mahalakshmi engineering college-trichy...

MAHALAKSHMI ENGINEERING COLLEGE-TRICHY ELECTROMAGNETIC FIELDS

KSRINIVASAN APECE

UNIT -V

PART-A

1Define skin depth (or)Depth of penetration (AU NOVDEC10)

It is defined as the wave has been attenuated to 37 of its original value is known

as skin depth

1

In a good conductor the rate of attenuation is very high and the wave penetrate only at a

very short distance before being reduced to negligible small percentage of its original

strengthThis term is known as skin depth

2Define skin effect (AU NOVDEC11)

The inductance of the inner region of the conductor is greater than the outer regionsince

flux linkages of the inner region is higher than the outer region So most of the current is

confined to outer region This is known as skin effect

3Define Polarisation (AU NOVDEC12)

The polarization of the uniform plane wave refers to the time varying behaviour of the

electric field strength vector at some fixed point in space

It can be classified into three typesThey are

(i)Linear polarization

(ii)Circular polarisation

(iii)Electrical polarisation

4Define propagation constant (AU APR 09)

The propagation constant can be expressed in terms of properties of the medium as

( )j j j

5Define Standing wave ratio (AU MAY 11)

It is the ratio of the maximum to minimum magnitude of voltage or current on a line

having standing wave ratio

S= max

min

1

1

KV

V K


KSRINIVASAN APECE Page 2

where K is the reflection coefficient

6What is Brewster angle (AU APRMAY 09)

Brewster angle is an incident angle at which there is no reflected wave for parallelly

polarized wave

1 2

1

tan

7State the poynting theorem (AU NOVDEC100911)

The vector product of the electric field intensity and the magnetic field intensity at any

point is a measure of the rate of energy flow at that point

P=EtimesH

8Mention the two properties of the uniform plane wave (AU NOVDEC11)

(i) At every point in space the electric field E and magnetic field H are perpendicular to each

other and to the direction of the travel

(ii)The fields vary harmonically with time and at the same frequency everywhere in space

9Write the equation of skin depth of conductor (AU NOVDEC09)

Skin depth 1

f

10Write the wave equation in conducting medium (AU NOVDEC08)

2

2

20

E HE

t t

22

20

E HH

t t

11What is meant by charteristic impedance (AU NOVDEC08)

It is the ratio of electric field to the magnetic field or it is the ratio of square root of

permeability to permittivity to the medium

η=EH

12What is the velocity of the electromagnetic wave in free space and in lossless

dielectric (AU NOVDEC 07)

The velocity in lossless dielectric is



Vp =1 =R R

C

PART-B

1Deduce the equation of the propagation of of the plane electromagnetic waves in free space ( AU NOVDEC 12)

The wave equation for a plane wave traveling in the x direction is

where v is the phase velocity of the wave and y represents the variable which is changing as the wave passes This is the form of the wave equation which applies to a stretched string or a plane electromagnetic wave The mathematical description of a wave makes use of partial derivatives

In two dimensions the wave equation takes the form

which could describe a wave on a stretched membrane

General Wave Equations

Electromagnetic (EM) wave can be ultimately described by three things 1) Maxwell equations material response to the EM fields and boundary conditions



HMEPt

DJH

EuJt

BE

HMHHBB

PEEEDD

me

v

m

rv

00

00

00

10

In charge free medium (v=0) linear isotopic ( is scalar) homogeneous and time invariant medium (P responds to E instantaneously) We have

t

EEH

t

HE

H

E

0

0

if the EM wave only contains one frequency component we can rewrite the Maxwell equation in phasor form using E = Re (Es e

jωt)

sss

ss

s

s

EjEH

HjE

H

E

0

0

Use the third equation (Faradayrsquos Law) and apply the curl on both sides of the full time-varying

equation

Ht

E

Use the last equation into the above equation and use a vector identity

2

22

t

E

t

EEE

t

EE

tE

Since we are considering a charge-free region the general wave equation (Helmholtz equation)

becomes



2

22

t

E

t

EE

Again if the E and H fields only have one frequency components we can get wirte a phasor

equation for the time-harmonic field

or ss EE

22 where the material properties which govern the wave propagation are

described by

jjj

Where characterizes materials absorption (gain) and characterize the propagation speed

(propagation constant)

Uniform plane waveguide (UPW) solution

Now letrsquos try to solve the waveguide equation The simplest solution beyond trial (0) is the UPW

solution While we have

a) Electrical filed only have one components (say Ex component)



b) This components is constant in two directions (say x-y plane) only change along z-axis

(propagation direction)

So the proposed solution is (For notational convenience we drop the subscript s)

xx azEE

How do we know this is a solution Just plug in the wave equation if we can find a Ex(x) which

satisfies the wave equation we have a solution

02

2

2

xx E

dz

Ed

The solution is

x

zz

x aeEeEzE

00

If we translate this solution back into a fully time-varying solution it is

x

ztjzztjz

x aeeEeeEzE

00

Re

The first term characterizes the forward-propagating UPW the second term characterizes the

backward-propagating one

By the help of the Maxwell equations we can also figure out other field such D B and H

ss HjE

Therefore the H field can be calculated as

ss Ej

H

1



(The subcript s is written here to remind you that all quanatities are phasor quantitites) Having

