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6.013/ESD.013J Electromagnetics and Applications, Fall 2005
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6.013/ESD.013J — Electromagnetics and Applications
Fall 2005
Problem Set 6 - Solutions
Prof. Markus Zahn
MIT OpenCourseWare
Problem 6.1
A
∂(Jf · Jf )/∂t
∂Jf /∂t

= 2Jf · ∂Jf /∂t 
= εωp2 E

∂
(Jf · Jf ) = 2Jf ·
εωp2 E
∂t
=⇒
Noting that E · Jf = ∂wk /∂t, we have that
∂
∂wk
∂wk
∂
⇐⇒
(Jf · Jf ) = 2εωp2
=
∂t
∂t
∂t
∂t
Jf · Jf
2εωp2
=⇒ wk =
Jf · Jf
mJf · Jf
q2 n
=
since ωp2 =
2
2
2q n
mε
2εωp
B
Jf = ρv = nqv ⇐⇒ v =
Jf
nq
C
wk =
mJf · Jf
m(nq |v|)2
1
=
= mn|v|2
2q 2 n
2q 2 n
2
Therefore wk is the kinetic energy of the plasma charges.
D
ωp2
∂
(Jf · Jf ) = εωp2 E =⇒ jωĴf = εωp2 Ê ⇐⇒ Ĵf = −j εÊ.
∂t
ω
Maxwell’s equations in complex form are:

ˆ
∇×E



ˆ
∇×H

∇ · εÊ


ˆ
∇ · µH
(1)

ˆ
∇×E
ˆ

= −jωµH
(2)



2
ˆf + jωεÊ (3) (1)
ˆ
ω
=J
∇×H
ˆ = −j p εE
ˆ + jωεE
ˆ ⇐⇒
⇐⇒ ∇ × H
 ∇ · εE
ω
=0
ˆ



=0
ˆ
∇ · µH
1
ˆ
= −jωµ
H
= jωε 1 −
=0
= 0.
ωp2
ω2
ˆ
E
Problem Set 6
6.013, Fall 2005
E
Using (2) and (3) we have
1
ˆ ∗ ) = 1 (H
ˆ∗ ·∇× E
ˆ − Ê · ∇ × H
ˆ ∗ ) = 1 [H
ˆ ∗ · (−jωµH)
ˆ − Ê · (Ĵf∗ − jωεÊ∗ )]
∇ (Ê × H
2
2
2
1 ˆ 2 1 ˆ2
1ˆ ∗
= −2jω
µ|H| − ε|E| − E
· Ĵf
4
4
2
1
⇐⇒ ∇ · Ŝ + 2jω�wEM � = − Ê · Ĵf∗
2
where
Ŝ =
1
ˆ ∗ , and �wEM � = 1 µ|H|
ˆ 2 − 1 ε|Ê|2
Ê × H
2
4
4
(4)
But from (1):
1
1
Ê · Ĵf∗ = Ê ·
2
2
ωp2
j Ê∗
ω
!
= 2jω
1 ωp2 2
ε |Ê|
4 ω2
!
,
so
∇ · Ŝ + 2jω(�wEM � + �wk �) = 0,
where
�wk � =
1 ωp2 2
ε |Ê| .
4 ω2
(5)
F
From (4) and (5):
1 ˆ 2 1 ˆ 2 1 ωp2 ˆ 2
µ|H| − ε|E| + ε 2 |E|
4
4
4 ω
1 ˆ 2 1
ˆ 2
,
⇐⇒ �wEM � + �wk � = µ|H| − ε(ω)|E|
4
4
�wEM � + �wk � =
where
ωp2
ε(ω) = ε 1 − 2
ω
!
which is the result we found in Problem 5.3b.
Problem 6.2
A
Since neither ε, µ nor the excitation depend on x, y, we can set
x̂
∂
∇ × E = − µH ⇐⇒ 0
∂t
Ex
ŷ
0
0
∂
∂x
=
∂
∂y
= 0 and then:
ẑ ∂
∂Ex
∂Hy
∂ = − µ(ŷHy ) ⇐⇒
= −µ
∂z ∂t
∂z
∂t
0
2
(1)
Problem Set 6
6.013, Fall 2005
x̂ ŷ
∂
∇ × H = εE ⇐⇒ 0 0
∂t
0 Hy
∂
∂t (1)
=⇒
∂
∂z (2)
=⇒
∂ 2 Ex
∂t∂z
∂
∂z
1 ∂Hy
ε(z) ∂z
ẑ ∂
∂Hy
∂Ex
∂ ∂z = ∂t ε(x̂Ex ) ⇐⇒ ∂z = −ε ∂t
0 = −µ(z)
2
Ex
= − ∂∂z∂t
B
∂ 2 Hy
∂t2
(2)



