This preview shows pages 1–3. Sign up to view the full content.
ECE 222A Final Exam
Dec 8, 2000
Name
Instructions.
This a is three hour, closed book, examination. Calculators are
permitted, but not needed.
1. (10 points) An exotic material has the properties
μ
=
∞
,
σ
c
= 0, and
²
= 0. What boundary conditions does such a surface impose on
~
E
,
~
H
,
~
B
, and
~
D
in an adjacent free space region at nonzero frequencies? Hint:
Do not allow ﬁelds with inﬁnite magnitudes. Also, do not allow currents
or charges other than those owing to conduction eﬀects.
Solution
: First, we know that since
μ
=
∞
, it follows that
~
H
= 0
inside
the material. We can’t say anything about
~
B
because
μ
~
H
=
∞ ·
0
is
indeterminant. We also have
~
D
= 0
inside the material, a consequence
of
²
= 0
. The electric ﬁeld inside the material is unrestricted. Finally,
there are no currents (or conduction electrons) since
σ
c
= 0
. The four
boundary conditions (valid for all materials) are
ˆ
n
·
h
~
D
2

~
D
1
i
= 0
ˆ
n
·
h
~
B
2

~
B
1
i
= 0
ˆ
n
×
h
~
E
2

~
E
1
i
= 0
ˆ
n
×
h
~
H
2

~
H
1
i
= 0
where region “1” refers to the exotic material and region“2” is the vacuum.
If we make use of the speciﬁc material properties of region “2,” we can
reﬁne the boundary conditions to
ˆ
n
·
~
D
2
= ˆ
n
·
~
E
2
= 0
ˆ
n
·
h
~
B
2

~
B
1
i
= 0
ˆ
n
×
h
~
E
2

~
E
1
i
= 0
ˆ
n
×
~
H
2
= ˆ
n
×
~
B
2
= 0
2. (15 points) A beam of light is propagating inside glass with an index of
refraction
n
i
=
p
²
g
/²
0
= 2, where
²
g
and
μ
g
are the permittivity and
permeability of the glass, respectively, and where we assume that
μ
g
=
μ
0
.
The beam is incident upon a planar interface with a vacuum (
n
0
= 1) at
an angle
θ
that
exceeds
the critical angle of total reﬂection
θ
c
(see Figure
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document2). The resulting evanescent wave in the vacuum decays as exp[

z/
Δ].
What is the smallest possible value for the penetration depth Δ? Your
answer should express Δ as a function of the wavenumber
β
g
≡
p
ω
2
²
g
μ
0
Solution
: For the incident wave in the glass, we know that
β
2
g
=
ω
2
²
g
μ
0
=
β
2
x
+
β
2
z
where
β
x
=
β
g
sin
θ
i
, and
β
z
=
β
g
cos
θ
i
, where the direction normal to
the surface is z. We assume that the incidence angle exceeds the critical
angle, so it follows that in the vacuum we have
β
2
0
=
β
2
x

Δ

2
where we have used Snell’s Law in equating
This is the end of the preview. Sign up
to
access the rest of the document.