E and H solution available the ratio between the E and H field can be taken Letrsquos take a

forward-propagation UPW as examples

j

jj

H

E

Since E and H fields have unit (Vm) and (Am) The ratio yields a unit of () This ratio is

referred as intrinsic impedance of the materials

The wave propagates along az we can use a vector to represent that

za

The unit vector to represent this direction is defined as ap (az in this case) E-field propagates

along ax and E-field propagates along ay Therefore these three vectors form a right-handed

rectangle coordinate system ie

sps

sps

HaE

EaH

1



Now we can see the propagation characteristics for UPW is ultimately determined by

jjj We then discuss the responses in various materials

a) Propagation in lossless charge-free region

In a charge free region with zero loss the propagation constant

jjj

=0 so jjj 0

1

]coscos[

Re

00

00

p

xx

x

ztjztj

x

u

ztEztEazE

aeEeEzE

A special example will be vacuum where =0 =0



smup 103

120

8

0

0

b) Propagation in lossy media

In a lossy media the loss comes from two parts a non-zero electric conductivity 0) a

polarization loss (energy required of the filed to flip reluctant dipoles dielectric loss) A complex

permittivity c is used to characterize this part

jc

Therefore the propagation constant can be written

jjjj

jjjjj

eff

c

Where eff we can solve for the absorption and the propagation constant of the

medium ieand = Re =Im

1

12

1

12

2

2

eff

eff

Loss Tangent a standard measure of lossiness in a dielectric is measured by the ratio of

tan

eff



Referred as loss-tangent

Low-loss dielectrics

For low loss dielectrics ltlt1 rdquoltltrsquo therefore

1

eff

2

2Describe the Skin effect in detail (AU MAYJUN 10)

The skin effect is the tendency of an alternating electric current (AC) to distribute itself within a conductor so that the current density near the surface of the conductor is greater than that at its core That is the electric current tends to flow at the skin of the conductor

The skin effect causes the effective resistance of the conductor to increase with the frequency of the current The skin effect has practical consequences in the design of radio-frequency and microwave circuits and to some extent in AC electrical power transmission and distribution systems Also it is of considerable importance when designing discharge tube circuits

The current density J in an infinitely thick plane conductor decreases exponentially with depth δ from the surface as follows

where d is a constant called the skin depth This is defined as the depth below the surface of the conductor at which the current density decays to 1e (about 037) of the current density at the surface (JS) It can be calculated as follows

where



ρ = resistivity of conductor

ω = angular frequency of current = 2π times frequency

μ = absolute magnetic permeability of conductor where μ0 is the

permeability of free space (4πtimes10minus7 NA2) and μr is the relative permeability of the

conductor

The resistance of a flat slab (much thicker than d) to alternating current is exactly equal to the resistance of a plate of thickness d to direct current For long cylindrical conductors such as wires with diameter D large compared to d the resistance is approximately that of a hollow tube with wall thickness d carrying direct current That is the AC resistance is approximately

where

L = length of conductor

D = diameter of conductor

The final approximation above is accurate if D gtgt d

A convenient formula (attributed to FE Terman) for the diameter DW of a wire of circular cross-section whose resistance will increase by 10 at frequency f is

The increase in ac resistance described above is accurate only for an isolated wire For a wire close to other wires eg in a cable or a coil the ac resistance is also affected by proximity effect which often causes a much more severe increase in ac resistance

3Derive the Poynting Theorem and Poynting Vector in Electromagnetic Fields

(AU NOVDEC 09101112)

Electromagnetic waves can transport energy from one point to another point The

electric and magnetic field intensities associated with a travelling electromagnetic wave can be

related to the rate of such energy transfer

Let us consider Maxwells Curl Equations



Using vector identity

the above curl equations we can write

(1)

In simple medium where and are constant we can write

and

Applying Divergence theorem we can write

(2)



The term represents the rate of change of energy stored in the electric

and magnetic fields and the term represents the power dissipation within the volume

Hence right hand side of the equation (636) represents the total decrease in power within the

volume under consideration

The left hand side of equation (636) can be written as where

(Wmt2) is called the Poynting vector and it represents the power density vector associated with the

electromagnetic field The integration of the Poynting vector over any closed surface gives the net

power flowing out of the surface Equation (636) is referred to as Poynting theorem and it states that

the net power flowing out of a given volume is equal to the time rate of decrease in the energy stored

within the volume minus the conduction losses

4In a lossless medium for which =60πmicror = 1and H=-01 cos (wt -z)ax +05 sin (wt -z)ay

AmCalculate εr (AU NOVDEC 11)

Answer

Given

=60πmicror = 1

H=-01 cos (wt -z)ax +05 sin (wt -z)ay

For lossless medium case

=0α=0β=1

To find

εr =

2

120

2

c

(i)Relative Permittivity(εr )



εr =

2

120

=22

εr =4

(ii)Angular Frequency ( )

2

c

C= Velocity of light =3times108 msec

81(3 10 )

2

=15times108 radsec

5A signal is a lossy dielectric medium has a loss tangent 02 at 550 kHZThe dielectric constant of a medium is 25Determine attenuation constant and phase constant

(AU NOVDEC11)

Given

Frequency f=550 kHz

loss tangent =02 tan θ=02

εr =25

To Find

Attenuation constant (α) Phase constant (β)

loss tangent for dielectric

tan

02

For given dielectric

1therefore it is practical dielectric

002 02 (2 )( )rf



9

3 1002 02 (2 550 10 )( 25)

36

5153 10 Sm

Attenuation constant(α)

For Practical dielectric

2

0

02

r

r

5 7

9

153 10 4 10 1

10225

36

α=182times10-3 NPm

Propagation Constant

β=

=(2πf) 0 0r r

β= 00182 radm

6A lossy dielectric is characterised by εr =25micror=4 =10-3 mhom at 10 MHzLet E=10 e-yz ax

Vmfind (i)α(ii)β(iii) (iv)Vp (v)η (AU NOVDEC 10)

Given

εr =25micror=4 =10-3 mhomf=10times106 Hz

To Find

Attenuation constant(α) Phase constant (β) Wave length Velocity of propagation(Vp) Intrinsic Impedance(η)



Solution

Attenuation constant(α) and Phase constant(β)

The Propagation constant for lossy dielectric

(j j

j

0 0(2 f)( )( (2 )r rj j f

6 7 3 6 12(2 10 10 )(4 10 4)(10 (2 10 10 )(8854 10 25j j )

=3 3(31582)[10 13907 10 ]j j

07355 7214 02255 07j

j

α=02255 Npm

β=07 radm

Wavelength

2

=2π07

8975m

Velocity of Propagation(Vp)

pV

2 f

=2πtimes10times106 07=8976times107 ms

Intrinsic impedance(η)



j

j

0

0

(2 )

(2 )

r

r

j f

j f

=

6 7

3 6 12

(2 10 10 ) 4 10 4

10 (2 10 10 )8854 10 25

j

(12)

