ ωεωµ ωε σ
= = = = + = −
++

iE
r
E
r
2
t
γ
r
1
i
γ
r
iH
r
tE
r
tH
r
iE
r
r
2 2 2, ,ε µ σ
rE
r
1 1 1, ,ε µ σ
1
r
γ
r
rH
r
rE
r
Fig. 6.3A plane EM wave is incident from region I or medium 1 for z<0

Convention:
a circle with a dot in the center an arrow pointing
perpendicularly out of the page and
a circle with a cross an arrow pointing perpendicularly into
the page)
the page)
Notation for fields:
subscript i incident, r reflected and t transmitted

Note that the incident wave from region I will be partially
reflected and transmitted at the media interface between two
regions
For reflected wave, z<0, wave direction is in the –z axis and
the reflected electric field is expressed as
the reflected electric field is expressed as
where the Γ is the reflection coefficient
1
ˆ z
r oE x E e γ+
= Γ
r

The reflection coefficient is defined as the ratio of amplitude
of the reflected electric field divided by amplitude of the
incident electric field as follows:
r
i
E
E
Γ =
The reflected magnetic field can be obtained from the
reflected electric field using the Maxwell’s curl equations as
iE

1
1
1 1 1
ˆ ˆ ˆ
0 0
r r
r r
r
z
o
E j H
x y z
E j E j
H
j x y z
E e γ
ωµ
ωµ ωµ ωµ
+
∇× = −
∇× ∇× ∂ ∂ ∂
⇒ = − = =
∂ ∂ ∂
Γ
r r
r r
r
( ) ( ) ( )1 1 11
1
1 1 1
0 0
ˆ ˆ ˆ
o
z z z
o o o
E e
j j
y E e y E e y E e
z j
γ γ γγ
γ
ωµ ωµ ωµ
+ + +
Γ
∂ 
= Γ = Γ = − Γ 
∂ 
1
1
ˆ z
r oH y E e γ
η
+Γ
∴ = −
r

Therefore, the Poynting vector for reflected wave of region I
is
shows power is traveling in the -z axis for the reflected wave
2 2* 2
*
1
ˆz
r r r o
z
S E H E e α
η
+
= × == − Γ
r r r
shows power is traveling in the -z axis for the reflected wave
For transmitted wave, z>0, wave is propagating in lossy
medium 2 (we will refer this region as region II),

where η is the wave impedance of lossy medium 2 and τ is
z
ot eExE 2ˆ γτ −
=
r
zo
t e
E
yH 2
2
ˆ γ
η
τ −
=
r
where η2 is the wave impedance of lossy medium 2 and τ is
the transmission coefficient
intrinsic wave impedance and propagation constant are

transmission coefficient is defined as the ratio of amplitude of
the transmitted electric field divided by amplitude of the
( )
2 2 2 2
2 2 2 2 2 2
2 2 2 22 2 2
; 1
j j j j
j j
jj j
ωµ ωµ ωµ σ
η γ α β ω µ ε
γ ωε σ ωεωµ ωε σ
= = = = + = −
++
the transmitted electric field divided by amplitude of the
incident electric field as follows:
t
i
E
E
τ =

Our intention here is to find the transmission and reflection
coefficients
Let us rewrite the fields in the two regions: region I (z<0)
and region II (z>0) (seeTable 6.1) and
Apply the boundary conditions to obtain the two unknown
Apply the boundary conditions to obtain the two unknown
coefficients

Region I (lossy medium 1) Region II (lossy medium 2)
Table 6.1 Fields in the two lossy regions (normal incidence)
1
ˆ z
i oE xE e γ−
=
r
1
1
1
ˆ z
i oH y E e γ
η
−
=
r
1
ˆ z
r oE x E e γ+
= Γ
r
1
1
ˆ z
r oH y E e γ
η
+Γ
= −
r
z
ot eExE 2ˆ γτ −
=
r zo
t e
E
yH 2
2
ˆ γ
η
τ −
=
r

This is basically boundary value problem with boundary
conditions at z=0
Note that total electric and magnetic fields (both incident
and reflected) are tangential to the interface at z=0
Similarly, the transmitted electric and magnetic field are also
Similarly, the transmitted electric and magnetic field are also
tangential to the interface at z=0
Also note that there are no surface current density at the
interface
Hence, the tangential components of electric and magnetic
fields must be continuous at the interface z=0

Therefore,
( ) 0 00
1 1i r t z
E E E E Eτ τ
=
+ = ⇒ + Γ = ⇒ + Γ =
r r r
0 0
0
1 2 1 2
1 1
i r t
z
H H H E E
τ τ
η η η η=
   − Γ − Γ
+ = ⇒ = ⇒ =   
   
r r r
Therefore,
Hence,
( ) ( ) 2 1
2 1
1 2 1 2 2 1
1 1 1
1 1
η ητ
η η
η η η η η η
−− Γ − Γ + Γ
= ⇒ = ⇒ − Γ = + Γ ⇒ Γ =
+
2 1 2
2 1 1 2
2
1 1
η η η
τ
η η η η
−
= + Γ = + =
+ +

