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COURSE MATERIAL

ELECTROMAGNETIC THEORY &

SUBJECT TRANSMISSION LINES (19A04401)

UNIT 4

COURSE B.TECH

ELECTRONICS & COMMUNICATION

DEPARTMENT ENGINEERING

SEMESTER 2-2

K.Upendra Raju
PREPARED BY D.Srilatha
(Faculty Name/s) K.R.Surendra
V.Omkar Naidu

Version V-1

PREPARED / REVISED DATE 30-03-2021

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TABLE OF CONTENTS – UNIT 1

S. NO CONTENTS PAGE NO.
1 COURSE OBJECTIVES 3
2 PREREQUISITES 3
3 SYLLABUS 3
4 COURSE OUTCOMES 3
5 CO - PO/PSO MAPPING 3
6 LESSON PLAN 3
7 ACTIVITY BASED LEARNING 4
8 LECTURE NOTES 4
1.1 Introduction 4
1.2 Reflection of a Plane wave at Normal Incidence 4
1.3 Reflection of A Plane Wave at Oblique Incidence 8
1.4 Critical Angle & Total Internal Reflection 15
1.5 Poynting Vector & Theorem 15
9 PRACTICE QUIZ 16
10 ASSIGNMENTS 18
11 PART A QUESTIONS & ANSWERS (2 MARKS QUESTIONS) 18
12 PART B QUESTIONS 19
13 SUPPORTIVE ONLINE CERTIFICATION COURSES 19
14 REAL TIME APPLICATIONS 19
15 CONTENTS BEYOND THE SYLLABUS 19
16 PRESCRIBED TEXT BOOKS & REFERENCE BOOKS 20
17 MINI PROJECT SUGGESTION 20

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1. Course Objective
The objectives of this course is to
1. To analyse reflection and refraction of electromagnetic waves propagated in
normal and oblique incidences

2. Prerequisites
Students should have knowledge on
1. Electric and Magnetic fields
2. Basic Mathematics

3. Syllabus
UNIT IV
Reflection and Refraction of Plane Waves – Normal and Oblique Incidences, for both
Perfect Conductor and Perfect Dielectrics, Brewster Angle, Critical Angle and Total
Internal Reflection, Surface Impedance, Poynting Vector, and Poynting Theorem –
Applications, Power Loss in a Plane Conductor, Illustrative Problems.
4. Course outcomes
1. Understand principles of reflections and refraction for different incidences (L1)
2. State concepts of power flow using Poynting vector (L2)
3. Calculate Brewster angle, power flow and surface impedance (L3)

5. CO-PO / PSO Mapping

PO1 PO2 PO3 PO4 PO5 PO6 PO7 PO8 PO9 P10 PO11 PO12 PSO1 PSO2

CO1 3 3 2 2 3 2 2

CO2 3 3 2 2 3 2 2

CO3 3 3 2 2 3 2 2

C04 3 3 2 2 3 2 2

6. LESSON PLAN

LECTURE WEEK TOPICS TO BE COVERED REFERENCES

1 Reflection of a Plane wave at normal incidence T1, T2, R2

2 Reflection of a plane wave at oblique incidence T1, T2, R2

1
3 Refractions of planes waves at normal incidence T1, T2, R2

4 Refractions of planes waves at oblique incidence T1, T2, R2

5 2 Reflection and Refraction for both perfect conductor T1, T2, R2

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and perfect dielectric

6 Brewster angle, Critical angle T1, T2, R2

7 Total internal reflections T1, T2, R2

8 Surface impedance T1, T2, R2

9 Poynting Theorem – Applications T1, T2, R2

10 3 Power loss in a Plane conductor T1, T2, R2

11 Problems T1, T2, R2

7. ACTIVITY BASED LEARNING

1. Technical Quiz
2. Group Discussion
3. Think Pair Share

8. LECTURE NOTES
1.1 INTRODUCTION
When a plane wave from one medium meets a different medium, it is partly
reflected and partly transmitted. The proportion position of the incident wave that is
reflected or transmitted depends on the parameters ε, μ, and σ of the two media
involved. The incident wave can meet the boundary between two media either
normally or with some angle (oblique)
1.2 REFLECTION OF A PLANE WAVE AT NORMAL INCIDENCE:
Suppose that a plane wave propagating along the +Z direction is incedent
normally on the boundary Z = 0 between medium 1(Z < 0) characterized by σ 1, ε1, μ1
and medium 2 (Z > 0) characterized by σ2, ε2, μ2 as shown in the figure (4.1). In the
figure subscripts i, r and t denotes incident, reflected and transmitted waves
respectively.

