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Physics 210A PS 8 Solutions
Anthony Karmis
March 13, 2009
Jackson 7.
Part (a)
We will need to match the tangential components of E and H at the boundaries.
Let 1, 2, and 3 denote the regions of the first slab, the gap of air, and the
second slab, respectively. Then, we have the following equations at the interface
between slab one and the air:
E 1 i + E 1 r = E 2 t + R 2 r
n cos i(Ei 1 − E 1 r ) = cos r(E 2 t − E 2 r )
At the interface between the gap and the second slab, we have:
E 2 te
iΦ
−iΦ = E 3 t
n cos r(E 2 te
iΦ
−iΦ ) = n cos iE 3 t
Solving the equations above yields:
E 3 t
E 1 i
2Γ cos Φ − i(1 + Γ 2 ) sin Φ
E 1 r
E 1 i
i(1 − Γ 2 ) sin Φ
2Γ cos Φ − i(1 + Γ 2 ) sin Φ
With:
n cos i √
1 − n 2 sin
2 i
ωd
c
1 − n 2 sin
2 i
Part (b)
Since i is now greater than the critical angle, both Γ and Φ will be purely
imaginary. Therefore, let’s define:
Γ = iγ, Φ = iφ
Then the ratio becomes:
E 3 t
E 1 i
2 iγ
2 iγ cosh φ + (1 − γ 2 ) sinh Φ
Or:
T =
E 3 t
E 1 i
2
=
4 γ 2
4 γ 2
2 Φ
For n = 1.5 and i = 45
◦ , we have:
2 4 6 8 10 12
d Ω
c
T
Jackson 7.
Part (a)
We simply evaluate the integral in Jackson 7.120 with the given imaginary part.
The result is:
Re
ǫ(ω)
ǫ 0
λ
π
log
ω
2 2 −^ ω
2
ω
2 1 −^ ω
2
Part (b)
This integral is a bit more involved:
Re
ǫ(ω)
ǫ 0
π
λγP
0
dω
′ ω ′ 2
((ω 2 0 − ω ′ 2 ) + γ 2 ω ′ 2 )(ω ′ 2 − ω 2 )
Let the integrand be defined as f (ω ′ ). Then, we have:
P
o
dω
′ f (ω
′ ) =
P
−∞
dω
′ f (ω
′ )
= πi
UpperHalf
Res(f ) −
lim ǫ→ 0 ,R→∞
Cǫ,±ω
CR
The integral at infinity vanishes because f ∼ 1 /R
2
. The ǫ-integral also
vanishes because the residues of f at the singularities ω and −ω are opposite.
The singularities in the upper-half plane are:
ω± =
iγ ±
4 ω 2 0 − γ 2
With the definition:
g(ω
′ ) =
ω
′ 2
(ω 2 0 − ω′^2 − iγω′)(ω′^2 − ω^2 )
We can rewrite f as:
f (ω
′ ) = g(ω
′ )
(ω ′ − ω+)(ω ′ − ω−)
Note that if γ
2 = 4ω
2 0 , then the singularity in the upper half is of second
order, thus killing the residues. Assuming the singularities are simple poles, the
residues are simply limω′ (^) →ω ± (ω ′ − ω±)f (ω ′ ). Thus, we only have to evaluate
the function g at the singularities. The final result is:
Re
ǫ(ω)
ǫ 0
λ(ω 2 − ω 2 0
(ω 2 − ω 2 0
2