6.2.2 Lossless medium
If the regions are lossless dielectric,
then, σ = 0 and µ and ε are real quantities
The propagation constant for this case is purely imaginary
and can be written as
and can be written as
γ = jβ =jω√(µε)
The wave impedance of the dielectric is
η= = = = =jωµ
γ
j
j
ωµ
β
ωµ
ω µε
µ
ε r
r
ε
µ
η0

For a lossless medium,
η is real, so,
both Γ and τ are real
Electric and magnetic fields are in phase with each other
The wavelength in the dielectric is
The wavelength in the dielectric is
= =
And the phase velocity is
= = =
where c is the speed of light in free space
λ 2π
β
2π
ω µε
pv ω
β µε
1
rr
c
εµ

Table 6.2 Fields in the two lossless regions (normal incidence)
Region I (lossless medium 1) Region II (lossless medium 2)
1
ˆ j z
i oE xE e β−
=
r 1
1
1
ˆ j z
i oH y E e β
η
−
=
r
1
ˆ j z
r oE x E e β+
= Γ
r 1
1
ˆ j z
r oH y E e β
η
+Γ
= −
r
2
ˆ j z
t oE x E e β
τ −
=
r
2
2
ˆ j zo
t
E
H y e β
τ
η
−
=
r

Applying the boundary conditions like before,
we can get the expression for
reflection and
transmission coefficients,
we will get the same expression as before
2 1
2 1
;
η η
η η
−
Γ =
+
2
1 2
2η
τ
η η
=
+

Poynting vector in region I
( ) ( ) ( ) ( )1 1 1 1
* *
* *
1 1 1 0 0
0 0
1 1
ˆ ˆ ˆ ˆj z j z j z j z
tot tot i r i r
E E
S E H E E H H E e x E e x e y e yβ β β β
η η
− + − 
= × = + × + = + Γ × −Γ 
 
r r r r r r r
2 2 2 2 2 22 2 2j z j zβ β− +   Γ Γ Γ Γ
( ) ( )
( ) ( )
1 1
1 1
2 2 2 2 2 22 2 2
0 0 0 0 0 0 2 22
1 1 1 1 1 1
2 2
0 02
1
1 1
ˆ ˆ1
ˆ1 2 sin 2
j z j z
j z j z
E E e E e E E E
z e e z
E E
j z z
β β
β β
η η η η η η
β
η η
− +
+ −
   Γ Γ Γ Γ   
= − + − = − Γ + −   
      
 Γ 
= − Γ − 
  

Poynting vector in region II
( )2 2
* 22
02 * 0
0
2 2
ˆ ˆ ˆj z j z
t t
EE
S E H x E e e y zβ β ττ
τ
η η
− − 
= × = × = 
 
r r r
( )
( ) ( ) ( )
2 2
2 12 2 1 2
2 2 2
12 1 2 1 2 1
4 41
1 1
η η η η η
ηη η η η η η
− −Γ
− Γ = − = ⇒ =
+ + +
Q
( )
( ) 222 2 2
02 22 2
2
2 1 1 2 12 1
12 4 1
ˆ
E
S z
η η τ
τ τ
η η η η ηη η
− Γ− Γ
= ⇒ = ⇒ = ∴ =
+ +
r

Power conservation
Compare the time average power flow in the two regions
For z <0, the time average power flow through 1m2 cross
section is
( )
2
1 1 1− Γr
And for z >0, the time average power flow through 1m2
cross section is
( )
2
21 1
0
1
1 1 1
ˆRe
2 2
avgS S z E
η
− Γ
= • =
r
( )
2
22 2
0
1
1 1 1
ˆRe
2 2
avgS S z E
η
− Γ
= • =
r

Hence,
So real power is conserved
6.2.3 Good conductor:
If the region II (z >0) is a good (but not perfect) conductor
1 2
avg avgS S=
If the region II (z >0) is a good (but not perfect) conductor
and
region I is a lossless medium like free space,
the propagation constant for medium 2 is

where δ is the skin depth for the medium 2,
which is a good conductor
Similarly, the intrinsic impedance of the conducting medium
( ) ( )2 2
2 2 2 2
1
1 1
2 s
j j j
ωµ σ
γ α β
δ
= + = + = +
2
s
Similarly, the intrinsic impedance of the conducting medium
2 (see also example 6.2) simplifies to
( )
( )
( )( )
( )
( )
( )
( )
2
22 2 2
2
2 22
2 2 2 2
2
2 2
1 1
(1 )
2
1 1 12 2 2
1
1 1 2 21
j j j j
j j
j
j j j j
j
j j j
ωµωµ ωµ ωµ
η
γ ωµ σ σωµ σ
ωµ ωµ ωµ ωµ
σ σ σ σ
= = = =
+ +
+
− + +
= = = = +
+ − −

Now, the wave impedance of medium 2 is complex, with a
phase angle of 45o so the electric
and magnetic field will be 45o out of phase and Γ and τ will
be complex.
Now, let us write down the field expressions in region I and
Now, let us write down the field expressions in region I and
II for lossless dielectric (region I) and good conductor
(region II) interface

