5th Chapter

8/3/2019 5th Chapter

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UNIT V

ELECTROMAGNETIC WAVES

GENERAL OBJECTIVES

1 To analyze the Wave equations.

2 To apply the wave equation for different medium.

3 To understand the concept of Polarization.

4 To understand the concept of reflection of plane waves incident on a perfect

conductor.

5 To apply the concept of reflection of a plane wave incident on a perfect dielectric.

5.1 ELECTROMAGNETIC WAVE EQUATION

Wave equation for electric field E

The application of Maxwells equations is the prediction of existence of electromagnetic

wave. Electromagnetic wave equation can be obtained from Maxwells equations.

The Maxwells equation from Faradays law in point form is given by

E =tB

=t

H

Taking curl on both sides, we get

E =t

H

.. (1)

But, Maxwells equation from Amperes law in point form is

H = tD

J

+

=t

EE

+

Differentiate

t

H

=t

H

=

+

t

EE

t

t

H

=

2

2

t

E

t

E

+

.. (2)


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Substitute the equation (2) in equation (1)

E+ =

+

2

2

t

E

t

E

=2

2

t

E

t

E

.. (3)

But according to the identity

E = E)E.( 2 .. (4)

But,

E. = D.1

Since there is no net charge within the conductor, the charge density = 0

D. = 0 and E. = 0

Then equation (4) becomes

E = E2 .. (5)

Comparing the equations (3) and (5)

E2 =2

2

t

E

t

E

+

E2 = 0t

E

t

E2

2

=

.. (6)

This is the wave equation for electric field E.

Wave equation for Magnetic field H

The Maxwells equation from Amperes law in point form is given

H =t

EE

+

Take curl on both sides

H = tE

E

+ .. (7)

But Maxwells equation from Faradays law

E =t

H

Differentiating with respect tot,

t

E

=2

2

t

H

Substituting the values of E andt

E

in equation (7)


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H =2

2

t

H

t

H

.. (8)

But the identity is

H = H)H.( 2

Since B =0H. =

We have, H = H2 .. (9)

On comparing equations (8) and (9),

H2 =2

2

t

H

t

H

+

2

22

t

H

t

HH

= 0 .. (10)

This is the wave equation for magnetic field H.

5.2 WAVE EQUATION FOR FREE SPACE

Wave equation for Electric field

For the free space (dielectric medium) the conductivity of the medium is zero. (i.e. = 0)

and there is no charge containing in it (i.e. = 0). The electromagnetic wave equations for free space

can be obtained from Maxwells equations.

The Maxwells equation from Faradays law for free space in point form is

E =t

B

=t

H


E =t

H

.. (1)

But Maxwells equation from Amperes law for free space in point form is

H =t

D

=t

E

Then

t

H

=t

H


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=

t

E

t

t

H

= 2

t

E2.. (2)

On substituting the equation (2) in equation (1), we get

E =2

2

t

E

.. (3)

But the identity is given by

E = E)E.( 2

But E = 0D.1

=

=

Then, E = E2 .. (4)


E2 =2

2

t

E

2

22

t

EE

= 0 .. (5)

This is the wave equation for of electric field in free space.

Wave equation for Magnetic field H

The Maxwells equation from Amperes law for free space in point form is given by

H =t

E


H =t

E

.. (6)

Maxwells equation from Faradays law was given as,

E =t

H

On differentiating,

t

E

=2

2

t

H

.. (7)

On substituting the equation (7) in equation (6), we get

H =2

2

t

H

.. (8)

But the identity is given by


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H = H)H.( 2 .. (9)

But, E. = 0B.

1=

Then, H = H2 .. (10)


H2 =2

2

t

H

H2 = 0t

H

2

2

=

.. (11)

This is the wave equation for free space in terms of H.

For free space )air(1and1 rr == and the wave equation becomes

0

t

HH

2

2

002 =

Here,

00 = 97

1036

1104

=16

109

1

00

1

= sec/m103 8

= v

Where, v is the velocity of light

Then the wave equation becomes,

0t

H

v

1H

2

2

2

2 =

or 0

t

E

v

1E

2

2

2

2 =

5.3 UNIFORM PLANE WAVE

If the phase of a wave is the same for all points on a plane surface it is called as plane wave.

If the amplitude is also constant in a plane wave, it is called uniform plane wave. The properties

uniform plane waves are given as follows.

1. At every point is space electric field (E) and magnetic field (H) are perpendicular to each

other and to the direction of travel.