0

3

31582 90

17219 10 5428

η=4294 179

η=4086+j132Ω

7Explain the Polarisation of plane wave in detail (AU MAYJUN 0809)

The polarisation of a plane wave can be defined as the orientation of the electric field vector as

a function of time at a fixed point in space For an electromagnetic wave the specification of the

orientation of the electric field is sufficent as the magnetic field components are related to

electric field vector by the Maxwells equations

Let us consider a plane wave travelling in the +z direction The wave has both Ex and Ey

components

(51)

The corresponding magnetic fields are given by



Depending upon the values of Eox and Eoy we can have several possibilities

1 If Eoy = 0 then the wave is linearly polarised in the x-direction

2 If Eoy = 0 then the wave is linearly polarised in the y-direction

3 If Eox and Eoy are both real (or complex with equal phase) once again we get a linearly

polarised wave with the axis of polarisation inclined at an angle with respect to the x-

axis This is shown in fig 51

Fig 51 Linear Polarisation

4 If Eox and Eoy are complex with different phase angles will not point to a single spatial

direction This is explained as follows Let

Then

and (52)



To keep the things simple let us consider a =0 and Further let us study the nature of the electric field on the z =0 plain

From equation we find that

(53)

and the electric field vector at z = 0 can be written as

(54)

Assuming the plot of for various values of t is hown in figure 52

Figure 52 Plot of E(ot)

From equation and figure we observe that the tip of the arrow representing electric field vector

traces qn ellipse and the field is said to be elliptically polarised



Figure 53 Polarisation ellipse

The polarization ellipse shown in figure 66 is defined by its axial ratio(MN the ratio of

semimajor to semi minor axis) tilt angle (orientation with respect to x-axis) and sense of rotation(ie CW or CCW)

Linear polarization can be treated as a special case of elliptical polarization for which the axial ratio is infinite

In our example if from equation the tip of the arrow representing electric field vector traces out a circle Such a case is referred to as Circular Polarization For circular polarization the axial ratio is unity

Figure 54 Circular Polarisation (RHCP)

Further the circular polarization is aside to be right handed circular polarisation (RHCP) if the electric field vector rotates in the direction of the fingers of the right hand when the thumb points in the direction of propagation-(same and CCW) If the electric field vector rotates in the



opposite direction the polarization is asid to be left hand circular polarization (LHCP) (same as CW)

In AM radio broadcast the radiated electromagnetic wave is linearly polarised with the field vertical to the ground( vertical polarization) where as TV signals are horizontally polarised waves FM broadcast is usually carried out using circularly polarized waves

In radio communication different information signals can be transmitted at the same frequency at orthogonal polarisation ( one signal as vertically polarized other horizontally polarised or one as RHCP while the other as LHCP) to increase capacity Otherwise same signal can be transmitted at orthogonal polarization to obtain diversity gain to improve reliability of transmission

8Explain the Angle Of Incidence and Brewster angle ( AU MAYJUN 11)

Angle of incidence is a measure of deviation of something from straight on for example in the

approach of a ray to a surface or the direction of an airfoil with respect to the direction of an

airplane

Optics

In geometric optics the angle of incidence is the angle between a ray incident on a surface and

the line perpendicular to the surface at the point of incidence called the normal The ray can be

formed by any wave optical acoustic microwave X-ray and so on In the figure above the red

line representing a ray makes an angle θ with the normal (dotted line) The angle of incidence at

which light is first totally internally reflected is known as the critical angle The angle of

reflection and angle of refraction are other angles related to beams

Grazing angle

When dealing with a beam that is nearly parallel to a surface it is sometimes more useful to refer

to the angle between the beam and the surface rather than that between the beam and the surface

normal in other words 90deg minus the angle of incidence This angle is called a glancing angle or

grazing angle Incidence at small grazing angle is called grazing incidence



Grazing incidence is used in X-ray spectroscopy and atom optics where significant reflection

can be achieved only at small values of the grazing angle Ridged mirrors are designed for

reflection of atoms coming at small grazing angle This angle is usually measured in milliradians

Brewsters angle

An illustration of the polarization of light which is incident on an interface at Brewsters angle

Brewsters angle (also known as the polarization angle) is an optical phenomenon named after

the Scottish physicist Sir David Brewster (1781ndash1868)

When light moves between two media of differing refractive index generally some of it is

reflected at the boundary At one particular angle of incidence however light with one particular

polarization cannot be reflected This angle of incidence is Brewsters angle θB The polarization

that cannot be reflected at this angle is the polarization for which the electric field of the light

waves lies in the same plane as the incident ray and the surface normal (ie the plane of

incidence) Light with this polarization is said to be p-polarized because it is parallel to the

plane Light with the perpendicular polarization is said to be s-polarized from the German

senkrechtmdashperpendicular When unpolarized light strikes a surface at Brewsters angle the

reflected light is always s-polarized

The physical mechanism for this can be qualitatively understood from the manner in which

electric dipoles in the media respond to p-polarized light One can imagine that light incident on

the surface is absorbed and then reradiated by oscillating electric dipoles at the interface

between the two media The polarization of freely propagating light is always perpendicular to

the direction in which the light is travelling The dipoles that produce the transmitted (refracted)

light oscillate in the polarization direction of that light These same oscillating dipoles also

generate the reflected light However dipoles do not radiate any energy in the direction along

which they oscillate Consequently if the direction of the refracted light is perpendicular to the

direction in which the light is predicted to be specularly reflected the dipoles will not create any

reflected light Since by definition the s-polarization is parallel to the interface the



corresponding oscillating dipoles will always be able to radiate in the specular-reflection

direction This is why there is no Brewsters angle for s-polarized light

With simple trigonometry this condition can be expressed as

where θ1 is the angle of incidence and θ2 is the angle of refraction

Using Snells law

we can calculate the incident angle θ1=θB at which no light is reflected

Rearranging we get

where n1 and n2 are the refractive indices of the two media This equation is known as