Table 6.4 Fields in the two regions (normal incidence)
Region I (lossless medium) Region II (good conductor)
1
ˆ j z
i oE xE e β−
=
r 1
1
1
ˆ j z
i oH y E e β
η
−
=
r
1
ˆ j z
r oE x E e β+
= Γ
r
1
1
ˆ j z
r oH y E e β
η
+Γ
= −
r
2
ˆ z
t oE x E e γ
τ −
=
r
2
2
ˆ zo
t
E
H y e γ
τ
η
−
=
r

Applying the boundary conditions like before,
we can get the expression for
reflection and
transmission coefficients,
2 1
2 1
;
η η
η η
−
Γ =
+
2
1 2
2η
τ
η η
=
+

Note that we have to use the previous expressions for
intrinsic wave impedances for lossless medium and good
conductor
1 ;
µ
η
ε
= ( ) 2
2 1
2
j
ωµ
η
σ
= +
Power conservation
let us find out the Poynting vector in the two regions
Noting that the field in region I comprise of incident and
reflected wave,
we can write the Poynting vector for z <0
ε ( )2
2
1
2
jη
σ
= +

For z >0, only transmitted fields exist and we can write
( ) ( ) ( ) ( )
( )
1 1 1 1
1 1
* *
* *
1 1 1 *0 0
0 0
1 1
2
20 2 2*
1
ˆ ˆ ˆ ˆ
ˆ1
j z j z j z j z
tot tot i r i r
j z j z
E E
S E H E E H H E e x E e x e y e y
E
e e z
β β β β
β β
η η
η
− + −
− +
 
= × = + × + = + Γ × − Γ 
 
= − Γ + Γ − Γ
r r r r r r r
For z >0, only transmitted fields exist and we can write
down the Poynting vector as
( )
*
2 2* *
02 * 20
0 * *
2 2
ˆ ˆ ˆz z z
t t
EE
S E H x E e e y z eγ γ αττ
τ
η η
− − − 
= × = × = 
 
r r r

At z=0,
( )
2
201 *
1
ˆ1
E
S z
η
= − Γ + Γ − Γ
r
( )
( )
2 2 2 2
20 0 2 02 *4
ˆ ˆ ˆ1
E E E
S z z z
τ η
= = = − Γ + Γ − Γ
r
Complex power is conserved
Then the time average power flow through a 1m2 cross
section can be calculated as follows
Since
( )
( )2*
2 11 2
ˆ ˆ ˆ1S z z z
η ηη η
= = = − Γ + Γ − Γ
+

is purely an imaginary number, therefore,
1 12 2* j z j z
e eβ β− +
−Γ + Γ
( )
2
21 1
0
11 1
ˆRe
2 2
avgS S z E
η
− Γ
= • =
r
which shows power balance at z=0 only
( ) 0
1
ˆRe
2 2
avgS S z E
η
= • =
( ) z
avg eEzSS α
η
2
2
2
2
0
22 1
2
1
ˆRe
2
1 −Γ−
=•=
r

Note that the Poynting vector for z>0,
this power density in the lossy conductor decays
exponentially according to the attenuation factor
6.2.4 Perfect conductor:
Now assume that the region II (z >0) contains a perfect
Now assume that the region II (z >0) contains a perfect
conductor and region I is a lossless medium
Let us rewrite the expressions for propagation constant and
wave impedance of a good conductor and deduce the
expressions for perfect conductor

Perfect conductor implies σ →∝, and
( ) ( )2 2
2 2 2 2
1
1 1
2 s
j j j
ωµ σ
γ α β
δ
= + = + = +
( ) 2
2
2
1
2
j
ωµ
η
σ
= +
Perfect conductor implies σ2→∝, and
correspondingly, α2→∝, δs
2→0, η2→0
Note that
Therefore, τ→0 and Γ→-1
2 2 1
1 2 2 1
2
,
η η η
τ
η η η η
−
= Γ =
+ +

Table 6.5 Fields in the two regions (normal incidence)
Region I (lossless medium) Region II (perfect conductor)
1
ˆ j z
i oE xE e β−
=
r
1
1
1
ˆ j z
i oH y E e β
η
−
=
r
1
ˆ j z
r oE x E e β+
= Γ
r 1
1
ˆ j z
r oH y E e β
η
+Γ
= −
r
0tE ≅
r
0tH ≅
r

The field for z >0 thus decay infinitely fast and
are identically zero in the perfect conductor as we have seen
in example 6.2 as well
The perfect conductor can be thought of as‘shorting out’ the
incident electric field
incident electric field
Now let us write down the field expressions in region I and II
for lossless dielectric and perfect conductor interface (see
Table 6.5)

Power conservation
In order to see the power conservation,
let us find out the Poynting vector in the two regions
Noting that the field in region I comprise of
incident and
incident and
reflected wave,
we can write the Poynting vector for z <0 (Γ=-1)
( )1 11
0 0 1
ˆ ˆ2 sinj z j z
tot i rE E E xE e e x jE zβ β
β−
= + = − = −
r r r

