Page 449 - Engineering Electromagnetics, 8th Edition

CHAPTER 12 Plane Wave Reflection and Dispersion 431

The boundary condition for a continuous tangential electric field now reads:
E zs1 + E − = E zs2 (at x = 0)
+
zs1
We now substitute Eqs. (58) through (60) into (61) and evaluate the result at x = 0
to obtain
E cos θ 1 e − jk 1 z sin θ 1 + E cos θ e − jk 1 z sin θ 1 = E 20 cos θ 2 e − jk 2 z sin θ 2 (61)
+

−
10
10
1
Note that E , E , and E 20 are all constants (independent of z). Further, we require
−
+
10
10
that (61) hold for all values of z (everywhere on the interface). For this to occur, it
must follow that all the phase terms appearing in (61) are equal. Specifically,

k 1 z sin θ 1 = k 1 z sin θ = k 2 z sin θ 2
1
From this, we see immediately that θ = θ 1 ,or the angle of reflection is equal to the

1
angle of incidence. We also find that
k 1 sin θ 1 = k 2 sin θ 2 (62)
Equation (62) is known as Snell’s law of refraction. Because, in general, k = nω/c,
we can rewrite (62) in terms of the refractive indices:
n 1 sin θ 1 = n 2 sin θ 2 (63)
Equation (63) is the form of Snell’s law that is most readily used for our present
case of nonmagnetic dielectrics. Equation (62) is a more general form which would
apply, for example, to cases involving materials with different permeabilities as well
√
as different permittivities. In general, we would have k 1 = (ω/c) µ r1 r1 and k 2 =
√
(ω/c) µ r2 r2 .
Having found the relations between angles, we next turn to our second objective,
which is to determine the relations between the amplitudes, E , E , and E 20 .To
+
−
10
10
accomplish this, we need to consider the other boundary condition, requiring tangen-
tial continuity of H at x = 0. The magnetic field vectors for the p-polarized wave are
all negative y-directed. At the boundary, the field amplitudes are related through
H + H 10 = H 20 (64)
+
−
10
Then, when we use the fact that θ = θ 1 and invoke Snell’s law, (61) becomes

1
+ − (65)
E cos θ 1 + E cos θ 1 = E 20 cos θ 2
10
10
Using the medium intrinsic impedances, we know, for example, that E /H 10 = η 1
+
+
10
and E /H 20 = η 2 . Eq. (64) can be written as follows:
+
+
20
+ − +
E cos θ 1 E cos θ 1 E cos θ 2
10
20
10
− = (66)
η 1p η 1p η 2p
Note the minus sign in front of the second term in (66), which results from the fact
that E cos θ 1 is negative (from Figure 12.7a), whereas H 10 is positive (again from
−
−
10
the figure). When we write Eq. (66), effective impedances, valid for p-polarization,