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Fig. (4.1) A plane wave incident normally

General case:
Incident Wave: (⃗⃗⃗ , ⃗⃗⃗⃗ ) is travelling along + ⃗⃗⃗⃗ in medium 1 without considering the
time factor ejωt. We can write
⃗⃗⃗ ( ) ⃗⃗⃗⃗ and …………… (4.1)
⃗⃗⃗⃗ ( ) ⃗⃗⃗⃗ …………… (4.2)

= ⃗⃗⃗⃗

Reflected Wave: (⃗⃗⃗⃗ , ⃗⃗⃗⃗ ) is travelling along - ⃗⃗⃗⃗ in medium 1 without considering the
time factor ejωt. We can write
⃗⃗⃗⃗ ( ) ⃗⃗⃗⃗ and …………… (4.3)
⃗⃗⃗⃗ ( ) ( ⃗⃗⃗⃗ ) …………… (4.4)

= ⃗⃗⃗⃗

Transmitted Wave: (⃗⃗⃗ , ⃗⃗⃗⃗ ) is travelling along + ⃗⃗⃗⃗ in medium 1 without considering
the time factor ejωt. We can write
⃗⃗⃗ ( ) ⃗⃗⃗⃗ and …………… (4.5)
⃗⃗⃗⃗ ( ) ( ⃗⃗⃗⃗ ) …………… (4.6)

= ⃗⃗⃗⃗

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Where Eio, Ero, Eto are respectively the magnitudes of the incident, reflected and
transmitted electric field at Z = 0. The total field in medium 1 consists of both the
incident and reflected fields, whereas medium 2 has only transmitted field. i.e.,
⃗⃗⃗⃗ ⃗⃗⃗ ⃗⃗⃗⃗ ⃗⃗⃗⃗ ⃗⃗⃗⃗ ⃗⃗⃗⃗
⃗⃗⃗⃗ ⃗⃗⃗ ⃗⃗⃗⃗ ⃗⃗⃗⃗

At the interface Z = 0, the boundary conditions require that the tangential

components of ⃗ and ⃗ fields must be continuous. Since the waves are already
transverse. ⃗ and ⃗ fields both must be tangential to the interface.
Hence at Z = 0 ⃗⃗⃗⃗ ⃗⃗⃗⃗ ⃗⃗⃗⃗ ⃗⃗⃗⃗

Or interms of amplitude or magnitude, we can write

Ei(0) + Er(0) = Et(0) → Ei0 + Er0 = Et0 …………… (4.7)
Hi(0) + Hr(0) = Ht(0) → (Ei0 - Er0) = …………… (4.8)

From equ.,(4.7) and (4.8)

…………… (4.9)

…………… (4.10)

coefficient from equ.,

(4.9) and (4.10)
…………… (4.11)

…………… (4.12)

Note that
1. 1 + =
2. Both and are dimensionless and may be complex
3. 0 ≤ l l ≤ 1
The above case is considered as the general case.
Let us consider a special case when medium 1 is a perfect dielectric (lossless
σ1=0) and medium 2 is a perfect conductor (σ2=∞). For this case η2=0; hence Γ = -1
and = 0, showing that the wave is totally reflected. This should be expected
because fields in a perfect conductor must vanish, so there can be no transmitted
wave (E2 = 0). The totally reflected wave combines with the incident wave to form a
standing wave.
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A standing wave does not travel, it consists of two travelling waves (E i and Er) of
equal amplitudes but in opposite directions. Combining equ.(4.1) and (4.3) gives the
standing wave in medium 1 is
E1s = Eis + Ers = + (4.13)