Therefore, the Poynting vector for z<0 is
( )1 11 0 0
1
1 1
ˆ ˆ2 cosj z j z
tot i r
E E
H H H y e e y zβ β
β
η η
−
= + = + =
r r
( )
2
4 Er r
which has a zero real part and
thus indicates that no real power is delivered to the perfect conductor
Fields in region II is anyway zero and hence power is also zero
( )
2
* 01 1 1
1 1
1
4
ˆ sin costot tot
E
S E H zj z zβ β
η
= × = −
r r

interface at oblique incidence
We will consider the problem of a plane wave
obliquely incident on a plane interface
between two lossy conducting regions
We will first consider two particular cases of this problem as
follows:
follows:
the electric field is in the xz plane (parallel polarization)
the electric field is in normal to the xz plane (perpendicular
polarization)

Any arbitrary incident plane wave can be expressed
as a linear combination of these two principal polarizations
The plane of incidence is that plane containing
the normal vector to the interface and
the direction of propagation vector of the incident wave
the direction of propagation vector of the incident wave

For Fig. 6.4, this is the xz plane
For perpendicular polarization (TE),
electric field is perpendicular to the plane of incidence
For parallel polarization (TM),
electric field is parallel to the plane of incidence
electric field is parallel to the plane of incidence

tt HE
rr
,
z
x
rr HE
rr
,
rθ tθ
Fig. 6.5 Oblique incidence of plane EM wave at a media interface
1 1 1
, ,ε µ σ 2 2 2, ,ε µ σ
z
Region II
ii HE
rr
,
Region I
iθ
rθ tθ
nˆ

6.3.1 Perpendicular polarization (TE):
In this case, electric field vector is perpendicular to the xz
plane,
Hence, it will have component along the y-axis
Since the electric field is transversal to the plane of incidence
Since the electric field is transversal to the plane of incidence
They are also known transverse electric (TE) waves

iθ rθnˆ nˆ zˆ
xˆxˆ
zˆ
iθγ cos1
iθγ sin1
rθγ cos1
rθγ sin11
i
γ
r 1
r
γ
r
xˆ
tθnˆ zˆ
2 cos tγ θ
2 sin tγ θ 2
t
γ
r
Fig. 6.6Wave propagation vector for (a) incident (b) reflected and (c) transmitted
EM waves at oblique incidence

Let us assume that the incident wave propagates in the first
quadrant of xz plane without loss of generality and
(incident propagation vector) makes an angle θi with
the normal (see Fig. 6.6 (a))
1
i
γ
r
( ) ( ) ( )'
1 1 1 1 1 1
ˆ ˆˆ ˆcos sin cos sin cos sini
i i i i i iz z x zz xx z x z xγ γ θ γ θ γ θ γ θ γ θ θ• = + • + = + = +
r r
( )1 cos sin
0
ˆi iz x
iE E e y
γ θ θ− +
=
r
1
1
i
i i i
E
E j H H
j
ωµ
ωµ
∇×
∇× = − ⇒ =
−
r
r r r
Q

( )
( ) ( )
( ) ( )
1 1
1
1 1
cos sin cos sin
0
1 1
cos sin
0
cos sin cos sin
0
ˆ ˆ ˆ
1
ˆ ˆ
0 0
ˆ ˆ
i i i i
i i
i i i i
z x z x
z x
z x z x
x y z
E e e
x z
j x y z j z x
E e
E e e
x z
γ θ θ γ θ θ
γ θ θ
γ θ θ γ θ θ
ωµ ωµ
− + − +
− +
− + − +
    ∂ ∂ ∂ ∂ ∂ 
= = − +       − ∂ ∂ ∂ − ∂ ∂     
    ∂ ∂ 
= −       
( )
{ }1 cos sin0 1
ˆ ˆcos sini iz xE
e x z
γ θ θγ
θ θ− +
= − +
0
1
ˆ ˆ
E e e
x z
j z xωµ
∂ ∂ 
= −       ∂ ∂    
( )
{ }
( )
( )
1
1
cos sin0 1
1
cos sin0
1
ˆ ˆcos sin
ˆ ˆcos sin
i i
i i
z x
i i
z x
i i
E
e x z
j
E
e x z
γ θ θ
γ θ θ
γ
θ θ
ωµ
θ θ
η
− +
− +
= − +

= − +

Let us assume that the reflected wave propagates in the
second quadrant of xz plane and
(reflected propagation vector) makes an angle θr with
the normal (see Fig. 6.6 (b))
1
r
γ
r
( ) ( ) ( )1 1 1 1 1 1
ˆ ˆˆ ˆ' cos sin cos sin cos sinr
r r r r r rz z x zz xx z x z xγ γ θ γ θ γ θ γ θ γ θ θ• = − + • + = − + = − +
r r
( )1 cos sin
0
ˆr rz x
r TEE E e y
γ θ θ− − +
= Γ
r

