2.4 Solutions to Problems
31
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
⎛
⎜
⎝
−q α E α
y
q α E α
x − pE α
z
pE α
y
⎞
⎟
⎠ , H
β
=
c
⎛
⎜
⎝
q β E
β
y
−q β E
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 summarized in the following five
equations, (2.38), (2.39), (2.40), (2.41), and (2.42).
• From Eq.(2.7), boundary condition for B z = μH z :
B z :
c
pE
β
y − pE
α
y
= 0
Therefore,
E
β
y − E
α
y = 0
(2.38)
• From Eq.(2.8), condition for D z = εE z :
D z :
ε
β E
β
z − ε
α E
α
z = −4πipP
S
x
(2.39)
• From Eq.(2.9), conditions for E x and E y :
E x :
E
β
x − E
α
x = −
4πip
ε P
S
z
(2.40)
E y :
E
β
y − E
α
y = 0
(same as Eq. (2.38))
• From Eq.(2.10), conditions for H x and H y :
H x :
c
q
β E
β
y
−
−q
α E
α
y
=
4π
c
(−ii)P
S
y
Therefore, q
β E
β
y + q
α E
α
y = −4πiK
2 P
S
y
(2.41)
H y :
c
−q
β E
β
x −pE
β
z
−
q
α E
α
x −pE
α
z
= −
4π
c
(−ii)P
S
x
Therefore, q
β E
β
x + q
α E
α
x + p
E
β
z − E
α
z
= −4πiK
2 P
S
x
(2.42)
(Note that the right-hand side of Eq. (2.42) includes an extra minus sign due to the
property of vector product x × z = −z × x.) Using Eqs. (2.38), (2.39), (2.40),
(2.41), and (2.42), both P and S components of E α , E β in Eq. (2.14) are obtained
as follows.