Brewsters law

Note that since all p-polarized light is refracted (ie transmitted) any light reflected from the

interface at this angle must be s-polarized A glass plate or a stack of plates placed at Brewsters

angle in a light beam can thus be used as a polarizer

For a glass medium (n2asymp15) in air (n1asymp1) Brewsters angle for visible light is approximately 56deg

to the normal while for an air-water interface (n2asymp133) its approximately 53deg Since the

refractive index for a given medium changes depending on the wavelength of light Brewsters

angle will also vary with wavelength

The phenomenon of light being polarized by reflection from a surface at a particular angle was

first observed by Etienne-Louis Malus in 1808 He attempted to relate the polarizing angle to the

refractive index of the material but was frustrated by the inconsistent quality of glasses available

at that time In 1815 Brewster experimented with higher-quality materials and showed that this

angle was a function of the refractive index defining Brewsters law

Although Brewsters angle is generally presented as a zero-reflection angle in textbooks from the

late 1950s onwards it truly is a polarizing angle The concept of a polarizing angle can be

extended to the concept of a Brewster wave number to cover planar interfaces between two

linear bianisotropic materials


KSRINIVASAN APECE

where K is the reflection coefficient

6What is Brewster angle (AU APRMAY 09)

Brewster angle is an incident angle at which there is no reflected wave for parallelly

polarized wave

1 2

1

tan

7State the poynting theorem (AU NOVDEC100911)

The vector product of the electric field intensity and the magnetic field intensity at any

point is a measure of the rate of energy flow at that point

P=EtimesH

8Mention the two properties of the uniform plane wave (AU NOVDEC11)

(i) At every point in space the electric field E and magnetic field H are perpendicular to each

other and to the direction of the travel

(ii)The fields vary harmonically with time and at the same frequency everywhere in space

9Write the equation of skin depth of conductor (AU NOVDEC09)

Skin depth 1

f

10Write the wave equation in conducting medium (AU NOVDEC08)

2

2

20

E HE

t t

22

20

E HH

t t

11What is meant by charteristic impedance (AU NOVDEC08)

It is the ratio of electric field to the magnetic field or it is the ratio of square root of

permeability to permittivity to the medium

η=EH

12What is the velocity of the electromagnetic wave in free space and in lossless

dielectric (AU NOVDEC 07)

The velocity in lossless dielectric is



Vp =1 =R R

C

PART-B










HMEPt

DJH

EuJt

BE

HMHHBB

PEEEDD

me

v

m

rv

00

00

00

10


t

EEH

t

HE

H

E

0

0


jωt)

sss

ss

s

s

EjEH

HjE

H

E

0

0


equation

Ht

E


2

22

t

E

t

EEE

t

EE

tE


becomes



2

22

t

E

t

EE



or ss EE


described by

jjj












xx azEE



02

2

2

xx E

dz

Ed

The solution is

x

zz

x aeEeEzE

00


x

ztjzztjz

x aeeEeeEzE

00

Re




ss HjE


ss Ej

H

1






j

jj

H

E




za




sps

sps

HaE

EaH

1







jjj

=0 so jjj 0

1

]coscos[

Re

00

00

p

xx

x

ztjztj

x

u

ztEztEazE

aeEeEzE




smup 103

120

8

0

0





jc


jjjj

jjjjj

eff

c



1

12

1

12

2

2

eff

eff


tan

eff






1

eff

2






where







conductor


where
















(1)


and


(2)















Answer

Given

=60πmicror = 1



=0α=0β=1

To find

εr =

2

120

2

c




εr =

2

120

=22

εr =4


2

c


81(3 10 )

2

=15times108 radsec


(AU NOVDEC11)

Given

Frequency f=550 kHz


εr =25

To Find



tan

02



002 02 (2 )( )rf



9

3 1002 02 (2 550 10 )( 25)

36

5153 10 Sm



2

0

02

r

r

5 7

9

153 10 4 10 1

10225

36

α=182times10-3 NPm


β=

=(2πf) 0 0r r

β= 00182 radm



Given


To Find




Solution



(j j

j

0 0(2 f)( )( (2 )r rj j f

6 7 3 6 12(2 10 10 )(4 10 4)(10 (2 10 10 )(8854 10 25j j )

=3 3(31582)[10 13907 10 ]j j

07355 7214 02255 07j

j

α=02255 Npm

β=07 radm

Wavelength

2

=2π07

8975m


pV

2 f





j

j

0

0

(2 )

(2 )

r

r

j f

j f

=

6 7

3 6 12

(2 10 10 ) 4 10 4

10 (2 10 10 )8854 10 25

j

(12)

0

3

31582 90

17219 10 5428

η=4294 179

η=4086+j132Ω







components

(51)













Then

and (52)





(53)


(54)






















airplane

Optics







Grazing angle










Brewsters angle





























Using Snells law


Rearranging we get


Brewsters law


















KSRINIVASAN APECE

Vp =1 =R R

C

PART-B










HMEPt

DJH

EuJt

BE

HMHHBB

PEEEDD

me

v

m

rv

00

00

00

10


t

EEH

t

HE

H

E

0

0


jωt)

sss

ss

s

s

EjEH

HjE

H

E

0

0


equation

Ht

E


2

22

t

E

t

EEE

t

EE

tE


becomes



2

22

t

E

t

EE



or ss EE


described by

jjj












xx azEE



02

2

2

xx E

dz

Ed

The solution is

x

zz

x aeEeEzE

00


x

ztjzztjz

x aeeEeeEzE

00

Re




ss HjE


ss Ej

H

1






j

jj

H

E




za




sps

sps

HaE

EaH

1







jjj

=0 so jjj 0

1

]coscos[

Re

00

00

p

xx

x

ztjztj

x

u

ztEztEazE

aeEeEzE




smup 103

120

8

0

0





jc


jjjj

jjjjj

eff

c



1

12

1

12

2

2

eff

eff


tan

eff






1

eff

2






where







conductor


where
















(1)


and


(2)















Answer

Given

=60πmicror = 1



=0α=0β=1

To find

εr =

2

120

2

c




εr =

2

120

=22

εr =4


2

c


81(3 10 )

