Boundary Conditions Lecture 141 1 1 1 and 1 me me 3/7/2013 v = Dr. Aparna Tripathi v = •Putting the value of v1 and v2 in eq1 2 sin sin 1 1 2 me me q q = t i •For perfect dielectric

Boundary Conditions

Lecture 14

3/7/2013 Dr. Aparna Tripathi

Oblique Incidence

1 1 , m e 2 2 , m e

0 = x

r i E E E + = 1 t E E = 2

t q r q

i q A

B

D C E

t q

•When a plane wave is incident upon a boundary surface that is not parallel to

the plane, then a part of incident wave will reflected at an angle 𝛉t and moves

into the 2nd medium.

•Rest is reflected at an angle 𝛉r into the same medium.

•The incident wave travels the

distance CB.

•Refracted wave travels the

distance AD.

•Reflected wave travels the

distance AE.

•If v1 and v2 be the speed of

wave in medium 1 and 2

respectively 3/7/2013 Dr. Aparna Tripathi

1 1 , m e 2 2 , m e

0 = x

r i E E E + = 1 t E E = 2

t q r q

i q A

B

D C E

t q 2

1

21

vAD

tvAD and

vCB

tvCB

=

==

...1 2

1

sin

sin

sin and sin

v

v

AD

CB

t

i

AB

ADt

AB

CBi

==

==

q

q

qq

tABD and

geometry of help by the

qq == iCAB

•If m1 andm2 are the permeabilities of the mediums the speed v1 and v2

22

2

11

1

1 and

1

emem== vv


•Putting the value of v1 and v2 in eq1

sin

sin

11

22

em

em

q

q=

t

i

•For perfect dielectric m1=m2= m

sin

sin

1

2

e

e

q

q=

t

i

•Form fig AE = CB

incidence of angle reflection of angle

sinsin

=

=

=

=

ir

ir

AB

CB

AB

AE

qq

qq


•If n1 and n2 are the refractive indices of medium 1 and 2 respectively then

2

2

1

1

2

2

1

1

vand v

v and

v

n

c

n

c

cn

cn

==

==

•Substituting v1 and v2 in eq 1

)refraction of law s(Snell' ...2 sin

sin

1

2

n

n

t

i=

q

q

•Thus concluded that em waves obey all the laws of reflection and refraction

at a surface separating two dielectric media.


•Energy transmitted per square meter

Z

ES

2

=

1

2cos

Z

ES ii

i

q=

•Energy transmitted per square meter by incident wave

•Energy transmitted per square meter by reflected wave

•Energy transmitted per square meter by transmitted wave

1

2cos

Z

ES ir

r

q=

2

2cos

Z

ES tt

t

q=

==

n to being ,0 2

EnEnZ

ES

Angle between E and n is 900


•By the law of conservation of energy

2

2

1

2

2

2

2

2

1

2

2

2

2

2

1

2

1

2

cos

cos1

cos

cos1

coscoscos

ZE

ZE

E

E

ZE

ZE

E

E

Z

E

Z

E

Z

E

ii

tt

i

r

ii

tt

i

r

ttirii

q

q

q

q

qqq

=

+=

+=

Energy associated with the incident wave = Energy associated with the

reflected wave+ Energy associated with the transmitted wave

=== mmm

e

e21 dielectricperfect for

1

2

2

1

Z

Z

...2 cos

cos1

2

1

2

2

2

2

ii

tt

i

r

E

E

E

E

qe

qe=


Oblique Incidence

1. Incident plane wave is linearly polarized with its electric

vector perpendicular to the plane of incident or parallel to the

boundary surface.

2.Electric vector is parallel to the plane of incidence.