∂
1 ∂Hy
∂ 2 Hy
=⇒ ε(z)
− ε(z)µ(z)
=0

∂z ε(z) ∂z
∂t2

Using now ε(z) = εa e+αz and µ(z) = µa e−αz we get:
∂ 2 Hy
d
1
∂Hy
∂Hy
(3) =⇒
+
ε(z)
− ε(z)µ(z)
=0
∂z 2
dz ε(z)
∂z
∂t
∂ 2 Hy
ε′ (z) ∂Hy
∂ 2 Hy
⇐⇒
−
− ε(z)µ(z)
=0
2
∂z
ε(z) ∂z
∂t2
∂ 2 Hy
εa αe+αz ∂Hy
∂ 2 Hy
− εa e+αz µa e−αz
⇐⇒
−
2
+αz
∂z
∂t2
∂z
εa e
2
2
∂ Hy
∂Hy
∂ Hy
⇐⇒
−α
− εa µa
=0
(4)
2
∂z
∂z
∂t2
Therefore β = α and γ = εa µa .
C
h
i
Trying now the solution H = Re Ĥy ej(ωt+κz) ŷ we get
(4) =⇒ (+κ)2 Ĥy − α(+κ)Ĥy − εa µa (jω)2 Ĥy = 0
⇐⇒ (κ2 − ακ + ω 2 εa µa )Ĥy = 0
p
p
α + α2 − 4ω 2 εa µa
α − α2 − 4ω 2 εa µa
=⇒ κ1 =
, κ2 =
2
2
(5)
D
Boundary Conditions:
�0 ˆ
�
ˆ y (z = 0+ ) = K0
ˆ = 0+ ) − H(z�
ˆ� =�
• @ z = 0 : ẑ × [H(z
0− )] = K
⇐⇒ −H
� 0
ˆ
�
ˆ y (z = d− ) = 0
ˆ� =�
• @ z = d : ẑ × [H(z�
d+ ) − H(z
= d− )] = 0 ⇐⇒ H
E
The general solution is of the form
Ĥy = Ĥ1 eκ1 z + Ĥ2 eκ2 z ,
so applying the boundary conditions we get
⇐⇒ Ĥ1 + Ĥ2 = −K0
(6) =⇒
−(Ĥ1 + Ĥ2 ) = K0
(7) =⇒
Ĥ1 eκ1 d + Ĥ2 eκ2 d = 0
⇐⇒ Ĥ2 = −Ĥ1 e(κ1 −κ2 )d
3
(6)
(7)
(3)
Problem Set 6
6.013, Fall 2005
Together, these give
Ĥ1 [1 − e(κ1 −κ2 )d ] = −K0 ⇐⇒ Ĥ1 =
and
−K0
1 − e(κ1 −κ2 )d
−K0
K0
K0
e(κ1 −κ2 )d = + −(κ −κ )d
=
,
(κ
−κ
)d
1
2
1
2
1−e
e
−1
1 − e−(κ1 −κ2 )d
p
where κ1 − κ2 = α2 − 4ω 2 εa µa ≡ Δ from (5).
Ĥ2 = +
F
Therefore
Ĥy = −
and from (2)
e κ1 z
e κ2 z
+
K0 ,
1 − e(κ1 −κ2 )d
1 − e−(κ1 −κ2 )d
∂ Ĥy
= −jωε(z)Êx
∂z
=⇒ Eˆx = −
−1
jωεa e+αz
κ 1 e κ1 z
κ 2 e κ2 z
K0 κ1 eΔz/2
κ2 e−Δz/2 −αz/2
+
K0 = +
+
e
jωεa 1 − eΔd
1 − e−Δd
1 − e(κ1 −κ2 )d
1 − e−(κ1 −κ2 )d
Problem 6.3
A
|k|2 = ω 2 ε0 µ0 =⇒ ky2 + kz2 = ω 2 ε0 µ0 ⇐⇒ kz = ±
q
ω 2 ε0 µ0 − ky2
(1)
B
ˆ = jωε0 E
ˆ ⇐⇒ ŷ
∇×H
=⇒ Ê =
∂Ĥx
∂Ĥx
ˆ
− ẑ
= jωε0 E
∂z
∂y
−Ĥ1
Ĥ1
[ŷ(−jkz ) − ˆ
z(−jky )] e−j(ky y+kz z) =
(ŷkz − ẑky )e−j(ky y+kz z) , z > 0
jωε0
ωε0
and
Ê =
Ĥ2
Ĥ2
[ŷ(jkz ) − ˆ
z(−jky )] e−j(ky y−kz z) =
(ˆ
ykz + ẑky )e−j(ky y−kz z) , z < 0.
jωε0
ωε0
C
Boundary conditions at z = 0 :
• Êy (z = 0+ ) = Êy (z = 0− ) ⇐⇒
Ĥ2
Ĥ1
ˆ1
kz e−jky y = −
kz e−jky y =⇒ Ĥ2 = −H
ωε0
ωε0
• ε0 Ê(z = 0+ )− ε0 Ê(z = 0− ) = σ̂0 e−jky y ⇐⇒ ε0
Ĥ2
−Ĥ1
ωσˆ0
ky − ε0
(−ky ) = σ̂0 =⇒ Ĥ2 − Ĥ1 =
ωε0
ωε0
ky
From (2) and (3), we have that
Ĥ2 = −Ĥ1 =
(2)
ω σ̂0
.
2ky
4
(3)
Problem Set 6
6.013, Fall 2005
D
For the fields to be evanescent in z, kz must be imaginary, so from (1)
ky
= cky .
ω 2 ε0 µ0 − ky2 < 0 =⇒ ω < √
ε0 µ0
E
h
i
ωσ̂0 −jky y