2

=15times108 radsec


(AU NOVDEC11)

Given

Frequency f=550 kHz


εr =25

To Find



tan

02



002 02 (2 )( )rf



9

3 1002 02 (2 550 10 )( 25)

36

5153 10 Sm



2

0

02

r

r

5 7

9

153 10 4 10 1

10225

36

α=182times10-3 NPm


β=

=(2πf) 0 0r r

β= 00182 radm



Given


To Find




Solution



(j j

j

0 0(2 f)( )( (2 )r rj j f

6 7 3 6 12(2 10 10 )(4 10 4)(10 (2 10 10 )(8854 10 25j j )

=3 3(31582)[10 13907 10 ]j j

07355 7214 02255 07j

j

α=02255 Npm

β=07 radm

Wavelength

2

=2π07

8975m


pV

2 f





j

j

0

0

(2 )

(2 )

r

r

j f

j f

=

6 7

3 6 12

(2 10 10 ) 4 10 4

10 (2 10 10 )8854 10 25

j

(12)

0

3

31582 90

17219 10 5428

η=4294 179

η=4086+j132Ω







components

(51)













Then

and (52)





(53)


(54)






















airplane

Optics







Grazing angle










Brewsters angle





























Using Snells law


Rearranging we get


Brewsters law


















KSRINIVASAN APECE

HMEPt

DJH

EuJt

BE

HMHHBB

PEEEDD

me

v

m

rv

00

00

00

10


t

EEH

t

HE

H

E

0

0


jωt)

sss

ss

s

s

EjEH

HjE

H

E

0

0


equation

Ht

E


2

22

t

E

t

EEE

t

EE

tE


becomes



2

22

t

E

t

EE



or ss EE


described by

jjj












xx azEE



02

2

2

xx E

dz

Ed

The solution is

x

zz

x aeEeEzE

00


x

ztjzztjz

x aeeEeeEzE

00

Re




ss HjE


ss Ej

H

1






j

jj

H

E




za




sps

sps

HaE

EaH

1







jjj

=0 so jjj 0

1

]coscos[

Re

00

00

p

xx

x

ztjztj

x

u

ztEztEazE

aeEeEzE




smup 103

120

8

0

0





jc


jjjj

jjjjj

eff

c



1

12

1

12

2

2

eff

eff


tan

eff






1

eff

2






where







conductor


where
















(1)


and


(2)















Answer

Given

=60πmicror = 1



=0α=0β=1

To find

εr =

2

120

2

c




εr =

2

120

=22

εr =4


2

c


81(3 10 )

2

=15times108 radsec


(AU NOVDEC11)

Given

Frequency f=550 kHz


εr =25

To Find



tan

02



002 02 (2 )( )rf



9

3 1002 02 (2 550 10 )( 25)

36

5153 10 Sm



2

0

02

r

r

5 7

9

153 10 4 10 1

10225

36

α=182times10-3 NPm


β=

=(2πf) 0 0r r

β= 00182 radm



Given


To Find




Solution



(j j

j

0 0(2 f)( )( (2 )r rj j f

6 7 3 6 12(2 10 10 )(4 10 4)(10 (2 10 10 )(8854 10 25j j )

=3 3(31582)[10 13907 10 ]j j

07355 7214 02255 07j

j

α=02255 Npm

β=07 radm

Wavelength

2

=2π07

8975m


pV

2 f





j

j

0

0

(2 )

(2 )

r

r

j f

j f

=

6 7

3 6 12

(2 10 10 ) 4 10 4

10 (2 10 10 )8854 10 25

j

(12)

0

3

31582 90

17219 10 5428

η=4294 179

η=4086+j132Ω







components

(51)













Then

and (52)





(53)


(54)






















airplane

Optics







Grazing angle










Brewsters angle





























Using Snells law


Rearranging we get


Brewsters law


















KSRINIVASAN APECE

2

22

t

E

t

EE



or ss EE


described by

jjj












xx azEE



02

2

2

xx E

dz

Ed

The solution is

x

zz

x aeEeEzE

00


x

ztjzztjz

x aeeEeeEzE

00

Re




ss HjE


ss Ej

H

1






j

jj

H

E




za




sps

sps

HaE

EaH

1







jjj

=0 so jjj 0

1

]coscos[

Re

00

00

p

xx

x

ztjztj

x

u

ztEztEazE

aeEeEzE




smup 103

120

8

0

0





jc


jjjj

jjjjj

eff

c



1

12

1

12

2

2

eff

eff


tan

eff






1

eff

2






where







conductor


where
















(1)


and


(2)















Answer

Given

=60πmicror = 1



=0α=0β=1

To find

εr =

2

120

2

c




εr =

2

120

=22

εr =4


2

c


81(3 10 )

2

=15times108 radsec


(AU NOVDEC11)

Given

Frequency f=550 kHz


εr =25

To Find



tan

02



002 02 (2 )( )rf



9

3 1002 02 (2 550 10 )( 25)

36

5153 10 Sm



2

0

02

r

r

5 7

9

153 10 4 10 1

10225

36

α=182times10-3 NPm


β=

=(2πf) 0 0r r

β= 00182 radm



Given


To Find




Solution



(j j

j

0 0(2 f)( )( (2 )r rj j f

6 7 3 6 12(2 10 10 )(4 10 4)(10 (2 10 10 )(8854 10 25j j )

=3 3(31582)[10 13907 10 ]j j

07355 7214 02255 07j

j

α=02255 Npm

β=07 radm

Wavelength

2

=2π07

8975m


pV

2 f





j

j

0

0

(2 )

(2 )

r

r

j f

j f

=

6 7

3 6 12

(2 10 10 ) 4 10 4

10 (2 10 10 )8854 10 25

j

(12)

0

3

31582 90

17219 10 5428

η=4294 179

η=4086+j132Ω







components

(51)













Then

and (52)





(53)