Note that and will have the same magnitude
since both the waves are still in the same region I,
only their direction changes
Since the Poynting vector must be negative like the previous
case of normal incidence,
1
r
γ
r
1
i
γ
r
case of normal incidence,
You could also use the Maxwell’s curl equation below to find
this
( )
( )1 cos sin0
1
ˆ ˆcos sinr rz x
r TE r r
E
H e x z
γ θ θ
θ θ
η
− − +
= Γ +
r
1ωµj
E
H r
r
−
×∇
=
r
r

The transmitted fields will have similar expression with the
incident fields except
that now the θi should be replaced by θt (angle that transmitted
propagation vector makes with the normal),
γ1 should be replaced by γ2 (wave is in region II now) and
γ1 should be replaced by γ2 (wave is in region II now) and
multiplication by (transmission coefficient)
The transmitted fields are ( )2 cos sin
0
ˆ t tz x
t TEE yE e
γ θ θ
τ − +
=
r
( )
( )2 cos sin0
2 2
ˆ ˆcos sint tz xt TE
t t t
E E
H e x z
j
γ θ θτ
θ θ
ωµ η
− +∇×
= = − +
−
r
r

Table 6.5 Fields in two regions (oblique incidence:
perpendicular polarization)
( )1 cos sin
0
ˆi iz x
iE E e y
γ θ θ− +
=
r
( )
( )1 cos sin0
1
ˆ ˆcos sini iz x
i i i
E
H e x z
γ θ θ
θ θ
η
− +
= − +
r
( )1 cos sin
0
ˆr rz x
r TEE E e y
γ θ θ− − +
= Γ
r
( )
( )1 cos sin0
1
ˆ ˆcos sini iz xTE
r r r
E
H e x z
γ θ θ
θ θ
η
− +Γ
= +
r
( )2 cos sin
0
ˆ t tz x
t TEE yE e
γ θ θ
τ − +
=
r
( )
( )2 cos sin0
2
ˆ ˆcos sint tz xTE
t t t
E
H e x z
γ θ θτ
θ θ
η
− +
= − +
r

Equating the tangential components of electric field
(electric field has only Ey component and it is tangential at the
interface z=0) and
magnetic field
(magnetic field has two components: Hx and Hz and only Hx is
(magnetic field has two components: Hx and Hz and only Hx is
tangential at the interface z=0)
at z=0 gives 1 21sin sinsini trx xx
TE TEe e eγ θ γ θγ θ
τ− −−
+ Γ =
1 21sin sinsin
1 1 2
1
cos cos cosi trx xxTE TE
i r te e eγ θ γ θγ θ τ
θ θ θ
η η η
− −−Γ−
+ = −

If Ex and Hy are to be continuous at the interface z = 0 for all
x,
then, this x variation must be the same on both sides of the
equations (also known as phase matching condition)
tri θγθγθγ sinsinsin 211 ==
1 2; sin sini r i tθ θ γ θ γ θ⇒ = =

The first is Snell’s law of reflection
which states that the angle of incidence equals the angle of
reflection
The second result is the Snell’s law of refraction
(refraction is the change in direction of a wave due to change in
(refraction is the change in direction of a wave due to change in
velocity from one medium to another medium)
Also note that refractive index of a medium is defined as
0 0
0 0
r r
r r
p
c
n
v
µ ε µ ε
µ ε
µ ε
= = =

hence, for a lossless dielectric media,
Now we can simplify above two equations by applying Snell’s
2 2 22 2 1 2
1 1 2 11 1 1
sin
sin
i
t
v n
v n
µ ε εθ γ β
θ γ β µ ε ε
= = = = = =
Now we can simplify above two equations by applying Snell’s
two laws as follows
1 TE TEτ+Γ =
1 1 2
cos cos
cosi r TE
TE t
θ θ τ
θ
η η η
− + Γ = −

The above two equations has two unknowns τTE and ГTE and
it can be solved easily as follows
Therefore,
2
1 1
cos cos
cos
i r
TE TE
t
θ θ η
τ
η η θ
 
= −Γ 
 
1 TE TEτ+Γ =
Therefore,
22 2 2
1 1 1 1
1 2 2 1 2 1
1 1 1 2
cos coscos cos
1 1 1
cos cos cos cos
cos cos cos cos cos cos
cos cos cos cos
i ir r
TE TE TE
t t t t
t r i t i t
TE TE
t t t r
θ η θη η θ η θ
η θ η θ η θ η θ
η θ η θ η θ η θ η θ η θ
η θ η θ η θ η θ
     
+ Γ = −Γ ⇒ Γ + = −     
     
   + − −
⇒ Γ = ⇒ Γ =   
+   

Hence, the reflection and transmission coefficients for
oblique incidence (Fresnel coefficients) for perpendicular
2 1 1 2 2 1 2
1 2 1 2 1 2
cos cos cos cos cos cos 2 cos
1 1
i t t r i t i
TE TE
t r t r t r
η θ η θ η θ η θ η θ η θ η θ
τ
η θ η θ η θ η θ η θ η θ
− + + −
∴ = + Γ = + = =
+ + +
oblique incidence (Fresnel coefficients) for perpendicular
polarization are given as follows:
For normal incidence, it is a particular case and put
2 1
2 1
cos cos
cos cos
i t
TE
i t
η θ η θ
η θ η θ
−
Γ =
+
2
2 1
2 cos
cos cos
i
TE
i t
η θ
τ
η θ η θ
=
+
0=== tri θθθ

