30
2 Electrodynamics at Interface
ˆ
e
β
P · E
β
=
1
K
qE
β
x + pE
β
z
=
2πi
K
qE
β
x + p
p
q
E
β
x
=
2πi
Kq
(q
2
+ p
2 )
−qP
S
x − pP
S
z
=
2πK 2
iq
ˆ
e
β
P · P
S
.
S components Eqs. (2.28) and (2.31) derive
E
α
y = E
β
y =
2πK 2
iq
P
S
y .
(2.36)
E α
y and E
β
y in Eq. (2.36) satisfy the condition of Eq. (2.16) in the S polarization;
ˆ
e
α
S · E
α
= E
α
y =
2πK 2
iq
P
S
y =
2πK 2
iq
ˆ
e
α
S · P
S
,
ˆ
e
β
S · E
β
= E
β
y =
2πK 2
iq
P
S
y =
2πK 2
iq
ˆ
e
β
S · P
S
.
2.4.4 Electric Field and Interfacial Polarization
[Problem 2.4] Derive Eq. (2.19) in the general dielectric constants ε α , ε β , ε on the
basis of the boundary conditions (2.7), (2.8), (2.9), and (2.10).
Let us generalize the solution of Problem 2.3 without resort to the previous
assumption ε α (() = ε β (() = ε (() = 1. The following derivation is rather similar
to that in Problem 2.3, and note the differences from the previous derivation.
For arbitrary values of ε α ((), ε β ((), ε ((), the wavevectors k
α ((), k
β (() for
the reflected and transmitted lights, respectively, become
k
α
=
⎛
⎝
p
0
q α
⎞
⎠ , k
β
=
⎛
⎝
p
0
−q β
⎞
⎠ ,
where
p = k
α
x (() = k
β
x (() and
q α =
ε α K 2 − p 2
q β =
ε β K 2 − p 2
(2.37)
The unit vectors for S and P polarizations are expressed by
ˆ
e
α
P =
1
√
ε α K
⎛
⎝
−q α
0
p
⎞
⎠ , ˆ
e
α
S =
⎛
⎝
0
1
0
⎞
⎠ , ˆ
e
β
P =
1
√
ε β K
⎛
⎝
q β
0
p
⎞
⎠ , ˆ
e
β
S =
⎛
⎝
0
1
0
⎞
⎠