(54)






















airplane

Optics







Grazing angle










Brewsters angle





























Using Snells law


Rearranging we get


Brewsters law


















KSRINIVASAN APECE




xx azEE



02

2

2

xx E

dz

Ed

The solution is

x

zz

x aeEeEzE

00


x

ztjzztjz

x aeeEeeEzE

00

Re




ss HjE


ss Ej

H

1






j

jj

H

E




za




sps

sps

HaE

EaH

1







jjj

=0 so jjj 0

1

]coscos[

Re

00

00

p

xx

x

ztjztj

x

u

ztEztEazE

aeEeEzE




smup 103

120

8

0

0





jc


jjjj

jjjjj

eff

c



1

12

1

12

2

2

eff

eff


tan

eff






1

eff

2






where







conductor


where
















(1)


and


(2)















Answer

Given

=60πmicror = 1



=0α=0β=1

To find

εr =

2

120

2

c




εr =

2

120

=22

εr =4


2

c


81(3 10 )

2

=15times108 radsec


(AU NOVDEC11)

Given

Frequency f=550 kHz


εr =25

To Find



tan

02



002 02 (2 )( )rf



9

3 1002 02 (2 550 10 )( 25)

36

5153 10 Sm



2

0

02

r

r

5 7

9

153 10 4 10 1

10225

36

α=182times10-3 NPm


β=

=(2πf) 0 0r r

β= 00182 radm



Given


To Find




Solution



(j j

j

0 0(2 f)( )( (2 )r rj j f

6 7 3 6 12(2 10 10 )(4 10 4)(10 (2 10 10 )(8854 10 25j j )

=3 3(31582)[10 13907 10 ]j j

07355 7214 02255 07j

j

α=02255 Npm

β=07 radm

Wavelength

2

=2π07

8975m


pV

2 f





j

j

0

0

(2 )

(2 )

r

r

j f

j f

=

6 7

3 6 12

(2 10 10 ) 4 10 4

10 (2 10 10 )8854 10 25

j

(12)

0

3

31582 90

17219 10 5428

η=4294 179

η=4086+j132Ω







components

(51)













Then

and (52)





(53)


(54)






















airplane

Optics







Grazing angle










Brewsters angle





























Using Snells law


Rearranging we get


Brewsters law


















KSRINIVASAN APECE




j

jj

H

E




za




sps

sps

HaE

EaH

1







jjj

=0 so jjj 0

1

]coscos[

Re

00

00

p

xx

x

ztjztj

x

u

ztEztEazE

aeEeEzE




smup 103

120

8

0

0





jc


jjjj

jjjjj

eff

c



1

12

1

12

2

2

eff

eff


tan

eff






1

eff

2






where







conductor


where
















(1)


and


(2)















Answer

Given

=60πmicror = 1



=0α=0β=1

To find

εr =

2

120

2

c




εr =

2

120

=22

εr =4


2

c


81(3 10 )

2

=15times108 radsec


(AU NOVDEC11)

Given

Frequency f=550 kHz


εr =25

To Find



tan

02



002 02 (2 )( )rf



9

3 1002 02 (2 550 10 )( 25)

36

5153 10 Sm



2

0

02

r

r

5 7

9

153 10 4 10 1

10225

36

α=182times10-3 NPm


β=

=(2πf) 0 0r r

β= 00182 radm



Given


To Find




Solution



(j j

j

0 0(2 f)( )( (2 )r rj j f

6 7 3 6 12(2 10 10 )(4 10 4)(10 (2 10 10 )(8854 10 25j j )

=3 3(31582)[10 13907 10 ]j j

07355 7214 02255 07j

j

α=02255 Npm

β=07 radm

Wavelength

2

=2π07

8975m


pV

2 f





j

j

0

0

(2 )

(2 )

r

r

j f

j f

=

6 7

3 6 12

(2 10 10 ) 4 10 4

10 (2 10 10 )8854 10 25

j

(12)

0

3

31582 90

17219 10 5428

η=4294 179

η=4086+j132Ω







components

(51)













Then

and (52)





(53)


(54)






















airplane

Optics







Grazing angle










Brewsters angle





























Using Snells law


Rearranging we get


Brewsters law


















KSRINIVASAN APECE





jjj

=0 so jjj 0

1

]coscos[

Re

00

00

p

xx

x

ztjztj

x

u

ztEztEazE

aeEeEzE




smup 103

120

8

0

0





jc


jjjj

jjjjj

eff

c



1

12

1

12

2

2

eff

eff


tan

eff






1

eff

2






where







conductor


where
















(1)


and


(2)















Answer

Given

=60πmicror = 1



=0α=0β=1

To find

εr =

2

120

2

c




εr =

2

120

=22

εr =4


2

c


81(3 10 )

2

=15times108 radsec


(AU NOVDEC11)

Given

Frequency f=550 kHz


εr =25

To Find



tan

02



002 02 (2 )( )rf



9

3 1002 02 (2 550 10 )( 25)

36

5153 10 Sm



2

0

02

r

r

5 7

9

153 10 4 10 1

10225

36

α=182times10-3 NPm


β=

=(2πf) 0 0r r

β= 00182 radm



Given


To Find




Solution



(j j

j

0 0(2 f)( )( (2 )r rj j f

6 7 3 6 12(2 10 10 )(4 10 4)(10 (2 10 10 )(8854 10 25j j )

=3 3(31582)[10 13907 10 ]j j

07355 7214 02255 07j

j

α=02255 Npm

β=07 radm

Wavelength

2

=2π07

8975m


pV

2 f





j

j

0

0

(2 )

(2 )

r

r

j f

j f

=

6 7

3 6 12

(2 10 10 ) 4 10 4

10 (2 10 10 )8854 10 25

j

(12)

0

3

31582 90

17219 10 5428

η=4294 179

η=4086+j132Ω







components

(51)













Then

and (52)





(53)


(54)






















airplane

Optics







Grazing angle










Brewsters angle





























Using Snells law


Rearranging we get


Brewsters law


















KSRINIVASAN APECE

smup 103

120

8

0

0





jc


jjjj

jjjjj

eff

c



1

12

1

12

2

2

eff

eff


tan

eff






1

eff

2






where







conductor


where
















(1)


and


(2)















Answer

Given

=60πmicror = 1



=0α=0β=1

To find

εr =

2

120

2

c




εr =

2

120

=22

εr =4


2

c


81(3 10 )