6.3.2 Parallel Polarization (TM):
In this case, electric field vector lies in the xz plane
Since the magnetic field is transversal to the plane of
incidence such waves are also called as transverse magnetic
(TM) waves
(TM) waves
So let us start with H vector which will have only y
component,
( )ii xz
i e
E
yH θθγ
η
sincos
1
0 1ˆ +−
=
r

1
1
i
i
H
H j E E
j
ωε
ωε
∇×
∇× = ⇒ =
r
r r r
Q
( ) ( )1 1cos sin cos sin
ˆ ˆ ˆ
1 i i i iz x z x
x y z
E e e
γ θ θ γ θ θ− + − +    ∂ ∂ ∂ ∂ ∂ 
( )
( ) ( )
( )
{ } ( )
1 1
1
1 1
cos sin cos sin
0
1 1 1
cos sin0
1
cos sin cos sin0 1
0
1 1
1
ˆ ˆ
0 0
ˆ ˆˆ ˆcos sin cos si
i i i i
i i
i i i i
z x z x
z x
z x z x
i i i
E e e
x z
j x y z j z x
E
e
E
e x z E e x z
j
γ θ θ γ θ θ
γ θ θ
γ θ θ γ θ θ
ωε ωε η
η
γ
θ θ θ
ωε η
− + − +
− +
− + − +
    ∂ ∂ ∂ ∂ ∂ 
= = − +       ∂ ∂ ∂ ∂ ∂     
= − = −( )n iθ

Similar to the previous case of parallel polarization, we can
write the reflected magnetic field vector as
( )1 cos sin0
1
ˆ ˆ r rz xTM
r
E
H y ye
γ θ θ
η
− − +Γ
= −
r
incident fields except that
now the θi should be replaced by θt (angle that transmitted
propagation vector makes with the normal),
γ1 should be replaced by γ2 (wave is in region II) and
multiplied by a factor (transmission coefficient forTM case)

Table 6.6 Fields in two regions (oblique incidence: parallel
( )
( )2
cos sin
0
ˆ ˆcos sint tz x
t TM t tE E e x z
γ θ θ
τ θ θ
− +
= −
r
( )2 cos sin0
2
ˆ t tz xTM
t
E
H y e
γ θ θτ
η
− +
=
r
Table 6.6 Fields in two regions (oblique incidence: parallel
polarization) {next page}

( )
( )1 cos sin
0
ˆ ˆcos sini iz x
i i iE E e x z
γ θ θ
θ θ− +
= −
r
( )Er
( )
( )2
cos sin
0
ˆ ˆcos sint tz x
t TM t tE E e x z
γ θ θ
τ θ θ
− +
= −
r
( )ii xz
i e
E
yH θθγ
η
sincos
1
0 1ˆ +−
=
r
( )
{ }1 cos sin
0
ˆ ˆcos sinr rz x
r TM r iE E e x z
γ θ θ
θ θ− − +
= Γ +
r
( )1 cos sin0
1
ˆ ˆ r rz xTM
r
E
H y ye
γ θ θ
η
− − +Γ
= −
r
( )2 cos sin0
2
ˆ t tz xTM
t
E
H y e
γ θ θτ
η
− +
=
r

Equating the tangential components of magnetic field
(magnetic field has only Hy component and it is tangential at the
interface z=0) and
electric field
(electric field has two components: Ex and Ez and only Ex is
(electric field has two components: Ex and Ez and only Ex is
tangential at the interface z=0)
at z=0 gives
1 21sin sinsin
cos cos cosi trx xx
i TM r TM te e eγ θ γ θγ θ
θ θ τ θ− −−
+ Γ =
21 1 sinsin sin
1 1 2
ti r xx x
TM TMe ee γ θγ θ γ θ
τ
η η η
−− −
Γ
− =

Therefore,
which is Snell’s law of reflection and refraction and this
tri θγθγθγ sinsinsin 211 ==
1 2; sin sini r i tθ θ γ θ γ θ⇒ = =
which is Snell’s law of reflection and refraction and this
implies that
cos cos cosi TM r TM tθ θ τ θ+ Γ =
( )
1 2
1
1 TM
TM
τ
η η
− Γ =