2. The fields vary harmonically with time and at the same frequency, everywhere in space.


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3. Each field has the same direction, magnitudes and phase at every point in any plane

perpendicular to the direction of wave travel. If the electric field is in x direction and the magnetic

field iny direction, then the wave is traveling inzdirection.

The wave equation for free space is given by

E2 = 2

2

t

E

2

2

2

2

2

2 EEE

zyx

+

+

=2

2

t

E

Consider electric field E, which varies in x direction. Then E is independent of directions y

andz.

Then wave equation becomes

2

2E

x

=

2

2

t

E

=

=

0EE2

2

2

2

zy

Since E is a vector quantity, it can be written as in terms of the components of E as,

2

2E

x

x

= 2

2

t

E

x

2

2E

x

y

=

2

2

t

E

y

2

2E

x

z

= 2

2

t

E

z

For free space there is no charge density. Therefore,

D. = 0E. =

E. = 0

zyx

zyx

+

+

EEE

= 0

For uniform plane wave, E is independent ofy andz.Then,

x

x

E

= 0

This equation shows that there is no variation of xE in the x direction

On differentiating the above equation with respect to x,

2

x

x E2

= 0


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The above equation is satisfied when xE becomes zero or constant.

If xE is a constant, then it is not a field. So xE must be zero. Therefore a uniform plane

wave propagating in thex direction has nox component of E.

xE = 0 for uniform plane wave

A similar analysis would show that there is nox component of H.

B. = 0H. =

H. = 0

zyx

zyx

+

+

HHH

= 0

Since H is propagatingx direction, it is independent ofy andz.

Then,

x

x

H

= 0 and 0H

2

=

2x

x

Since xH is not a constant, xH must be zero.

xH = 0 for uniform plane wave.

5.4 Characteristic impedance or intrinsic impedance )( 0

Consider the plane wave propagating inx direction. The wave equation for free space is

2

2E

x

=2

2

00t

E

The general solution of this differential equation is in the form.

E = )()( 0201 tvxftvxf ++

Where,

0v =00

1

f1 and f2 are functions of )( 0tvx and )( 0tvx+ respectively

The solution of wave equation consists of two waves; one traveling in positive direction and

other traveling in negative direction. Consider the wave travel in positive direction alone.

)( 02 tvxf + = 0 (negative direction)

The general solution of wave equation becomes

E = )( 0tvxf +


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E =

zyx EEEzyx

zyx

=

+

+

yxz

xzy

zxx xyzxyz EEEEEE

Since the wave is traveling inx direction, E and H are both independent ofy andz. Therefore,

0EE

and0HE =

=

==zy

xx

E = zx

yx

yz

+

EE

Similarly,

H = zx

yx

yz

+

HH

But H =t

E

On comparing the above two equations we get,

zx

yx

yz

+

HH

=

+

zy zy

t

E

t

E]0[ =xE

Equating y and z terms

x

z

H

=t

E

y

x

y

H=

t

E

z

From Maxwells second equation for free space,

E =t

H

But

E = zx

yx

yz

+

EE

Equating the above two equations, we get

zx

yx

yz

+

EE

=

+

zy z

y

t

H

t

H ]0H[ = x

Equating y and z terms

xz

E = tH

y


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x

y

E=

t

H

z

Let the solution of this equation is given by

yE = )( 0tvxf

On differentiating,tE y =

t)(

)t(0

o

tvxvx

f

= )()(' 00 vtvxf

Simply )(' 0tvxf can be written as 'f

t

E

y= '0 fv

But

x

z

H= t

E

y

x

z

H

= )'( 0 fv

= '0 fv

= '

1f

x

z

H

= 'f

On integrating, Hz =

dxf'

Hz = f

= yE

z

y

H

E=

Similarly, it can be shown that

y

z

H

E=

If E is the total electric field and H is the total magnetic field, then

E = 22 EE zy +

H = 22 HH zy + and

H

E=


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The above ratio is referred as characteristics impedance or intrinsic impedance of the

medium. It is the ratio of square root of permeability to the dielectric constant of the medium and it is

denoted by

=H

E=

For free space 1 rr == (air) and the characteristic impedance or intrinsic impedance for free

space is given by

0 =0

0

=9

7

1036

1

104

= 210364

0 = 120

or 0 = 377

5.4.1 Dot product of E and H

The dot product of E and H is given by

E.H = zzyy HEHE +

But

=y

z

z

y

H

E

H

E =

E.H = zyzy HHHH

E.H = 0

Thus in a uniform plane wave, E and H are at right angles to each other.