2

=15times108 radsec


(AU NOVDEC11)

Given

Frequency f=550 kHz


εr =25

To Find



tan

02



002 02 (2 )( )rf



9

3 1002 02 (2 550 10 )( 25)

36

5153 10 Sm



2

0

02

r

r

5 7

9

153 10 4 10 1

10225

36

α=182times10-3 NPm


β=

=(2πf) 0 0r r

β= 00182 radm



Given


To Find




Solution



(j j

j

0 0(2 f)( )( (2 )r rj j f

6 7 3 6 12(2 10 10 )(4 10 4)(10 (2 10 10 )(8854 10 25j j )

=3 3(31582)[10 13907 10 ]j j

07355 7214 02255 07j

j

α=02255 Npm

β=07 radm

Wavelength

2

=2π07

8975m


pV

2 f





j

j

0

0

(2 )

(2 )

r

r

j f

j f

=

6 7

3 6 12

(2 10 10 ) 4 10 4

10 (2 10 10 )8854 10 25

j

(12)

0

3

31582 90

17219 10 5428

η=4294 179

η=4086+j132Ω







components

(51)













Then

and (52)





(53)


(54)






















airplane

Optics







Grazing angle










Brewsters angle





























Using Snells law


Rearranging we get


Brewsters law


















KSRINIVASAN APECE




1

eff

2






where







conductor


where
















(1)


and


(2)















Answer

Given

=60πmicror = 1



=0α=0β=1

To find

εr =

2

120

2

c




εr =

2

120

=22

εr =4


2

c


81(3 10 )

2

=15times108 radsec


(AU NOVDEC11)

Given

Frequency f=550 kHz


εr =25

To Find



tan

02



002 02 (2 )( )rf



9

3 1002 02 (2 550 10 )( 25)

36

5153 10 Sm



2

0

02

r

r

5 7

9

153 10 4 10 1

10225

36

α=182times10-3 NPm


β=

=(2πf) 0 0r r

β= 00182 radm



Given


To Find




Solution



(j j

j

0 0(2 f)( )( (2 )r rj j f

6 7 3 6 12(2 10 10 )(4 10 4)(10 (2 10 10 )(8854 10 25j j )

=3 3(31582)[10 13907 10 ]j j

07355 7214 02255 07j

j

α=02255 Npm

β=07 radm

Wavelength

2

=2π07

8975m


pV

2 f





j

j

0

0

(2 )

(2 )

r

r

j f

j f

=

6 7

3 6 12

(2 10 10 ) 4 10 4

10 (2 10 10 )8854 10 25

j

(12)

0

3

31582 90

17219 10 5428

η=4294 179

η=4086+j132Ω







components

(51)













Then

and (52)





(53)


(54)






















airplane

Optics







Grazing angle










Brewsters angle





























Using Snells law


Rearranging we get


Brewsters law


















KSRINIVASAN APECE





conductor


where
















(1)


and


(2)















Answer

Given

=60πmicror = 1



=0α=0β=1

To find

εr =

2

120

2

c




εr =

2

120

=22

εr =4


2

c


81(3 10 )

2

=15times108 radsec


(AU NOVDEC11)

Given

Frequency f=550 kHz


εr =25

To Find



tan

02



002 02 (2 )( )rf



9

3 1002 02 (2 550 10 )( 25)

36

5153 10 Sm



2

0

02

r

r

5 7

9

153 10 4 10 1

10225

36

α=182times10-3 NPm


β=

=(2πf) 0 0r r

β= 00182 radm



Given


To Find




Solution



(j j

j

0 0(2 f)( )( (2 )r rj j f

6 7 3 6 12(2 10 10 )(4 10 4)(10 (2 10 10 )(8854 10 25j j )

=3 3(31582)[10 13907 10 ]j j

07355 7214 02255 07j

j

α=02255 Npm

β=07 radm

Wavelength

2

=2π07

8975m


pV

2 f





j

j

0

0

(2 )

(2 )

r

r

j f

j f

=

6 7

3 6 12

(2 10 10 ) 4 10 4

10 (2 10 10 )8854 10 25

j

(12)

0

3

31582 90

17219 10 5428

η=4294 179

η=4086+j132Ω







components

(51)













Then

and (52)





(53)


(54)






















airplane

Optics







Grazing angle










Brewsters angle





























Using Snells law


Rearranging we get


Brewsters law


















KSRINIVASAN APECE



(1)


and


(2)















Answer

Given

=60πmicror = 1



=0α=0β=1

To find

εr =

2

120

2

c




εr =

2

120

=22

εr =4


2

c


81(3 10 )

2

=15times108 radsec


(AU NOVDEC11)

Given

Frequency f=550 kHz


εr =25

To Find



tan

02



002 02 (2 )( )rf



9

3 1002 02 (2 550 10 )( 25)

36

5153 10 Sm



2

0

02

r

r

5 7

9

153 10 4 10 1

10225

36

α=182times10-3 NPm


β=

=(2πf) 0 0r r

β= 00182 radm



Given


To Find




Solution



(j j

j

0 0(2 f)( )( (2 )r rj j f

6 7 3 6 12(2 10 10 )(4 10 4)(10 (2 10 10 )(8854 10 25j j )

=3 3(31582)[10 13907 10 ]j j

07355 7214 02255 07j

j

α=02255 Npm

β=07 radm

Wavelength

2

=2π07

8975m


pV

2 f





j

j

0

0

(2 )

(2 )

r

r

j f

j f

=

6 7

3 6 12

(2 10 10 ) 4 10 4

10 (2 10 10 )8854 10 25

j

(12)

0

3

31582 90

17219 10 5428

η=4294 179

η=4086+j132Ω







components

(51)













Then

and (52)





(53)


(54)






















airplane

Optics







Grazing angle










Brewsters angle





























Using Snells law


Rearranging we get


Brewsters law


















KSRINIVASAN APECE













Answer

Given

=60πmicror = 1



=0α=0β=1

To find

εr =

2

120

2

c




εr =

2

120

=22

εr =4


2

c


81(3 10 )

