Chapter 5

Power
= Pavg × Area
=
900
900
× p=
(2)2
W = 15 W
240p
60
Average power crossing through the given area = 15W
WAVE POLARIZATION
Polarization is defined as electric fields orientation and the possible
planner components satisfying the basic transverse nature. It is study of
time behaviour of EM Wave.
The polarization of a uniform plane wave defines the locus drawn by electric
field vector as wave propagates.
Linear Polarization
In linear polarization the electric field of light is confined to a single plane
along the direction of propagation. i.e. phase difference between different
components of electric field vector is nπ i.e. ∆φ = ± nπ, n = 0, 1, 2, ……
ˆ x + Ey sin(ωt − βz)a
ˆ y E(z,
=
t) Ex sin(ωt − βz)a
…(1)
Here, Ex and Ey are the maximum magnitude of the x and y components
respectively and φx and φyare initial phase difference of x, and y components
respectively.
As wave is propagating in positive z direction
ˆx × a
ˆy =
ˆz
a
a
Dφ = φy − φx
Example 11:
The polarization of the wave whose electric
field vector is given by
ˆ y + 3 sin(ωt − βx − p)a
ˆZ V/ m
E 2 sin(ωt − βx)a
=
Solution:
Wave is propagating in positive x direction
ˆy × a
ˆz =
ˆx
a
a
Previous Year’s Questions
The time averaged Poynting vector, in


W/m2, for a wave with E = 24e( ωt +βz)a y
V/m in free space is
2.4 
2.4 
(B)
(A) −
az
az
p
p
2.8 
2.8 
(C)
(D) −
az
az
p
p
= φz − φy
225.
Electromagnetic Waves
In the given equation,
∆φ = 0
Hence, it is an example of linear polarization.
Chapter 5
∆φ = π
This satisfy the condition of linear
polarization.
Hence, the given wave is linearly polarized.
Circular Polarization
Circular polarization is achieved when the
two electric field components are having
equal amplitude’s and out of phase by
p
±(2n +=
1) , n 0, 1, 2, 3 ……
2
y
Let wave is propagating in positive
z-direction and electric field components
are in x and y direction. Then condition
for circular polarization is

Ex = E y


( Ex and E y are maximum amplitude of x
and y components respectively)
p
Dφ = φy − φx = (2n + 1)
2
 Left circularly

 polarized
 or clockwise






 Right circularly 


polarized or


counter − clockwise 


p
Dφ = −(2n + 1)
2
Rack your Brain
Ey
What is the cause of polarization of
electromagnetic wave?
Electromagnetic Waves
Ex
Figure: Left circularly polarized (or) clockwise
226.
Chapter 5
Ey
Ex
Previous Year’s Questions
Figure: Right circularly polarized (or) counter
clockwise
Circular polarization is preferred
antennas of unknown polarization.
for
Elliptical Polarization
In general, all the wave which are not linear
or circular are elliptical.
A plane electromagnetic wave traveling
along the +z direction, has its electric
field given by Ex = 2cos(ωt) and Ey =
2cos(ωt + 90°), the wave is
(A) linearly polarized
(B) right circularly polarized
(C) left circularly polarized
(D) elliptically polarized
Condition for elliptical polarization

Ex ≠ E y
Dφ = φy − φx = (2n + 1)
Dφ = −(2n + 1)
p
2
p
2
 Left elliptically 


polarized 

 Right elliptically 


polarized


Figure: Right elliptically polarized (or)
counter clockwise
227.
Electromagnetic Waves
Depends also on the direction of wave
propagation
Chapter 5
Rack your Brain
What are orthogonal sets of polarization?
Figure: Left elliptically polarized (or) clockwise
Example 12:
The electric field vector of a plane wave is
given by