But, = -1, σ1=0, α1 = 0, γ1 = jβ1

Hence E1s = -Eio( - )ax

or E1s = -2jEio sin β1z ax (4.14)
Thus E1 = Re(Eis )
or E1 = 2Eio sin β1z sin ωt ax (4.15)
By taking the similar procedure, the magnetic field component of the wave is

H1 = cos β1z cos ωt ay

when media 1 and 2 are both lossless we consider another special case (σ1 = 0 =
σ2). In this case η1 and η2 are real so that Γ and . Let us consider the following cases:
CASE A

If η2 > η1, Γ > 0, Again there is a standing wave in medium 1 but there is also a
transmitted wave in medium 2. However, the incident and reflected waves have
amplitudes that are not equal in magnitude. It can be shown that the maximum values of
l E1l occur at
- β1 zmax = nπ

or = , n = 0, 1, 2, …….. (4.16)

and the minimum values of lE1l occurs at

- β1 zmin = (2n + 1)

( ) ( )
= n = 0, 1, 2, …….. (4.17)

CASE B
If η2 < η1, Γ < 0. For this case, the locations of lE1l maximum are given by eq. (4.17)
whereas those of lE1l minimum are given by eq. (4.16). All these are illustrated in Fig. (4.2)
Note that

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Fig. (4.2) Standing waves due to reflection at an interface between two lossless media.

1. lH1l minimum occurs whenever there is lE1l maximum and vice versa.
2. The transmitted wave in medium 2 is a purely traveling wave and consequently there
are no maxima and minima in this region.
The ratio of lE1lmax to lE1lmin (or lH1lmax to lH1lmin) is called the standing – wave ratio s; that
is,
| | | | | |
| |
= | |
= | |
(4.18)

or
| | (4.19)

Since | | ≤ 1 it follows that 1 ≤ s ≤ ∞. The standing - wave ratio is dimensionless and it is

customarily expressed in decibels (dB) as
S in dB = 20 log10 s

1.3 REFLECTION OF A PLANE WAVE AT OBLIQUE INCIDENCE

We now consider a more general situation than that in normal incidence. To simplify
the analysis, we will assume that we are dealing with lossless media. In this section we shall
consider the case of oblique incidence. So, we consider two cases.

1. When the second medium is a perfect conductor.

2. when the second medium is a perfect dielectric.

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A plane incidence is defined as the plane containing the vector indicating the
direction of propagation of the incident wave and normal to the interface.

A uniform plane wave takes the general form of

( )
E (r, t) = Eo cos (k . r – ωt) = Re [ ] (4.20)

Where r = xax + yay + zaz is the radius or position vector and k = kxax + kyay + kzaz is the
wave number vector or the propagation vector. k is always in the direction of wave
propagation. The magnitude of k is related to ω according to the dispersion relation

(4.21)
Thus for lossless media, k is essentially the same as β. With the general form of E as in
equ.(4.20), Maxwell’s equations reduce to
k x E = ωμH (4.22a)
k x E = - ωεE (4.22b)
k.H = 0 (4.22c)
k.E = 0 (4.22d)
showing that (i) E, H, and k are mutually orthogonal and
(ii) E and H lie on the plane
K . r = kxx + kyy + kzz = constant
from equ.(4.22a) the H field corresponding to the E field in equ. (4.20)

(4.23)

This expression E and H in the general form, we can now consider the oblique
incidence of a uniform plane wave at a plane boundary as shown in fig. (4.3). The plane
defined by the propagation vector k and a unit normal vector an to the boundary is
called the plane of incidence. The angle θi between k and an is the angle of incidence.
Again both the incident and the reflected waves are in medium 1 and transmitted
(or refracted wave) is in medium 2. Let
( ) (4.24a)
( ) (4.24b)
( ) (4.24c)

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Fig. (4.3) Oblique incidence of a plane wave