5.4.2 Cross product of E and H

The cross product of E and H is given by

E H =

zyx

zyx

zyx

HHH

EEE

= ]0[]0[]HEHE[ zyx yzzy ++ ]0HE[ == xx

E H = ]HEHE[ yzzyx

= ]HH[ 22 yzx +

E H = 2Hx ]HHH[Where,222zy +=


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The cross product of E and H gives the direction in which the wave travels.

The following figure 5.1 shows a uniform plane wave. Here it is found that, E and H fields

are perpendicular to each other and z indicates the direction of propagation of the wave.

Figure: 5.1 Uniform Plane wave moving along z direction

5.5 MAXWELLS EQUATION IN PHASOR FORM

Maxwells equation can be rewritten for phasors, with the time derivatives transformed into

linear terms

( )

( )2

22 tEofphasorE-

tEofphasorEj

t

t

=

=

Now, the Maxwells equation can be rewritten in phasor form as,

Hj-E =

EjJH +=

0B.

D.

=

=

5.6 WAVE EQUATION IN PHASOR FORM

Wave equation in phasor form for Electric and Magnetic field was written as follows:

0HH

0EE

00

22

00

22

=+

=+

5.7 WAVE PROPAGATION IN A LOSSLESS MEDIUM

The wave equation for free space (lossless medium) is

E2 =2

2

t

E

The phasor value of E is


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E (x, t) = ])(ERe[( tjex

Apply to the wave equation

][EeRetj2 = ]E[Re

t

2

2tj

e

][EeRe

tj2

= ]E[Re2 tj

e

]eE)E[Retj22 + = 0

EE 22 + = 0

This is the wave equation for lossless medium in phasor form and it is called Vector Helmholtz

equation.

EE 22 + = 0

Where,

2 = 2

and

is called as Phase shift constant

The velocity of propagation is

v = =

1

The wave propagates inx direction i.e. no variation iny andz.

EE 22

2

+x

= 0

The solution of the equation isE = xjxj ee + 21 CC

5.8 WAVE PROPAGATION IN A CONDUCTING MEDIUM

The wave equation for conducting medium is

t

E

t

E-E

2

22

= 0

The phasor form of wave equation is

EEE22 j = 0

E)(E 22 + jj = 0

E)(E2 + jj = 0

EE 22 = 0

Where,

)(2 += jj and

was called as propagation constant ( It has both real and imaginary parts)Let us take the value of propagation constant as,


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+= j

Where,

is attenuation constant and

is Phase shift constant

Now,+= j

= )( + jj

Squaring on both sides

+ j222 = 22j

Equating real and imaginary parts

22 = 2 and

2 =

We have to solve these two equations to find the value of and .

We know that,

22 = 22222 4)( +

But, 222 )( = 22 )( and

2)2( = ( )

2

2 2 = 222224 +

2

2= 2

Adding these two equations

22 = 2222242 ++

2 =22

222

122

+

++

= 112 22

22

+

+

Attenuation factor is given by

=

+

11

2 22

2

By subtracting 2

2 from 2+

2, the value of the becomes

= 112 22

22

+

+


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5.9 WAVE PROPAGATION IN GOOD DIELECTRICS

The ratio of conduction current density to displacement current density in the medium is

. Hence

= 1 can be considered to mark the dividing line between conductor and dielectrics. For

good conductors

is much greater than unity. For good dielectrics

is very much less than

unity.

For dielectrics,

< < 1

22

21

+ =

2

1

22

2

12

+

=

+22

2

21

The attenuation factor is

=

+ 11

222

2

2

2

4

2

~

The phase shift constant is given by,

=

++ 11

222

2

_~ + 222

22

2

_~

+

22

2

41

2

_~ 2

1

22

2

41

+

=

+

22

2

81

The velocity of the wave in the dielectric is


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v =

=

+

22

2

81

_~

2281

1

=

10v

v_~

220

81v

The intrinsic or characteristic impedance of medium is given by

=+

j

j

=

1

1

1

+=

+

j

jj

j

=

j1

=2

1

1I

+

j

=

+

21

j

5.10 WAVE PROPAGATION IN GOOD CONDUCTOR

For good conductor 1>>

= )( + jj

=

+j

j 1

= j )1(


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v =

2

=

v =

2

The intrinsic impedance of the conductor

=

+

jj

j

1

=

jj

j

.

= j

=

45

It is found that in good conductors, and are large since is large. Therefore,

the wave is attenuated greatly as it progresses through the conductor. But the velocity and

characteristic impedance are reduced considerably.

