32
2 Electrodynamics at Interface
P components Equation (2.42) is modified with Eqs. (2.39) and (2.40) so as to
eliminate E α ,
q
β E
β
x + q
α
E
β
x +
4πip
ε P
S
z
+ pE
β
z − p
1
ε α
ε
β E
β
z + 4πipP
S
x
= −4πiK
2 P
S
x .
Therefore, using Eq. (2.37),
(q
α
+ q
β )E
β
x + p
1 −
ε β
ε α
E
β
z = −4πi
K
2
−
p 2
ε α
P
S
x − 4πip
q α
ε P
S
z
= −
4πi(q α )
2
ε α
P
S
x +
ε α p
ε q α P
S
z
.
Using the relation k
β
· E β = pE
β
x − q β E
β
z = 0, we insert E
β
x = q β E
β
z /p into the
above equation and obtains
(q
α
+ q
β )
q β
p
+ p
1 −
ε β
ε α
E
β
z = −
4πi(q α )
2
ε α
P
S
x +
ε α p
ε q α P
S
z
.
Therefore,
E
β
z = −
4πipq α
ε α q β + ε β q α
P
S
x +
ε α p
ε q α P
S
z
,
(2.43)
E
β
x =
q β
p
E
β
z = −
4πiq α q β
ε α q β + ε β q α
P
S
x +
ε α p
ε q α P
S
z
.
(2.44)
Consequently, using Eqs. (2.39) and (2.40), we also obtain
E
α
z =
4πipq β
ε α q β + ε β q α
P
S
x −
ε β p
ε q β P
S
z
,
(2.45)
E
α
x = −
4πiq α q β
ε α q β + ε β q α
P
S
x −
ε β p
ε q β P
S
z
.
(2.46)
It is confirmed that Eqs. (2.43) and (2.44) satisfy the condition of Eq. (2.19) in the P
polarization as follows.
ˆ
e
β
P · E
β
=
1
√
ε β K
q
β E
β
x +pE
β
z
=
1
√
ε β K
q
β q β
p
E
β
z +pE
β
z
=
√
ε β K
p
E
β
z
= −
ε β K
4πiq α
ε α q β + ε β q α
P
S
x +
ε α p
ε q α P
S
z
,