Where ki, kr, kt with their normal and tangential components are shown in fig. (4.3b).
since the tangential component of E must be continuous at the boundary z = 0.
Ei (z = 0) + Er (z = 0) = Et (z = 0) (4.25)
The only way this boundary condition will be satisfied by the waves in equ. (4.24) for
all x and y is that
1.
2.
3.
Condition 1 implies that the frequency is unchanged. Conditions 2 and 3 require that
the tangential components of the propagation vectors be continuous. This means
that the propagation vectors ki, kt and kr must all lie in the plane of incidence. Thus,
by conditions 2 and 3
(4.26)
(4.27)

Where θr is the angle of reflection and θt is the angle of transmission. But, for lossless
media,
√ (4.28a)
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√ (4.28b)
From equ. (4.26) and (4.28a) it is clear that
(4.29)
So, that the angle of reflection equals the angle of incidence as in optics. Also
from equ. (4.27) and (4.28)

√ (4.30)

The equ. (4.30) is known as Snell’s Law. Which can be written as

(4.31)
Where √ and √ are the refractive indices of the
media.
Based on these general preliminaries on oblique incidence, we will consider two
special cases: one is with the ⃗⃗⃗ is perpendicular to the plane of incidence (perpendicular
polarization) and ⃗⃗⃗ is parallel to the plane of incidence (parallel polarization).

(a) Parallel Polarization:

Fig. (4.4) Oblique incidence with E parallel to the plane of incidence.

From the fig.(4.4) where E field lies in the xz – plane, the plane of incidence. In
medium 1, we have both incident and reflected fields given by
( )
( ) (4.32a)
( )
(4.32b)
( )
( ) (4.33a)

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( )
(4.33b)

Where √ . To derive how to arrive at each field component is to get the

polarization vector k as shown in fig. (4.3b) for incident, reflected and transmitted waves.
Once k is known we define Es such that ▼ . Es = 0; or k . Es = 0 and Hs is obtained from

The transmitted field exist in medium 2 are given by

( )
( ) (4.34a)
( )
(4.34b)

Where √ .
Requiring that and that the tangential components of E and H be continuous
at the boundary z = 0, we obtain
( ) (4.35a)

( ) (4.35b)

Expressing Ero and Eto in terms of Eio, we obtain

|| (4.36a)

or || (4.36b)

|| (4.37a)

or || (4.37b)
Equations (4.36) and (4.37) are called Fresnel’s Equations. Note that the equations
reduce to equ. (4.11) and (4.12) when as expected. Since and are
related according to Snell’s law of equ.(4.30), (4.36) and (4.37) can be written in terms of
by substituting

√ √ ( ) (4.38)

From equ.(4.36) and (4.37) it is easily shown that

|| || ( ) (4.39)

From equ.(4.36a) it is evident that it is possible that || because the numerator is

the difference of two terms. Under this condition, there is no reflection (E ro = 0) and the
incident angle at which this takes place is called the Brewster angle ||
. The Brewster

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angle also known as the Polarizing angle because an arbitrarily polarized incident wave
will be reflected with only the component of E perpendicular to the plane of incidence.
The Brewster angle is obtained by setting ||
when || in equ.(4.36)

||

or ( ) ( ||
)

introducing equ.(4.30) and (4.31) gives

|| ( )
(4.40)

It is of practical value to consider the case when the dielectric media are not only
lossless, but nonmagnetic as well – i.e., = = . For this the equ.(4.40) becomes

||
→ || √

or || √ (4.41)

showing that there is a Brewster angle for any combination of and

(b) Perpendicular Polarization:

In this case, the E field is perpendicular to the plane of incidence as shown in Figure
(4.5).

Fig. (4.5) Oblique incidence with E perpendicular to the plane of incidence.