The above two equations has two unknowns τTM and ГTM and
it can be easily solved as follows
Therefore,
cos cos
cos
i TM r
TM
t
θ θ
τ
θ
+ Γ
= ( )2
1
1TM TM
η
τ
η
= − Γ
Therefore,
( ) ( )2 2
1 1
2 1 1 22 2
1 1 1 1
2 1
2
cos cos cos cos
1 1
cos cos cos
cos cos cos cos coscos
cos cos cos cos
cos cos
cos
i TM r i TM r
TM TM
t t t
i t r i tr
TM TM
t t t t
t i
TM
θ θ θη η θ
θ η η θ θ
θ η θ η θ η θ η θη θ η
η θ θ η η θ η θ
η θ η θ
η θ
+Γ Γ
= −Γ ⇒ −Γ = +
       − − −
⇒ Γ − − = − ⇒ Γ =       
       
−
⇒ Γ = 2 1
1 2 1
cos cos
cos cos cos
t i
t r t i
η θ η θ
η θ η θ η θ
−
=
+ +

Hence, the reflection and transmission coefficients for oblique
( ) 2 1 2 1 2 12 2 2
1 1 2 1 1 2 1
1 22
1 2 1 2 1
1 1
cos cos cos cos
2 cos 2 cos
cos cos cos cos
t i t i t i
TM TM
t i t i
i i
t i t i
η θ η θ η θ η θ η θ η θη η η
τ
η η η θ η θ η η θ η θ
η θ η θη
η η θ η θ η θ η θ
   − + − +
∴ = −Γ = − =   
+ +   
 
= = 
+ + 
Hence, the reflection and transmission coefficients for oblique
incidence (Fresnel coefficients) for parallel polarization are given as
follows:
Note that cosine terms multiplication in the above equations is
different from the previous expression forTE waves
2 1
2 1
cos cos
cos cos
t i
TM
t i
η θ η θ
η θ η θ
−
∴Γ =
+
2
2 1
2 cos
cos cos
i
TM
t i
η θ
τ
η θ η θ
=
+

6.3.3 Brewster angle:
For Γ=0, the angle of incidence is known as Brewster angle
(θB)
This means at this angle of incidence of the EM wave from
region I,
region I,
there will be no reflection from region II,
all the EM waves will be absorbed

For parallel polarization,
2 1
2 1
cos cos
0
cos cos
t i
TM
t i
η θ η θ
η θ η θ
−
Γ = =
+
TM
i Bθ θ=
Also we have from 2nd Snell’s law
2 1cos cos TM
t Bη θ η θ⇒ =
t
TM
B θγθγ sinsin 21 =

Using the above two equations, we can get (see textbook)










−
= 22
2
2
2
1
2
1
sin
γη
η
η
θTM
B
For lossless media, we have,
1
1 1 1 1
1 1
2
2 2 22 2
2 2 2 2 1 21 1 1 2
2 2 2 2 2
2 12 1 2 1 1 2 1 2 2 2 2
1 2 2 2
1
; sin TM
B
µ ε
µ γ µ µ ε µ ε µ εη γ ε µ
θ
µ ε µ εη γ µ µ ε µ ε µ γ ε µ
ε µ ε µ
−
= = = = ∴ =
−
Q








− 2
2
2
1
2
2
2
1
γ
γ
η
ηB

For magnetic media,
is of the form 1/0 and Brewster angle does not
exist
21 εε =
TM
Bθ2
sin
exist
But, for dielectric media, it exist and can be calculated as
follows
2
1 2 2 2
2 1 2 1 1 1
1 2
1
sin tanTM TM
B B
n
n
ε
ε ε ε
θ θ
ε ε ε ε ε
ε ε
−
⇒ = = ⇒ = =
+−

For perpendicular polarization,
Similarly, Brewster’s angle doesn’t exist for perpendicular
polarization at oblique incidence for a non-magnetic or
dielectric medium (see textbook for derivation)
For magnetic medium
For magnetic medium
( )
( ) ( )
2
2 1 22 2 2 2 2
2 2
1 2 1 2 1 11 2
sin sin ;tanTE TE TE
B B B
n
n
µ µ µ µ µ µ
θ θ θ
µ µ µ µ µµ µ
−
∴ = = ⇒ = = =
+ +−

Brewster angle exist for oblique incidence (perpendicular
polarization) for interface between two magnetic media
An incident wave consisting of both polarizations will have
reflected wave for only one polarization for Brewster angle
Brewster angle is also known as Polarizing angle (used in
optics and lasers)
Polarizing angle
optics and lasers)
In other words, an incident wave which is composed of both
TE andTM waves will have reflectedTE (TM) waves only for
dielectric (magnetic) media interface
Thus, a circularly polarized incident wave at polarizing angle
will become a linearly polarized wave

6.3.4Total internal reflection:
Used in optical fibers
Light strikes an angle of incidence greater than critical angle,
all of the light energy will be reflected back, and
none of them will be absorbed
none of them will be absorbed
This critical angle is related to the refractive index of the two
media and is given by
where are the refractive index of medium 2 and 1
respectively
1 2
1
sinc
n
n
θ −  
=  
 

In fact, this can be observed for both the polarizations we
have discussed above
From Snell’s Second law, we have,
2
1 2
sin
sin sin sin t
i t i
γ θ
γ θ γ θ θ
γ
= ⇒ =
If we consider non-magnetic materials (dielectrics) of , in
that case,
1 2
1
sin sin sini t iγ θ γ θ θ
γ
= ⇒ =
2
1
sin
sin t
i
ε θ
θ
ε
⇒ =