5.11 SKIN EFFECT AND DEPTH OF PENETRATION

If a plane wave is incident on a highly-conducting surface, then the electric field and

also the current density was found to be concentrated at the surface of the conductor. The

same phenomenon occurs for a current carrying conductor also. This effect depends on the

frequency and is known as skin effect.

In a good conductor the wave is attenuated as it progress. At radio frequencies the

rate of attenuation is very large and the wave may penetrate only a very short distance before

being reduced to a negligibly small value. The depth of penetration ( ) is defined as that

depth at which the wave has been attenuated toe

1or approximately 37 percent of its

original value. It is also known as skin depth.

The amplitude of the wave was decreased by the factorx

e

, as it is propagating

through the conductor at a distancex.

By definition of skin depth,

xe =e

1


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e = 1e

= 1

=1

This is the depth of penetration orskin depth.

=1

=11

2

1

22

2

+

For a good conductor the depth of penetration is

=

=

21

5.12 POLARIZATION

The polarization of a uniform plane wave refers to the time-varying behavior of the

electric field strength vector at some fixed point in space. It represents the orientation of the

electric and magnetic field in the three dimensional space.

Consider a uniform plane wave traveling in the zdirection, with E and H lying inx -

y plane. If 0E =x and only yE is present, the wave is said to be polarized in the y

direction.

5.12.1 Linear Polarization

Figure: 5.2. Linear Polarization


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If both xE and yE are present and are in phase, the resultant electric field has a

direction at an angle of )E/E(tan 1 xy . If the direction of the resultant vector is constant with

time, the wave is said to be linearly polarized. Figure 5.2 shows the linear polarization.

5.12.2 Circular Polarization

Let xE and yE have the same magnitude aE and differ 90 in phase.

The resultant electric field in vector form was given as,

E = ayax aja EE +

The corresponding time varying field is

E =

tsinEtcosE + ayax aa

The components are,

Ex = tcosE a and

Ey = tsinE a

Then, Figure 5.3 Circular Polarization

222EEE ayx =+

This equation shows that the locus of the resultant E is a circle whose radius is aE as

shown in figure 5.3. According to the phase angle difference between Ex and Ey, the

polarization is Left circular polarized or Right circular polarized as shown in the figure 5.4

Figure: 5.4 Left and Right circular polarization

5.12.3 Elliptical Polarization


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If xE and yE have different amplitudes and 2/ phase difference, then the locus of

the resultant E is a ellipse and the wave is said to be elliptically polarized.

Let xE has the magnitude A and yE has the magnitude B and differ 90 in phase.

Then the resultant electric field in vector form is,

E = BA yx aja +

The corresponding time varying field is

E =

tsinBtcosA + yx aa

The components are,

xE = A cos t and

yE = B sin t

Then, tcosA

E=x

tsinB

E=y

1B

E

A

E

2

2

2=+ yy Fig.5.5 Elliptical

Polarization

This equation shows that the locus of the resultant E is an ellipse. It is shown in

Figure 5.5.

5.13 REFLECTION OF PLANE WAVE FROM A CONDUCTOR

When the electromagnetic wave traveling in one medium strikes upon a second

medium, the wave will be partially transmitted and partially reflected. It depends upon types

of wave incidence. The two types of incidence are

1. Normal incidence and

2. Oblique incidence

Normal incidence

Whenever the plane wave was incident normally upon the surface of a perfect

conductor, the wave is entirely reflected back. Since there can be no loss within a perfect

conductor, none of the energy is absorbed. Normal incidence is shown in figure 5.6


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Figure: 5.6 Normal Incident

As a result, the amplitudes of E and H in the incident wave are the same as in the

reflected wave and differ by (out of phase.)

i.e. ri EE = .

Let the electric field of the incident wave isx

ieE . Since attenuation is zero ( =0),

the propagation constant becomes = j . Then incident wave is xjieE (opposite

direction).

The resultant electric field is the sum of the electric field of incident and reflected

waves.

)(E xT =xj

r

xj

iee

+EE

But, iE = rE

)(E xT = ][E xjxji ee xjee

xjxj

=

sin

2

= xj i sinE2

Expressing in time variation form,),(E txT =

tji exj

sinE2

If iE is chosen to be real.

),(E txT = txi sinsinE2

This equation shows that the incident and reflected waves combine to produce a

standing wave, which does not progress.

In order to maintain the reversal of direction of energy propagation magnetic field

must be reflected without reversal of phase. So, the incident iH and reflected rH are in the

same phase.