This may also be viewed as the case where H field is parallel to the plane of
incidence. The incident ans reflected fields in medium 1 are given by
( )
(4.42a)
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( )
( ) (4.42b)
( )
(4.43a)
( )
( ) (4.43b)

While the transmitted fields in medium 2 are given by

( )
(4.44a)
( )
( ) (4.44b)

Note that in defining the field components in equ.(4.42) to (4.44), Maxwell’s

equations (4.22)are always satisfied. Again, requiring that the tangential components of E
and H be continuous at z = 0 and setting equal to , we get
( ) (4.46a)

( ) (4.46b)

Expressing Ero and Eto in terms of Eio, leads to

(4.47a)

or (4.47b)

(4.48a)

or (4.48b)
which are the Fresnel’s equations for perpendicular polarization. From equ.(4.47)
and (4.48), it is easy to show that
(4.49)
When = = 0 equ.(4.47) and equ.(4.48) becomes equ. (4.11) and (4.12).
For no reflection = 0 (or Er = 0). This is the same as the case of total transmission
( = 0). By replacing with the corresponding Brewster angle

or ( ) ( )
incorporating equ.(4.31)

( )
(4.50)

Note that for nonmagnetic media ( = = ), → ∞ in equ.(4.50), so does

not exist because the sine of an angle is never greater than unity. Also if ≠ and
equ. (4.50) reduces to
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or √ (4.51)

1.4 CRITICAL ANGLE & TOTAL INTERNAL REFLECTION

Total internal reflection refers to the complete reflection of a ray of light within an
optically-denser medium from the surrounding surfaces of optically less dense media
back into the denser medium.
 Light ray travel from an optically denser medium to a less dense medium.
 The angle of incidence must be greater than a certain angle, called the critical
angle.
For a ray of light passing from an optically denser to a less dense medium, critical
angle, c, is the angle of incidence at which the angle of refraction is 90o.
 When the angle of incidence is less than the critical angle, the ray passes out into
the less dense medium.
 When the angle of incidence is greater than the critical angle, the ray is reflected
back into the denser medium.
The equation relating critical angle, c and the refractive index, n is:
Sin c=1/n (4.52)

1.5 POYNTING VECTOR & THEOREM

The energy can be transported from one point (where a transmitter is located) to
another point (with a receiver) by means of EM waves. The rate of such energy
transportation can be obtained from Maxwell’s equations:
(4.53a)
(4.53b)
Dotting both sides of equ.(4.53b) with E gives
( ) (4.54)

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But for any vector fields A and B

( ) ( ) ( )
Applying this vector identity to equ.(4.54) (Let A = H and B = E)

( ) ( ) (4.55)

From equ.(4.53a)

( ) ( ) ( ) (4.56)

and thus equ.(4.55) becomes

( )

Rearranging terms and taking the volume integral of both sides

∫ ( ) ∫[ ] ∫ (4.57)

Applying the divergence theorem to the left hand side gives

∮( ) ∫[ ] ∫ (4.58)

Total Power Rate of decrease Ohmic power

leaving the in energy stored dissipated
volume in electric and
magnetic fields

Equation (4.58) is referred to as Poynting theorem. The quantity E x H on the left hand
side of equ.(4.58) is known as the Poynting vector P in watts per square meter (W/m2).
P=ExH (4.59)

9. PRACTICE QUIZ
1. Electromagnetic waves travel in conductors than in dielectrics?
a) True
b) False

2. In a good conductor, E and H are in time phase

a) True
b) False

3. The incident wave that is reflected or transmitted depends on the

constitutive on ________ parameters

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a) ε
b) μ
c) σ
d) All
4. If the medium 1 is a perfect dielectric, then σ1 = _________
a) -1
b) 1
c) 0
d) ∞

5. The Poynting vector physically denotes the power density leaving or

entering a given volume in a time-varying field.
a) True
b) False

6. If the medium 1 is a perfect conductor, then σ2 = _________

a) -1
b) 1
c) 0
d) ∞
7. The total reflected wave combines with the incident wave to form a ___
a) Reflection coefficient
b) Transmission coefficient
c) Standing wave
d) Incident wave
8. The standing wave ratio is lies between_____
a) -1 ≤ s ≤ 1
b) 0 ≤ s ≤ 1
c) 1 ≤ s ≤ ∞
d) -∞ ≤ s ≤ ∞
9. Poynting vector p is measured in
a) Wm2
b) W
c) W/m3
d) W/m2
10. In good conductors E leads H by
a) 90o
b) 180o
c) 45o
d) 0o