p

ˆ x + cos(ωt + βz)a
ˆ y V/ m
E 2 cos  ωt + βz −  a
=
2

The polarization of wave is?
Solution:
Given
p

ˆ x + cos(ωt + βz)a
ˆ y V/ m
=
E 2 cos  ωt + βz −  a
2


Ex = 2

Ey = 1


Ex ≠ E y
Previous Year’s Questions
The polarization of a wave with electric



=
field
vector E E0 e( ωt −βz) (a x + a y ) is
(A) linear
(B) elliptical
(C) left hand circular
(D) right hand circular
As direction of wave is negative z-direction.
ˆy × a
ˆx =
ˆz
a
−a
Hence,
and
p
φx − φ y = − 
2
 Right elliptical polarization


| Ex | ≠ | E y | 

Electromagnetic Waves
Hence, the wave is right elliptically polarized.
Example 13:
The electric field vector of a plane wave is given by

ˆ y + 2ja
ˆz ) e j( ωt −βx) V/ m
=
E (2a
228.
Wave is propagating in positive x direction
ˆy × a
ˆz =
ˆx
a
a
Chapter 5
Solution:
Given
ˆ y + 2ja
ˆz ) e j( ωt −βx) V/ m
=
E (2a

Ey = 2

Ez = 2
Gray Matter Alert!!!
General Case: Dφ ≠ 0, Dφ ≠
p
2
In general, EM wave is elliptically
polarized.
p 
φz − φy = 

2  Left circular
| E y | = | Ez |

Hence, the given wave is left circularly
polarized.
Example 14:
The electric field vector of a plane wave is
given by

ˆ x + 3 sin(ωt + βz)a
ˆ y V/ m
E 3 cos(ωt + βz − 60°)a
=
Find the polariztion of the plane wave.
Solution:
Given
ˆ x + 3 sin(ωt + βz)a
ˆ y V/ m
E 3 cos(ωt + βz − 60°)a
=

Ex = 3

Ey = 3

ˆ x + 3 sin(ωt + βz)a
ˆ y V/ m
=
E 3 cos(ωt + βz − 60°)a

ˆ x + 3 cos(ωt + βz − 90°)a
ˆ y V/ m
E 3 cos(ωt + βz − 60°)a
=
As, wave is propagating in negative z direction
ˆy × a
ˆx =
ˆz
a
−a
229.
Electromagnetic Waves


Ex = E y
Chapter 5
φx − φy = −60° + 90° = 30°


 Left elliptically
| Ex | = | E y |

Hence, the wave is left elliptically polarized.
REFLECTION AND TRANSMISSION OF EM WAVE
Figure: Reflection and transmission
When a plane wave propagating in a uniform medium faces a different
medium, a portion of wave is reflected from the interface and remaining
wave is transmitted to other medium. For finding reflected and transmitted
waves, electromagnetic field boundary conditions at the medium interface
are used.
Reflection coefficient is the ratio of amplitude of reflected wave to the
incidence wave.
Er

= ΓE 
Ei

 Reflection Coefficient
Hr
= ΓH 

Hi
Ei = h1Hi
Et = h2Ht
Electromagnetic Waves
Er = –h1Hr
When reflection takes place, it changes the direction of propagation and
one of the fields has a phase shift of 180°.
For normal incidence of wave on BoundaryEt1 = Et2 
 Tangential Field
Ht1 = Ht2 
230.
1+
.... (i)
.... (ii)
Er Et
=
Ei
Ei
1 + Γ = τ
where
Chapter 5
Et1 = Ei + Er = Et2
Now, Ei + Er = Et
Hi + Hr = Ht
And, Ei = h1Hi
Er = –h1Hr
Et = h2Ht
Divide equation (i) By Ei
.... (iii)
Et
= t transmission coefficient
Ei
Transmission coefficient is the ratio of
amplitude of transmitted wave to incident
wave.
Now if we replace H in equation (ii) By E,
we get-
Gray Matter Alert!!!
Et
Ei Er
−
=
η1 η1
η2
η
Ei − Er =1 Et η2
Now divide equation (iv) By Ei
η1
1−=
Γ
t
η2
.... (iv)
Transmitted wave never undergo any
phase change for dielectric media
interface. While reflected wave may
undergoes phase change.
.... (v)
Solving equation (iii) & (v)
Er
η2 − η1
ΓE =
=
Ei
η2 + η1
Et
2η2
= 1 + ΓE =
Ei
η2 + η1
Now for magnetic fieldHr
−Er
ΓH=
=
E
Hi
η1 i
η1
ΓH = −ΓE =
η1 − η2
η1 + η2
231.
Electromagnetic Waves
and tE =
Chapter 5
Example 15:
An EM wave is incident from air to a non-magnetic dielectric of er = 16
(normal). Find the following(a) The transmitted
 power fraction into dielectric.
(b) The reflected E field fraction.

