Page 291 - Electromagnetics

respectively. We assume that a linearly-polarized plane-wave field of the form (4.216) is
created within region 1 bya process that we shall not studyhere. We take this field to
be the known “incident wave” produced byan impressed source, and wish to compute
the total field in regions 1 and 2. Here we shall assume that the incident field is that of a
uniform plane wave, and shall extend the analysis to certain types of nonuniform plane
waves subsequently.
Since the incident field is uniform, we maywrite the wave vector associated with this
field as
i
ˆ i i
ˆ i
k = k k = k (k + jk )
i
i
where
i 2 2 c
[k (ω)] = ω ˜µ 1 (ω)˜ (ω).
1
ˆ i
We can assume without loss of generalitythat k lies in the xz-plane and makes an angle
θ i with the interface normal as shown in Figure 4.18. We refer to θ i as the incidence angle
of the incident field, and note that it is the angle between the direction of propagation
of the planar phase fronts and the normal to the interface. With this we have
i
i
i
k = ˆ xk 1 sin θ i + ˆ zk 1 cos θ i = ˆ xk + ˆ zk .
z
x
Using k 1 = β 1 − jα 1 we also have
i
k = (β 1 − jα 1 ) sin θ i .
x
i
The term k is written in a somewhat different form in order to make the result easily
z
applicable to reflections from multiple interfaces. We write
i − jγ
i
i
i
i
i
k = (β 1 − jα 1 ) cos θ i = τ e i = τ cos γ − jτ sin γ .
z
Thus,

i 2 2 i −1
τ = β + α cos θ i , γ = tan (α 1 /β 1 ).
1 1
We solve for the fields in each region of space directlyin the frequencydomain. The
incident electric field has the form of (4.216),
i
˜ i
˜ i
E (r,ω) = E (ω)e − jk (ω)·r , (4.268)
0
while the magnetic field is found from (4.219) to be
i
˜ i
k × E
˜ i
H = . (4.269)
ω ˜µ 1
The incident field maybe decomposed into two orthogonal components, one parallel
ˆ
to the plane of incidence (the plane containing k and the interface normal ˆ z) and one
perpendicular to this plane. We seek unique solutions for the fields in both regions, first
for the case in which the incident electric field has onlya parallel component, and then
for the case in which it has onlya perpendicular component. The total field is then
determined bysuperposition of the individual solutions. For perpendicular polarization
we have from (4.268) and (4.269)
i
i
˜ i
˜ i
E = ˆ yE e − j(k x x+k z z) , (4.270)
⊥ ⊥
i
i ˜ i
−ˆ xk + ˆ zk E
i
i
˜ i z x ⊥ − j(k x x+k z z)
H = e , (4.271)
⊥
k 1 η 1
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