2.4 Solutions to Problems
33
2πK 2
iq β e P · P
S
=
2πK 2
iq β
1
√
ε β K
F
β→α
xx
q
β P
S
x + F
β→α
zz
pP
S
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
,
where e P = F β→α · ˆ
e
β
P , and F β→α is given in Eq. (2.18). Therefore, Eq. (2.19) is
proved for the β phase,
ˆ
e
β
P · E
β
=
2πK 2
iq β e P · P
S .
(2.47)
The same argument can be applied to E α in Eqs. (2.45) and (2.46);
ˆ
e
α
P · E
α
=
1
√
ε α K
−q
α E
α
x +pE
α
z
=
1
√
ε α K
q
α q α
p
E
α
z +pE
α
z
=
√
ε α K
p
E
α
z
=
√
ε α K
4πiq β
ε α q β + ε β q α
P
S
x −
ε β p
ε q β P
S
z
,
2πK 2
iq α e P · P
S
=
2πK 2
iq α
1
√
ε α K
−F
α→β
xx
q
α P
S
x + F
α→β
zz
pP
S
z
=
√
ε α K
4πiq β
ε α q β + ε β q α
P
S
x −
ε β p
ε q β P
S
z
,
where e P = F α→β · ˆ
e
α
P . F α→β is also given in Eq. (2.18). Therefore,
ˆ
e
α
P · E
α
=
2πK 2
iq α e P · P
S .
(2.48)
S components Equations (2.38) and (2.41) derive
E
α
y = E
β
y =
4πK 2
i(q α + q β )
P
S
y
(2.49)
E
β
y in Eq. (2.49) satisfy the condition of Eq. (2.19) in the S polarization.
ˆ
e
β
S · E
β
= E
β
y =
2πK 2
iq β
2q β
q α + q β P
S
y =
2πK 2
iq β F
β→α
yy
P
S
y =
2πK 2
iq β e S · P
S .
(2.50)
The same argument holds for E α
y in Eq. (2.49),
ˆ
e
α
S · E
α
= E
α
y =
2πK 2
iq α
2q α
q α + q β P
S
y =
2πK 2
iq α F
α→β
yy
P
S
y =
2πK 2
iq α e S · P
S .
(2.51)