ˆ 1 )e−jky y (5)
ˆ = ẑ × H(z
ˆ = 0+ ) − H(z
ˆ = 0− ) = ŷ(Ĥ2 − H
K
= ŷ
e
.
ky
Problem 6.4
A
The TM reflection coefficient is
ηi cos θi − ηt cos θt
R=
,
ηi cos θi + ηt cos θt
so for no reflection:
R = 0 ⇐⇒ ηi cos θi = ηt cos θt
Since µ1 = µ2 = µ, we have
r
r
µ
µ
ηi =
, and ηt =
.
ε1
ε2
Together, these imply
√
√
ε2 cos θi = ε1 cos θt .
(1)
From Snell’s law:
ni sin θi = nt sin θt .
√
√
Since µ1 = µ2 = µ, ni = ε1 µ and nt = ε2 µ,
√
√
(2)
ε1 sin θi = ε2 sin θt .
From (1) · (2),
√
√
ε1 ε2 sin θi cos θi = ε1 ε2 sin θt cos θt =⇒ sin 2θi = sin 2θt
π
θi = θt (cannot satisfy (1), (2))
=⇒
=⇒ θi + θt =
2θi = π − 2θt
2
(3)
B
From (2),
√
π
√
√
− θi = ε2 cos θi =⇒ tan θi = tan θB =
ε1 sin θi = ε2 sin
2
r
ε2
.
ε1
(4)
Note: Physical explanation for zero reflection: Physically, the transmitted wave excites dipoles inside the
transmission medium into oscillation. These dipoles cannot radiate along their axis of oscillation, which lies,
due to (3), in the direction of the reflected beam. Therefore the reflected beam cannot be excited and thus
total transmission occurs.
5
Problem Set 6
6.013, Fall 2005
ε2,μ
ε1,μ
radiating dipoles
θt
θi
Figure 1: Dipole radiation. (Image by MIT OpenCourseWare.)
C
From (2) we see that for total internal reflection we must have:
sin θt ≥ 1 =⇒
r
ε1
sin θi ≥ 1 =⇒ sin θc =
ε2
r
ε2
.
ε1
(5)
For the critical angle to exist:
√
√
sin θc ≤ 1 =⇒ ε2 ≤ ε1 ⇐⇒ n2 ≤ n1 .
D
(d)
θ
θ1 =θB
θt
n2
n1
Figure 2: Refraction. (Image by MIT OpenCourseWare.)
From the small triangle:
π
θ+2
− θt = π.
2
But from (3):
θB =
π
− θt .
2
Hence,
θ + 2θB = π =⇒ θ = π − 2θB .
(6)
(6)
For n1 = 1 and n2 = 1.45, we get tan θB = 1.45 =⇒ θB ≈ 0.31π = 55.8◦ =⇒ θ ≈ π − 2 · 0.31π = 0.38π =
68.4◦ and then θt = π/2 − θB = 0.19π = 34.2◦.
6
Problem Set 6
6.013, Fall 2005
E
Clearly for TM light, the transmitted power equals the incident one. For TE light, the reflection coefficient
is (using µ1 = µ2 = µ)
R=
ηt cos θi − ηi cos θt
ni cos θi − nt cos θt
cos 55.8◦ − 1.45 cos 34.2◦
=
=
= −0.3618
ηt cos θi + ηi cos θt
ni cos θi + nt cos θt
cos 55.8◦ + 1.45 cos 34.2◦
at the input surface and −R = 0.3618 at the output. Therefore the reflected power at each surface is
|R|2 = 0.131
and thus the transmitted power is
1 − |R|2 = 0.869.
In total, the transmitted power from both surfaces
St /Si = (1 − |R|2 )2 = 0.755 = 75.5%.
7