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10. ASSIGNMENTS

S.No Question BL CO
1 Write short notes on: (i) Surface impedance. (ii) Brewster angle 2 4
Explain reflection of uniform plane wave by a perfect conductor in
2 2 4
the case of oblique incidence for parallel polarization
3 State and prove pointing theorem? 2 4
Discuss the determination of the reflected and wave fields of a
4 uniform plane wave incident normally onto a plane boundary 1 4
between two material media.

11. PART A QUESTIONS & ANSWERS (2 MARKS QUESTIONS)

S.No Question & Answers BL CO
1 Define reflection coefficient? 1 4
Ans.

2 Define transmission coefficient? 1 4

Ans.
3 Define Standing wave ratio? 1 4
Ans. The ratio of lE1lmax to lE1lmin (or lH1lmax to lH1lmin) is called the
standing – wave ratio s; that is,
| | | | | |
| |
= | |
= | |
4 Define reflection coefficient interms of s? 1 4
Ans. | |
5 What is plane of incidence? 1 4
Ans. The plane defined by the propagation vector k and a unit
vector an to the boundary is called the plane of incidence.
6 Define angle of incidence? 1 4
Ans. The angle θi between propagation vector k and a unit vector
an is the angle of incidence.
7 Define Poynting theorem? 1 4
Ans. States that the net power flowing out of a given volume v is
equal to the time rate of decrease in the energy stored within v
minus the conduction losses.
8 What is Poynting vector? 1 4
Ans. The instantaneous power density vector associated with the EM
field at a given point.
9 Write the Fresnel’s equations? 1 4
Ans. ||

or ||

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||

or ||
10 Define Brewster angle? 1 4
Ans. ||
From the above equation that ||=0 because the numerator is the
difference of two terms. Under this conditions, there is no reflection
(Ero = 0) and the incident angle at which this takes place is called
the Brewster angle.

12. PART B QUESTIONS

S.No Question BL CO
1 Explain the reflection of uniform plane wave with normal incidence 2 3
at plane dielectric boundary?
2 Explain the reflection of uniform plane wave with oblique incidence 2 3
at plane dielectric boundary?
3 Determine the resultant electric and magnetic fields of plane wave 2 3
when it is incident on a perfect conductor normally?
4 Explain the wave motion in perfect dielectric medium? 1 2
5 Discuss about the concept of Poynting theorem 2 3

13. SUPPORTIVE ONLINE CERTIFICATION COURSES

1. https://nptel.ac.in/courses/115/106/115106122/
2. https://onlinecourses.nptel.ac.in/noc20_ee93/preview
3. https://onlinecourses.nptel.ac.in/noc21_ee43/preview

14. REAL TIME APPLICATIONS

S.No Application CO
1 Reflections in Light waves 4
2 Fiber Optics for communications 4
3 Endoscope (Fiber Optics)

15. CONTENTS BEYOND THE SYLLABUS

 Plane waves in Free space medium

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16. PRESCRIBED TEXT BOOKS & REFERENCE BOOKS

Text Book:
1. Matthew N.O. Sadiku, “Elements of Electromagnetic”, 4th edition, Oxford Univ.
Press 2008.
2. William H. Hayt Jr. and John A. Buck, “Engineering Electromagnetics”, 7th edition.,
TMH 2006
References:
1. E.C Jordan and K.G Balman, “Electromagnetic waves and Radiating Systems”,
2nd Edition, PHI, 2000.
2. John D. Krauss, “Electromagnetics”, 4th Edition, McGraw-Hill Publication 1999
3. Electromagnetics, Schaum’s outline series, 2nd Edition, Tata McGraw-Hi;;
Publications, 2006.

17. MINI PROJECT SUGGESTION

 Find the Brewster angle when dielectric medium is not lossless?
 Find the reflection and refraction angle of light waves in different medium?

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