28
2 Electrodynamics at Interface
2.4.3 Boundary Condition at Interface (3)
[Problem 2.3] Derive Eq. (2.16) from the boundary conditions (2.7), (2.8), (2.9),
and (2.10).
Here we have assumed ε α (() = ε β (() = ε (() = 1. The wave vectors of sum
frequency of reflection k
α (() and transmission k
β (() in Eq. (2.13) are
k
α (() =
⎛
⎝
p
0
q
⎞
⎠ and k
β (() =
⎛
⎝
p
0
−q
⎞
⎠ ,
respectively, where p = k α
x (() = k
β
x ((), and q = q α (() = q β (() =
K 2 − p 2 . In the following we only treat the sum frequency components of
field and polarization, and thus omit “(()” from the notations of k
α ((), E α ((),
P S ((), etc.
Each emitted light with the wave vector k
i (i = α, β) may have two polarizations, namely P and S, which are represented with unit vectors of the electric fields,
ˆ
e
i
P and ˆ
e
i
S , respectively. ˆ
e
i
P , ˆ
e
i
S , and k
i are orthogonal each other, and ˆ
e
i
S is parallel
to the y axis in Fig. 2.1. Accordingly, these polarization vectors are
ˆ
e
α
P =
1
K
⎛
⎝
−q
0
p
⎞
⎠ , ˆ
e
α
S =
⎛
⎝
0
1
0
⎞
⎠ , ˆ
e
β
P =
1
K
⎛
⎝
q
0
p
⎞
⎠ , ˆ
e
β
S =
⎛
⎝
0
1
0
⎞
⎠ .
Then the amplitudes of electric and magnetic fields, E i and H i = (c//) k
i
×E i ,
in Eq. (2.14) are represented by
E
i
=
⎛
⎝
E i
x
E i
y
E i
z
⎞
⎠
(i = α, β),
H
α
=
c
⎛
⎜
⎝
−qE α
y
qE α
x − pE α
z
pE α
y
⎞
⎟
⎠ , H
β
=
c
⎛
⎜
⎝
qE
β
y
−qE
β
x − pE
β
z
pE
β
y
⎞
⎟
⎠ .
The boundary conditions for the electromagnetic fields at z = 0 are derived from
Eqs. (2.7), (2.8), (2.9), and (2.10), and expressed using the electric field amplitudes
in the following five equations, (2.28), (2.29), (2.30), (2.31), and (2.32).
• From Eq.(2.7), condition for B z = μH z :
B z :
μ
c
pE
β
y − pE
α
y
= 0
Therefore, E
β
y − E
α
y = 0
(2.28)