Page 297 - Electromagnetics

t
that if A = 0 then τ = 0 and from (4.293) we have θ t = π/2. This defines the critical
angle, which from (4.295) is
2
β 2 −1 µ 2 2
−1
θ c = sin 2 = sin .
β µ 1 1
1
Therefore

0, θ i <θ c ,
t
γ =
π/2,θ i >θ c .
Using these we can write down the transmitted wave vector from (4.291):
√

ˆ xβ 1 sin θ i + ˆ z |A|, θ i <θ c ,
t t t
k = k + jk = √ (4.296)
ˆ xβ 1 sin θ i − j ˆ z |A|,θ i >θ c .
By(4.293) we have
β 1 sin θ i β 1 sin θ i
=
sin θ t =
2 2 2 2 2 β 2
β sin θ i + β − β sin θ i
1 2 1
or
β 2 sin θ t = β 1 sin θ i . (4.297)
This is known as Snell’s law of refraction. With this we can write for θ i <θ c
2 2 2 2 2
A = β − β sin θ i = β cos θ t .
2 1 2
Using this and substituting β 2 sin θ t for β 1 sin θ i , we mayrewrite (4.296) for θ i <θ c as
t
t
t
k = k + jk = ˆ xβ 2 sin θ t + ˆ zβ 2 cos θ t . (4.298)
Hence the transmitted plane wave is uniform with k = 0. When θ i >θ c we have from
t
(4.296)

2
2
2
t
t
k = ˆ xβ 1 sin θ i , k =−ˆ z β sin θ i − β .
1 2
t
Since k and k t are not collinear, the plane wave is nonuniform. Let us examine the
cases θ i <θ c and θ i >θ c in greater detail.
Case 1: θ i <θ c . By(4.289)–(4.290) and (4.298) the wave vectors are
i
k = ˆ xβ 1 sin θ i + ˆ zβ 1 cos θ i ,
r
k = ˆ xβ 1 sin θ i − ˆ zβ 1 cos θ i ,
t
k = ˆ xβ 2 sin θ t + ˆ zβ 2 cos θ t ,
and the wave impedances are
η 1 η 2
Z 1⊥ = , Z 2⊥ = ,
cos θ i cos θ t
Z 1 = η 1 cos θ i , Z 2 = η 2 cos θ t .
The reflection coefficients are
˜ η 2 cos θ i − η 1 cos θ t ˜ η 2 cos θ t − η 1 cos θ i
⊥ = , = . (4.299)
η 2 cos θ i + η 1 cos θ t η 2 cos θ t + η 1 cos θ i

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