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)(H xT =xj

rxj

i ee +HH

= xjxji ee +(H ]HH[ ri =

= xi cos2H

iH is real

),(HT tx = ]cos[2HRetj

i ex

),(HT tx = txi coscos2H

This equation shows that the magnetic field H has a standing wave distribution. These

two equations indicate that the E and H are differ /2 in phase.

5.14 REFLECTION OF PLANE WAVES BY A PERFECT DIELECTRIC

Oblique Incidence

When a plane electromagnetic wave is incident obliquely on the boundary, a part of

the wave transmitted and part of it reflected. However, the transmitted wave will be

refracted. i.e. the direction of propagation will be changed.

When the wave is incident obliquely at an angle of i with normal, then a part of the

wave was reflected at an angle r in the same medium and part of it was transmitted

(refracted) at angle of t in second medium as shown in figure 5.7.

Figure: 5.7 Oblique incidence on a perfect dielectric

By Snells law,


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r

i

sin

sin=

2

1

v

v

Where,

1v is the velocity of wave in medium 1

2v is the velocity of wave in medium 2

1v =11

1

and 2v =

22

22

r

i

sin

sin=

11

22

Since the permeability of the dielectrics do not vary much from that of free space.

1 = 02 =

Substitute these values in above equation

r

i

sin

sin=

1

2

Since there is no loss of power in perfect dielectric, the incident power must be equal to the

sum of reflected power and transmitted power.

By the conservation of energy

iP = tr PP +

The power/unit area P = E H

= E.H sin2

=E.H =

2E.

iP = iii cosHE

iP = ii

cosE

1

2

rP =r

r

cosE

1

2

tP = tt

cosE

2

2

ii cos

E

1

2

= tt

rr

+

cos

Ecos

E

2

2

1

2

By law of reflection, the angle of incidence is equal to the angle of reflection. That is,

i = r

ii cos

E

1

2

= tt

rr

+

cos

Ecos

E

2

2

1

2


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]EE[cos 22

1

iri

= t

t

cosE

2

2

On dividing by 2Ei on both sides of the above equation,

2

2

1 EE1cos

i

ri = ti

t cos

EE1 2

2

2

2

2

E

E1

i

r =i

t

i

t

cos

cos

E

E

2

2

2

1

2

2

E

E

i

r

=ii

tt

cosE

cosE

1 22

21

1 =1

1

and 2 =2

2

1 =1

0

and 2 =2

0

][ 021 ==

2

2

E

E

i

r=

ii

tt

cosE

cosE1

21

22

There are two elementary orientations (polarizations) for the electromagnetic fields

1. Perpendicular Polarization

Here the electric field is perpendicular to the plane of incidence and the magnetic field

is parallel to the plane of incidence. The fields are configured as in the Transverse Electric

(TE) modes.

2. Parallel Polarization


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Here the magnetic field is perpendicular to the plane of incidence and the electric field

is parallel to the plane of incidence. The fields are configured as in the Transverse Magnetic

(TM) modes.

Any plane wave with general field orientation can be obtained by superposition of

two waves with perpendicular and parallel polarization.

5.14.1 Horizontal Polarization (Perpendicular Polarization)

In this case, electric field E is perpendicular to the plane of incidence and parallel to

the reflecting surface. This is shown in figure 5.8.

Figure: 5.8 Perpendicular Polarizations

By applying the boundary condition, that the tangential component of E is continuous

across the boundary i.e. Ei one medium is same as Erother medium.

ri EE + = tE

i

r

E

E1+ =

i

t

E

E

But

2

2

E

E

i

r =ii

tt

cosE

cosE1 2

1

22


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=i

t

i

r

+

cos

cos

E

E11

2

1

2

i

t

i

r

i

r

cos

cos

E

E1

E

E1

2

1

2

2

+=

+

i

r

i

r

E

E1

E

E1 =

i

t

i

r

+

cos

cos

E

E1

2

1

2

i

r

E

E1 =

i

t

i

r

cos

cos

E

E1

1

2

+

+i

t

i

r

cos

cos1

E

E

1

2=

i

t

cos

cos1

1

2

i

r

E

E=

i

t

i

t

+

cos

cos1

coscos1

1

2

1

2

=ti

ti

+

coscos

coscos

21

21

Now t cos2 can be written as,

t cos2 = t2

2 sin1

Butt

i

sin

sin=

1

2

t2

sin =2

21 sin

i

Substituting this value in above equation

t cos2 =2

21

2

sin1

t

= i2

12 sin

Substituting this value ini

r

E

Eequation

i

r

E

E=

ii

ii

+

2

121

2121

sincos

sincos

Reflection co-efficient is given by


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i

r

E

E=

ii

ii

+

2

1

2

2

1

2

sincos

sincos

This gives the ratio of reflected to incident electric field for horizontally polarized

wave. It is the reflection co-efficient for horizontal polarization.