(c) Transmitted H field fraction.
Solution:
η=
1
m0
= 120p
e0
η
=
2
m0
120p
=
4
e0
η − η1
ΓE = 2
η1 + η2
120p
− 120p
= 4
120p
120p +
4
1
−1
3
ΓE =4
=−
1
5
1+
4
(a) Fraction of power transmitted = 1–|ΓE|2
9
16
=−
1
=
25 25
Pt
16
=
Pi
25
Previous Year’s Questions
A uniform plane wave travelling in
air is incident on the plane boundary
between air and another dielectric
medium with ∈r = 4. The reflection
coefficient for the normal incidence, is
(A) zero
(B) 0.5∠0°
(C) 0.333∠180°
(D) 0.333∠90°
Electromagnetic Waves

(b) Reflected E field fraction
Er
3
=ΓE =− (as calculated above)
Ei
5

(c) Transmitted H field fraction

Ht
3
 = 1 + (ΓH ) = 1 − ΓE = 1 +
5
Hi

Ht
8
 =
5
Hi
232.
Chapter 5
Reflection at perfect conductor
Figure: Reflection at perfect conductor
y
For perfect conductor
σ→∞
=
η
jωm
= 0
σ
Hence, for normal incidence
η − η1
ΓE = 2
η1 + η2
0 − 120p
ΓE =
0 + 120p
ΓE = –1
Rack your Brain
The incident wave amplitude is 17units.
Find the reflected wave amplitude if
the transmission coefficient is 0.6.
ΓH = –ΓE = 1
and
ΓP = –1
This implies that there is complete reflection at conductor surface.
As
ΓE = –1
tE = 1 + ΓE = 1 − 1
tE = 0
i.e. zero transmission of wave into perfect conducting medium.
and
tH = 1 + ΓH = 1 + 1 = 2
and
Hi + Hr = Ht = Hmax
233.
Electromagnetic Waves
i.e. Ei + Er = Et = 0
Chapter 5
y
y
when ΓE is very close to –1, ΓH is very close to +1.
ΓE being –ive means the reflected electric field is changing its direction
and hence cancels with the incident electric field. Due to this the
electric field near the boundary is zero and magnetic field is maximum.
Oblique-incidence/Inclined incidence
The general form of uniform plane wave in any arbitrary direction is given
by
 
E = E0e j( ωt −k⋅R) …(1)
where

ˆ + βy yˆ + βzz
ˆ = Propagation vector
k = βx x
And

ˆ + yyˆ + zz
ˆ = Position vector
R = xx
Rack your Brain
βx = βcosφx
βy = βcosφy
βz = βcosφz
What is the phase velocity of a wave in
a perfect conductor?
Electromagnetic Waves
φx, φy and φzare the angle between direction of propagation and axis x, y, z
respectively.
Now equation (1) can be written as  j( ωt −βx x −βy y −βzz)
E = E0e
This form is known as phase constant form and in this phase constant β is
given by magnitude of propagation vector k

i.e. β = | k |=
β2x + β2y + βz2
234.