Page 426 - Engineering Electromagnetics, 8th Edition

408 ENGINEERING ELECTROMAGNETICS

This wave, which moves away from the boundary surface into region 2, is called the
transmitted wave. Note the use of the different propagation constant k 2 and intrinsic
impedance η 2 .
Now we must satisfy the boundary conditions at z = 0 with these assumed fields.
With E polarized along x, the field is tangent to the interface, and therefore the E
fields in regions l and 2 must be equal at z = 0. Setting z = 0in (1) and (3) would
require that E + = E + . H, being y-directed, is also a tangential field, and must be
x10
x20
continuous across the boundary (no current sheets are present in real media). When
we let z = 0in (2) and (4), we find that we must have E + /η 1 = E x20 /η 2 . Since
+
x10
E + = E + , then η 1 = η 2 . But this is a very special condition that does not fit the
x10
x20
facts in general, and we are therefore unable to satisfy the boundary conditions with
only an incident and a transmitted wave. We require a wave traveling away from the
boundary in region 1, as shown in Figure 12.1; this is the reflected wave,
E − (z) = E − e jk 1 z (5)
x10
xs1
E −
H − (z) =− x10 e jk 1 z (6)
xs1
η 1
where E − may be a complex quantity. Because this field is traveling in the −z
x10
direction, E xs1 =−η 1 H ys1 for the Poynting vector shows that E × H must be in
−
−
−
−
1
1
the −a z direction.
The boundary conditions are now easily satisfied, and in the process the ampli-
tudes of the transmitted and reflected waves may be found in terms of E + . The total
x10
electric field intensity is continuous at z = 0,
(z = 0)
E xs1 = E xs2
or
E + + E − = E xs2 (z = 0)
+
xs1
xs1
Therefore
E + + E x10 = E x20 (7)
+
−
x10
Furthermore,
H ys1 = H ys2 (z = 0)
or
H ys1 + H ys1 = H + (z = 0)
−
+
ys2
and therefore
E + E − E +
x10
− x10 = x20 (8)
η 1 η 1 η 2
Solving (8) for E + and substituting into (7), we find
x20
η 2 η 2
E + + E − = E + − E −
x10
x10
η 1 x10 η 1 x10