5.14.2 Vertical Polarization (Parallel Polarization)

In this case electric field E is parallel to the plane of incidence as shown in the figure

5.9. By applying the boundary conditions that the tangential component of E is continuous

across the boundary.

Figure: 5.9 Vertical Polarization

From the figure 5.9

iri cos)EE( = ttE cos

Divide iE on both sides

i

r

E

E1 =

i

t

i

t

cos

cos

E

E

i

t

E

E=

ti

r i

cos

cos

E

E1

But

2

2

E

E

i

r=

ii

tt

cosE

cosE1

21

22


27/30

Substituting the value ofi

t

E

Ein above equation

2

2

E

E

i

r=

t

i

i

r

cos

cos

E

E11

2

1

2

2

E

E1

i

r =

2

1

2

E

E1

cos

cos

i

r

t

i

+

i

r

i

r

E

E1

E

E1 =

2

1

2

E

E1

cos

cos

i

r

t

i

i

r

E

E1+ =

i

r

t

i

E

E1

cos

cos

1

2

+ t

ii

r

coscos1

EE

1

2 = 1coscos

1

2

t

i

+

i

it

i

r

cos

coscos

E

E

1

21=

+

i

ti

cos

coscos

1

12

i

r

E

E=

+

ti

ti

coscos

coscos

12

12

But tcos = t2sin1 ,

i

r

E

E=

+

ti

ti

212

212

sin1(cos

sin1(cos

But,t

i

sin

sin=

1

2

tsin =2

1

isin

t2

sin =2

1

i

2sin

Substituting

i

r

E

E=

+

ii

ii

2

2

21

12

2

2

21

12

sincos

sincos


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i

r

E

E=

+

ii

ii

22121

2

2

22121

2

2

sin1

cos

sin1

cos

+

ii

ii

2

1

2

2

12

2

1

2

2

12

sincos

sincos

Dividing numerator and denominator by1

2

, we get the reflection coefficient as,

tC o e f f i c i

R e f l e c t i=

ii

ii

i

r

E

2

1

2

1

2

2

1

2

1

2

sincos

sincosE

+

=

This equation gives the ratio of reflected to incident electric field for vertically

polarized wave. It is nothing but a reflection coefficient for parallel to vertical polarization.

5.14.3 Brewster Angle (Total Refraction)

It is the angle at which no reflection takes place. This is shown in figure 5.10. This

occurs, when the numerator of the reflection coefficient is zero.

0sincos2

1

2

1

2 =

ii

0sincos2

1

2

1

2 =

ii

ii

= 2

1

22

1

2 sinsin1

On squaring both side of the above equation, we get

( )i

22

1

22 sin1 = i

2

1

2 sin Figure: 5.10 Total

refraction

i2

21

2

22

1

2

2 sin = i2

1

2 sin


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21

222 1sin

i = 21

22

1

2

)(sin 222

12

i= 2221

= )( 212

i2

sin = 22

21

212 )(

i2

sin = 221

2

+

i2cos =

21

1

+

i2tan =1

2

itan =1

2

i =1

21tan

When the incident wave is parallel polarized, this angle is called as Brewster angle

and at this angle there is no reflection takes place.

5.14.4 Total Internal Reflection:

When a wave is incident from the denser medium into rarer medium at an angle equal

to or greater than the critical angle, the wave will be totally, internally reflected back. This

phenomenon is called as total internal reflection and shown in figure 5.11.

Figure: 5.11 Total Internal Reflection


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If 1 is greater than 2 , then both the reflection coefficients for vertical and

horizontal polarizations become complex when

1

2sin

>i

At critical angle the reflection coefficients has unity value.

i

r

E

E=

ii

ii

+

2

1

2

2

1

2

sincos

sincos

But, ci = (Critical angle)

cc

2

1

2 sincos = cc

+ 2

1

2 sincos

= c 21

2 sin = 0

c2sin =1

2

and

c =1

21sin

This is the critical angle at which no refraction (transmission) takes places. That is,

the wave traversed along the boundary surface.

5th Chapter

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