Page 293 - Electromagnetics

r
E ˜ r r r ˆ xk − ˆ zk r r r
˜ r − j(k x x+k z z)
˜ r
˜ r
H = ˆ y − j(k x x+k z z) , E = z x E e ,
e

η 1 k 1
r 2
r 2
2
˜ t
where (k ) + (k ) = k . Similarly, letting E be the amplitude of the transmitted field
x z 1
we have
t
t ˜ t
−ˆ xk + ˆ zk E ⊥ − j(k x x+k z z)
t
t
t
t
˜ t
− j(k x x+k z z)
˜ t
˜ t
z
x
E = ˆ yE e , H = e ,
⊥ ⊥ ⊥
k 2 η 2
˜ t
E t t ˆ xk − ˆ zk t t t
t
˜ t
˜ t − j(k x x+k z z)
˜ t
e
H = ˆ y − j(k x x+k z z) , E = z x E e ,

η 2 k 2
t 2
t 2
2
where (k ) + (k ) = k .
x z 2
˜ i ˜ r ˜ t
The relationships between the field amplitudes E , E , E , and between the components
r
t
of the reflected and transmitted wave vectors k and k , can be found byapplying the
boundaryconditions. The tangential electric and magnetic fields are continuous across
the interface at z = 0:
˜ t
˜ r
˜ i
ˆ z × (E + E )| z=0 = ˆ z × E | z=0 ,
˜ i
˜ r
˜ t
ˆ z × (H + H )| z=0 = ˆ z × H | z=0 .
Substituting the field expressions, we find that for perpendicular polarization the two
boundaryconditions require
r
t
i
˜ r
˜ t
˜ i
E e − jk x x + E e − jk x x = E e − jk x x , (4.274)
⊥ ⊥ ⊥
r ˜ r
i ˜ i
t ˜ t
k E k E k E
i
r
t
z ⊥ − jk x x z ⊥ − jk x x z ⊥ − jk x x
e + e = e , (4.275)
k 1 η 1 k 1 η 1 k 2 η 2
while for parallel polarization theyrequire
k i k r k t
r
i
t
z ˜ i − jk x x z ˜ r − jk x x z ˜ t − jk x x
E e + E e = E e , (4.276)

k 1 k 1 k 2
˜ t
˜ i
E E ˜ r E
i
r
t
− jk x x − jk x x − jk x x
e + e = e . (4.277)
η 1 η 1 η 2
For the above to hold for all x we must have the exponential terms equal. This requires
t
r
i
k = k = k , (4.278)
x
x
x
i r t i 2 i 2 r 2 r 2
and also establishes a relation between k , k , and k . Since (k ) +(k ) = (k ) +(k ) =
z z z x z x z
r
2
i
k , we must have k =±k . In order to make the reflected wavefronts propagate away
1 z z
i
i
t
r
i
r
r
from the interface we select k =−k . Letting k = k = k = k 1x and k =−k = k 1z ,
z z x x x z z
we maywrite the wave vectors in region 1 as
r
i
k = ˆ xk 1x + ˆ zk 1z , k = ˆ xk 1x − ˆ zk 1z .
t 2
2
t 2
Since (k ) + (k ) = k , letting k 2 = β 2 − jα 2 we have
x z 2

t 2 2 2 2 2 t − jγ t
k = k − k = (β 2 − jα 2 ) − (β 1 − jα 1 ) sin θ i = τ e .
z 2 1x
Squaring out the above relation, we have
t
t 2
t 2
A − jB = (τ ) cos 2γ − j(τ ) sin 2γ t
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