2.4 Solutions to Problems
29
• From Eq.(2.8), condition for D z = E z :
D z :
E
β
z − E
α
z = −4πipP
S
x
(2.29)
• From Eq.(2.9), conditions for E x and E y :
E x :
E
β
x − E
α
x = −4πipP
S
z
(2.30)
E y :
E
β
y − E
α
y = 0 (same as Eq. (2.28))
• From Eq.(2.10), conditions for H x and H y :
H x :
c
qE
β
y
−
−qE
α
y
=
4π
c
(−ii)P
S
y
Therefore, q
E
β
y + E
α
y
= −4πiK
2 P
S
y
(2.31)
H y :
c
−qE
β
x − pE
β
z
−
qE
α
x − pE
α
z
= −
4π
c
(−ii)P
S
x
Therefore, q
E
β
x + E
α
x
+ p
E
β
z − E
α
z
= −4πiK
2 P
S
x
(2.32)
(Note the right-hand side of Eq. (2.32) includes an extra minus sign due to the
property of vector product, x × z = −z × x.) Using Eqs. (2.28), (2.29), (2.30),
(2.31), and (2.32), both P and S components of E α , E β in Eq. (2.14) are obtained
as follows.
P components Eqs. (2.29), (2.30), and (2.32) derive
E
β
x = 2πi
−qP
S
x − pP
S
z
.
(2.33)
This equation, along with the relation k
β
· E β = pE
β
x − qE
β
z = 0, leads to
E
β
z =
p
q
E
β
x = 2πi
p
q
−qP
S
x − pP
S
z
.
(2.34)
Therefore, E α
x and E α
z are also determined from Eqs. (2.29), (2.30).
E
α
x = 2πi
−qP
S
x + pP
S
z
, E
α
z = −2πi
p
q
−qP
S
x + pP
S
z
.
(2.35)
It is readily confirmed that Eqs. (2.33), (2.34), and (2.35) satisfy the condition of
Eq. (2.16) for the P polarization;
ˆ
e
α
P · E
α
=
1
K
−qE
α
x +pE
α
z
=
2πi
K
−q(−qP
S
x +pP
S
z ) − p
p
q
(−qP
S
x + pP
S
z )
= −
2πi
Kq
(q
2
+ p
2 )
−qP
S
x + pP
S
z
=
2πK 2
iq
ˆ
e
α
P · P
S
,