Hence, for
(it means the transmitted wave travels along the interface, for
our case, it is x-axis) which implies that
,
2
i c t
π
θ θ θ= =
our case, it is x-axis) which implies that
2 1 2
11
sin sinc c
ε ε
θ θ
εε
−
 
⇒ = ⇒ =   
 

We can write the above relation as
Note that to have a real value of critical angle
1 1 1 10 2 0 22 2 2 2
1 0 1 0 1 1 1 1
sin sin sin sinr r r r
c
r r r r
n
n
µ µ ε εε µ ε
θ
ε µ µ ε ε µ ε
− − − −
      
= = = =          
     
ε ε>
Note that to have a real value of critical angle
Besides, we have assumed that for a dielectric,
1 2ε ε>
1 2 1r rµ µ= =

What will happen when the angle of incidence is greater than
the critical angle?
For all angles greater than the critical angle,
using Snell’s second law, and
noting that waves are moving from a denser to a rarer dielectric
medium,
medium,
The value of should be imaginary
1 1 2
2 2
sin
sin 1 cos 1 sini
i c t t t
µ ε θ
θ θ θ θ θ
µ ε
> ∴ = > ⇒ = ± −Q
cos tθ
1 2 1r rµ µ= = 1 2ε ε>

It can be shown that (see textbook for derivation)
For z>0, since is an imaginary number, is
also purely imaginary
2 2
2 2
0 02 cos 2 cos* *
2 2
ˆ ˆsin cost tj z j z
t t
E E
S E H x e z eβ θ β θτ τ
θ θ
η η
− −
∴ = × = +
r r r
cos tθ *
cos tθ
also purely imaginary
Hence, the second term is imaginary, therefore, no real
power is flowing along z-axis in region II
The first term which has direction along x-axis is real and is
exponentially decaying with z

Hence,
When a wave is incident at an angle of incidence greater than
or equal to the critical angle from region I,
the wave will be total internally reflected in region I
Surface waves exist in the region II,
Surface waves exist in the region II,
which is propagating along x-axis (direction of power flow),
and
attenuating very fast along z-axis
People have used this for communications using submarines

6.3.5 Effects on polarization:
Since any arbitrary wave incident obliquely from region I can
be decomposed into linear combinations of TE andTM
waves and
we also know that the expressions for reflection and
we also know that the expressions for reflection and
transmission coefficients for parallel and perpendicular
polarizations are different
2 1
2 1
cos cos
;
cos cos
i t
TE
i t
η θ η θ
η θ η θ
−
Γ =
+
2 1
2 1
cos cos
cos cos
t i
TM
t i
η θ η θ
η θ η θ
−
Γ =
+

So, any circularly polarized incident wave will
become elliptically polarized after reflection or transmission
Similarly, a linearly polarized wave (note that linear
polarization could be expressed as linear combination of
RHCP and LHCP) may be
RHCP and LHCP) may be
rotated after reflection

For instance,
If the region II is a perfect conductor, η2=0, Г=-1 for both
TE andTM cases,
this means a RHCP wave after reflection will become LHCP
this means a RHCP wave after reflection will become LHCP
wave and vice versa
This is why circular polarization is not used for indoor
communication systems since we have many reflections inside a
room

6.4 Summary
Plane waves reflection from a media interface
Normal incidence
TE
Lossless
Oblique incidence
Lossy conducting
medium
Perfect
conductor
TM
Brewster
angle
Effect on
polarization
2 1
2 1
;
η η
η η
−
Γ =
+
2
1 2
2η
τ
η η
=
+
τ→0 and Γ→-1
2 1
2 1
cos cos
cos cos
i t
TE
i t
η θ η θ
η θ η θ
−
Γ =
+
2
2 1
2 cos
cos cos
i
TE
i t
η θ
τ
η θ η θ
=
+
cos cosη θ η θ−
CP EP
Lossless
medium
Good conductor
Fig. 6.7 Plane waves reflection from media interface in a nutshell
angle Total internal
reflection
( )
2
21 1
0
1
1 1 1
ˆRe
2 2
avgS S z E
η
− Γ
= • =
r
1 2
avg avgS S=
2 1
2 1
cos cos
cos cos
t i
TM
t i
η θ η θ
η θ η θ
−
∴Γ =
+
2
2 1
2 cos
cos cos
i
TM
t i
η θ
τ
η θ η θ
=
+
CP EP








= −
1
21
sin
ε
ε
θc
1
2
1
2
tan
n
nTM
B ==
ε
ε
θ
1
2
1
2
tan
n
nTE
B ==
µ
µ
θ
1 1 2
2 2
sin
sin 1 cos 1 sini
i c t t t
µ ε θ
θ θ θ θ θ
µ ε
> ∴ = > ⇒ = ± −Q( ) z
avg eEzSS α
η
2
2
2
2
0
2 1
2
1
ˆRe
2
1 −+ Γ−
=•=
r

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