1.Electric vector is perpendicular to the plane of

incidence

•E electric vector lie normal to the plane of

incidence

•H magnetic vector to the plane of incidence

Applying boundary conditions, tangential components of

E is continuous across the boundary

...3 1i

t

i

r

tri

E

E

E

E

EEE

=+

=+

Putting the value of Et/Ei in eq 2

1cos

cos1

2

1

2

2

2

+=

i

r

i

t

i

r

E

E

E

E

qe

qe...2

cos

cos1

2

1

2

2

2

2

=

ii

tt

i

r

E

E

E

E

qe

qe


2

1

2

2

2

1cos

cos1

+=

i

r

i

t

i

r

E

E

E

E

qe

qe

1cos

cos1

1

2

+=

i

r

i

t

i

r

E

E

E

E

qe

qe

1cos

cos11

2

1

2

+=

+

i

r

i

t

i

r

i

r

E

E

E

E

E

E

qe

qe

itirti

ritrii

EE

EEEE

qeqeqeqe

qeqe

coscoscoscos

coscos

2121

21

=+

+=

=+

= R

E

E

ti

ti

i

r

qeqe

qeqe

coscos

coscos

21

21 Reflection coefficient for

horizontal polarization 3/7/2013 Dr. Aparna Tripathi

Similarly the transmission coefficient for horizontal polarization

ti

iT

qeqe

qe

coscos

cos2

21

1

+=

ti

ti

i

r

i

t

T

E

E

E

ET

qeqe

qeqe

coscos

coscos1

1

21

21

+

+=

+==

But

sin

sin

sin

sin21

2

1

i

t

i

t

q

qee

e

e

q

q==


ti

ti

tiit

tiit

R

R

qq

qq

qqqq

qqqq

+

=

+

=

sin

sin

cossincossin

cossincossin

ti

it

tiit

it

T

T

qq

qq

qqqq

qq

+=

+=

sin

cossin2

cossincossin

cossin2


1.Electric vector is parallel to the plane of incidence

•E electric vector lie in the plane of incidence

•H magnetic vector to the plane of incidence

Applying boundary conditions, tangential components of

E is continuous across the boundary

=

=

=

t

i

i

r

i

t

i

t

i

t

i

r

ttirii

E

E

E

E

E

E

E

E

EEE

q

q

q

q

qqq

cos

cos 1

cos

cos 1

coscoscos

Putting the value of Et/Ei in eq 2

coscos

coscos 11

2

22

1

2

2

2

it

ti

i

r

i

r

E

E

E

E

qq

qq

e

e

= ...2

cos

cos1

2

1

2

2

2

2

=

ii

tt

i

r

E

E

E

E

qe

qe3/7/2013 Dr. Aparna Tripathi

t

i

i

r

i

r

E

E

E

E

q

q

e

e

cos

cos11

2

1

2

2

2

=

1cos

cos1

1

2

=

+

i

r

t

i

i

r

E

E

E

E

qe

qe

cos

cos111

2

1

2

t

i

i

r

i

r

i

r

E

E

E

E

E

E

q

q

e

e

=

+

itirit

riirit

EE

EEEE

qeqeqeqe

qeqe

coscoscoscos

coscos

1221

21

=+

=+

=+

= R

E

E

it

ti

i

r

qeqe

qeqe

coscos

coscos

21

12 Reflection coefficient for

vertical polarization 3/7/2013 Dr. Aparna Tripathi

Similarly the transmission coefficient for verticalpolarization

it

iT

qeqe

qe

coscos

cos2

21

2

+=

it

ti

i

r

i

t

T

E

E

E

ET

qeqe

qeqe

coscos

coscos1

1

21

12

+

+=

+==

But

sin

sin

sin

sin21

2

1

i

t

i

t

q

qee

e

e

q

q==


)tan(

)tan(

2sin2sin

2sin2sin

cossincossin

cossincossin

ti

ti

ti

ti

iitt

ttii

R

R

qq

qq

qq

qq

qqqq

qqqq

+

=

+

=

+

=

itti

ii

ti

ii

iiit

ii

T

T

tT

qqqq

qq

qq

qq

qqqq

qq

+=

+=

+=

cossin

cossin2

2sin2sin

cossin2

cossincossin

cossin2




Hη E =



…..(A)

…….(B) 3/7/2013 Dr. Aparna Tripathi

reflection coefficient

transmission coefficient


Electrostatic Boundary Conditions

Consider a rectangular closed path abcda of infinitesimal area in the plane

normal to the boundary and with its side ab and cd parallel to and on either

side of th boundary.

Since the line integral E.dl for closed path is zero.

Here considering the closed path abcd then

0

negligible are ad and bc parts then the

0h lettingby shrunk is loop theif

12 =+=

+++=

d

c

b

a

a

d

d

c

c

b

b

a

dlEdlEdlE

dlEdlEdlEdlEdlE

a b

d c

K2

K1

e2

e1

A B

E2t

Δl

E1t


E1t and E2t are components of electric field intensities then

tt

tt

EE

lElEdlE

21

21 0

=

==

Hence any point on the boundary the components of E1 and E2 tangential to

the boundary are equal

If medium 1 is conductor then E1t =0 then

02 =tE


Boundary Conditions for Displacement vector D

Consider a cylindrical bax ABCD which intersect the two dielectric media

having permittivities e1 and e2.

Supposing dS is the area of each plane and which is also the area of the

boundary enclosed by the cylinder.

Now the charge enclosed by cylinder will be 𝝆= ds

For finding the boundary conditions for D, apply Gauss’s law

Here cylinder will behave like Gaussian surface.

= dsdsD

dsDdsDdsDcurvedlowerupper

... ++=


D1

A B

D C

D2

q2

q1

Δs

D2n

D1n

e2

e1

2

1

Let h⟶ 0 then the curved path vanishes

=+slowerupper

dsdsDdsD ..


Documents

Boundary Conditions Lecture 141 1 1 1 and 1 me me 3/7/2013 v = Dr. Aparna Tripathi v = •Putting the value of v1 and v2 in eq1 2 sin sin 1 1 2 me me q q = t i •For perfect dielectric