2
JACOB LEWIS BOURJAILY
Problem 7.16
Let us consider plane waves propagating in a homogenous, nonpermeable but anisotropic dielectric.
The dielectric is characterized by the tensor ²ij . We will assume that the coordinate axes have been
chosen so that Di = ²i Ei where ²i are the eigenvalues of the tensor.
a) We are to show that plane waves with frequency ω and wave vector k satisfy
k × (k × E) + µ0 ω 2 D = 0.
This is relatively obvious enough. We know from our general studies of electromagnetic waves that in this principle axis system, E ∝ ²i ei(k·x−ωt) . Using simple vector
identities we have
k × (k × E) = (k · E)k − (k · k)E.
But k · E—the wave vector is always orthogonal to the field. Hence the first term
above vanishes.
Also notice that k · k = µ0 ω 2 ²i for the ith component of E. Therefore,
k × (k × E) = −µ0 ω 2 ²i Ei = −µ0 ω 2 D.
Hence,
∴ k × (k × E) + µ0 ω 2 D = 0.
‘
’
óπ²ρ
²́δ²ι
δ²ιξαι
c) We are to show that Da · Db = 0 where Da , Db are the displacements associated with the two
modes of propagation.
This is quite obvious: the two modes of propagation are naturally orthogonal—we choose
orthogonal bases—and so it is natural that Da · Db = 0.
Problem 7.2
Let us consider a plane wave that is normally incident on two parallel layers with refraction indices
of each layer being n1 , n2 . We are to determine the thickness d of the second layer such that no wave is
reflected from the system at the frequency ω in terms of the refractive indices.
Let us work in the system where the z-axis is normal to the planes—and hence parallel to the
direction of motion of the wave—with the origin at the surface of the first surface. From rather general
considerations the fields E1 , E2 , and E3 for the first medium, second medium and the air on the right
are given by
E1 = aeik1 z + be−ik1 z ,
E2 = αeik2 z + βe−ik2 z ,
E3 = ηeikz ,
where k1 = n1 c/ω, k2 = n2 c/ω and κ = c/ω and a, b, α, β, and η are constants.
From continuity requirements of the electric and magnetic field at the interfaces, we have the constraints
a+b=α+β
n2
(α − β)
a−b=
n1
Notice that the second two imply that
µ
1
α=
1+
2
µ
1
β=
1−
2
αeik2 d + βeik2 d = ηeikd ;
1 ikd
αeik2 d − βeik2 d =
ηe .
n2
¶
1
ηeid(k−k2 ) ;
n2
¶
1
ηeid(k+k2 ) .
n2
Now, the case in which there is no reflection is that where b = 0. We can solve for b using the above
system trivially and see that
µ
¶
µ
¶
n2
n2
2b = 1 −
α+ 1+
β.
n1
n1
PHYSICS 505: CLASSICAL ELECTRODYNAMICS HOMEWORK 12
3
Substituting our expressions for α and β and setting the above to zero, we have
µ
¶
µ
¶
n2
n2
0= 1−
α+ 1+
β,
n1
n1
µ
¶µ
¶
µ
¶µ
¶
n2
1
n2
1
id(k−k2 )
= 1−
1+
e
+ 1+
1−
eid(k+k2 ) ,
n1
n2
n1
n2
µ
¶µ
¶
µ
¶µ
¶
n2
1
n2
1
−idk2
= 1−
1+
e
+ 1+
1−
eidk2 ,
n1
n2
n1
n2
¶
¶
µ
µ
n2
1
n2
1
1
1
idk2
−
−
e
+
−
eidk2 ,
= 1+
+ 1−
n2
n1
n1
n2
n1
n1
n1 − 1
n2 − n1
=
cos (dk2 ) + 2
i sin (dk2 ) = 0.
n2
n1 n2
For this to vanish, each of the two contributions must vanish separately. Because n1 − 1 ≥ 0 and sin(dk2 )
can be nonzero in general, this implies that cos (dk2 ) = 0 and that n22 − n1 = 0. The cosine vanishes if
and only if d = π(i+1/2)
for some i ∈ Z. Therefore, we have that
k2
∴ n2 =
√
n1
and
d=
π(i + 1/2)c
| i ∈ Z.
n2 ω
‘
’
óπ²ρ
²́δ²ι
δ²ιξαι
Problem 7.3
Let us consider two parallel semi-infinite slabs of uniform, isotropic, nonpermeable, lossless dielectrics
with index n that are separated by an air gap (n = 1) of width d, and a linearly polarized electromagnetic
plane wave incident on the slabs with an angle ϕ to the normal of the air gap with its polarization
perpendicular to the plane of incidence.
a) We are to calculate the transmission and reflection from the gap.
We will assess this situation with the same notation as problem 7.2 above where we have
the various electric fields given by
E1 = aeik1 x + be−ik1 x ,
E2 = αeikx + βe−ikx ,
E3 = ηeik1 x ,
where k1 = nc/ω and κ = c/ω and a, b, α, β, and η are constants. Notice that the
transmission fraction is given by T = |η|2 /|a|2 , the ratio of the final wave magnitude
to the initial, incoming wave magnitude. Also, the reflection fraction is obviously
R = 1 − T . Therefore it will be very useful for us to solve for a in terms of η (or
vice versa).
The analysis is similar to that of problem 7.2 except that we must now allow for waves
which are nor normal to the surface. The conditions are spelled out rather explicitly
in Jackson’s section 7.3 and it is obvious that our continuity requirements become
the following system of constraints,
αeikλ + βeikλ = ηeik1 λ ;
a+b=α+β
a−b=
cos ϕ0
(α − β)
n cos ϕ
αeikλ − βeikλ =
n cos ϕ ik1 λ
ηe
;
cos ϕ0
where λ ≡ d/ cos ϕ0 , the distance travelled by the wave in going between the two
slabs. For the sake of simplicity, set ξ ≡ n cos ϕ/ cos ϕ0 . Solving for α, β in terms of
η, we see that
1
(1 + ξ) ηeiλ(k1 −k) ;
2
1
β = (1 − ξ) ηeid(k1 +k) .
2
α=