boundary conditions lecture 141 1 1 1 and 1 me me 3/7/2013 v = dr. aparna tripathi v = •putting...
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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
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
3/7/2013 Dr. Aparna Tripathi
•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
3/7/2013 Dr. Aparna Tripathi
•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.
3/7/2013 Dr. Aparna Tripathi
•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
3/7/2013 Dr. Aparna Tripathi
•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=
3/7/2013 Dr. Aparna Tripathi
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.
3/7/2013 Dr. Aparna Tripathi
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
3/7/2013 Dr. Aparna Tripathi
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==
3/7/2013 Dr. Aparna Tripathi
ti
ti
tiit
tiit
R
R
qqqq
qqqq
+
=
+
=
sin
sin
cossincossin
cossincossin
ti
it
tiit
it
T
T
qqqq
+=
+=
sin
cossin2
cossincossin
cossin2
3/7/2013 Dr. Aparna Tripathi
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
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==
3/7/2013 Dr. Aparna Tripathi
)tan(
)tan(
2sin2sin
2sin2sin
cossincossin
cossincossin
ti
ti
ti
ti
iitt
ttii
R
R
qqqq
qqqq
+
=
+
=
+
=
itti
ii
ti
ii
iiit
ii
T
T
tT
qqqq
qqqq
+=
+=
+=
cossin
cossin2
2sin2sin
cossin2
cossincossin
cossin2
3/7/2013 Dr. Aparna Tripathi
3/7/2013 Dr. Aparna Tripathi
3/7/2013 Dr. Aparna Tripathi
Hη E =
3/7/2013 Dr. Aparna Tripathi
3/7/2013 Dr. Aparna Tripathi
…..(A)
…….(B) 3/7/2013 Dr. Aparna Tripathi
reflection coefficient
transmission coefficient
3/7/2013 Dr. Aparna Tripathi
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
3/7/2013 Dr. Aparna Tripathi
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
3/7/2013 Dr. Aparna Tripathi
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
... ++=
3/7/2013 Dr. Aparna Tripathi
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 ..
3/7/2013 Dr. Aparna Tripathi