2

=15times108 radsec


(AU NOVDEC11)

Given

Frequency f=550 kHz


εr =25

To Find



tan

02



002 02 (2 )( )rf



9

3 1002 02 (2 550 10 )( 25)

36

5153 10 Sm



2

0

02

r

r

5 7

9

153 10 4 10 1

10225

36

α=182times10-3 NPm


β=

=(2πf) 0 0r r

β= 00182 radm



Given


To Find




Solution



(j j

j

0 0(2 f)( )( (2 )r rj j f

6 7 3 6 12(2 10 10 )(4 10 4)(10 (2 10 10 )(8854 10 25j j )

=3 3(31582)[10 13907 10 ]j j

07355 7214 02255 07j

j

α=02255 Npm

β=07 radm

Wavelength

2

=2π07

8975m


pV

2 f





j

j

0

0

(2 )

(2 )

r

r

j f

j f

=

6 7

3 6 12

(2 10 10 ) 4 10 4

10 (2 10 10 )8854 10 25

j

(12)

0

3

31582 90

17219 10 5428

η=4294 179

η=4086+j132Ω







components

(51)













Then

and (52)





(53)


(54)






















airplane

Optics







Grazing angle










Brewsters angle





























Using Snells law


Rearranging we get


Brewsters law


















KSRINIVASAN APECE

εr =

2

120

=22

εr =4


2

c


81(3 10 )

2

=15times108 radsec


(AU NOVDEC11)

Given

Frequency f=550 kHz


εr =25

To Find



tan

02



002 02 (2 )( )rf



9

3 1002 02 (2 550 10 )( 25)

36

5153 10 Sm



2

0

02

r

r

5 7

9

153 10 4 10 1

10225

36

α=182times10-3 NPm


β=

=(2πf) 0 0r r

β= 00182 radm



Given


To Find




Solution



(j j

j

0 0(2 f)( )( (2 )r rj j f

6 7 3 6 12(2 10 10 )(4 10 4)(10 (2 10 10 )(8854 10 25j j )

=3 3(31582)[10 13907 10 ]j j

07355 7214 02255 07j

j

α=02255 Npm

β=07 radm

Wavelength

2

=2π07

8975m


pV

2 f





j

j

0

0

(2 )

(2 )

r

r

j f

j f

=

6 7

3 6 12

(2 10 10 ) 4 10 4

10 (2 10 10 )8854 10 25

j

(12)

0

3

31582 90

17219 10 5428

η=4294 179

η=4086+j132Ω







components

(51)













Then

and (52)





(53)


(54)






















airplane

Optics







Grazing angle










Brewsters angle





























Using Snells law


Rearranging we get


Brewsters law


















KSRINIVASAN APECE

9

3 1002 02 (2 550 10 )( 25)

36

5153 10 Sm



2

0

02

r

r

5 7

9

153 10 4 10 1

10225

36

α=182times10-3 NPm


β=

=(2πf) 0 0r r

β= 00182 radm



Given


To Find




Solution



(j j

j

0 0(2 f)( )( (2 )r rj j f

6 7 3 6 12(2 10 10 )(4 10 4)(10 (2 10 10 )(8854 10 25j j )

=3 3(31582)[10 13907 10 ]j j

07355 7214 02255 07j

j

α=02255 Npm

β=07 radm

Wavelength

2

=2π07

8975m


pV

2 f





j

j

0

0

(2 )

(2 )

r

r

j f

j f

=

6 7

3 6 12

(2 10 10 ) 4 10 4

10 (2 10 10 )8854 10 25

j

(12)

0

3

31582 90

17219 10 5428

η=4294 179

η=4086+j132Ω







components

(51)













Then

and (52)





(53)


(54)






















airplane

Optics







Grazing angle










Brewsters angle





























Using Snells law


Rearranging we get


Brewsters law


















KSRINIVASAN APECE

Solution



(j j

j

0 0(2 f)( )( (2 )r rj j f

6 7 3 6 12(2 10 10 )(4 10 4)(10 (2 10 10 )(8854 10 25j j )

=3 3(31582)[10 13907 10 ]j j

07355 7214 02255 07j

j

α=02255 Npm

β=07 radm

Wavelength

2

=2π07

8975m


pV

2 f





j

j

0

0

(2 )

(2 )

r

r

j f

j f

=

6 7

3 6 12

(2 10 10 ) 4 10 4

10 (2 10 10 )8854 10 25

j

(12)

0

3

31582 90

17219 10 5428

η=4294 179

η=4086+j132Ω







components

(51)













Then

and (52)





(53)


(54)






















airplane

Optics







Grazing angle










Brewsters angle





























Using Snells law


Rearranging we get


Brewsters law


















KSRINIVASAN APECE

j

j

0

0

(2 )

(2 )

r

r

j f

j f

=

6 7

3 6 12

(2 10 10 ) 4 10 4

10 (2 10 10 )8854 10 25

j

(12)

0

3

31582 90

17219 10 5428

η=4294 179

η=4086+j132Ω







components

(51)













Then

and (52)





(53)


(54)






















airplane

Optics







Grazing angle










Brewsters angle





























Using Snells law


Rearranging we get


Brewsters law


















KSRINIVASAN APECE










Then

and (52)





(53)


(54)






















airplane

Optics







Grazing angle










Brewsters angle





























Using Snells law


Rearranging we get


Brewsters law


















KSRINIVASAN APECE



(53)


(54)






















airplane

Optics







Grazing angle










Brewsters angle





























Using Snells law


Rearranging we get


Brewsters law


















KSRINIVASAN APECE
















airplane

Optics







Grazing angle










Brewsters angle





























Using Snells law


Rearranging we get


Brewsters law


















KSRINIVASAN APECE







airplane

Optics







Grazing angle










Brewsters angle





























Using Snells law


Rearranging we get


Brewsters law


















KSRINIVASAN APECE




Brewsters angle





























Using Snells law


Rearranging we get


Brewsters law


















KSRINIVASAN APECE





Using Snells law


Rearranging we get


Brewsters law

















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