Page 451 - Engineering Electromagnetics, 8th Edition

CHAPTER 12 Plane Wave Reflection and Dispersion 433

Solution. First, we apply Snell’s law to find the transmission angle. Using n 1 = 1
for air, we use (63) to find
sin 30

θ 2 = sin −1 = 20.2 ◦
1.45
Now, for p-polarization:
η 1p = η 1 cos 30 = (377)(.866) = 326
377
η 2p = η 2 cos 20.2 = (.938) = 244
1.45
Then, using (69), we find
244 − 326
p = =−0.144
244 + 326
The fraction of the incident power that is reflected is

P r 2
=| p | = .021
P inc
The transmitted fraction is then
P t
2
= 1 −| p | = .979
P inc
For s-polarization, we have
η 1s = η 1 sec 30 = 377/.866 = 435
377
η 2s = η 2 sec 20.2 = = 277
1.45(.938)
Then, using (71):
277 − 435
s = =−.222
277 + 435
The reflected power fraction is thus
2
| s | = .049
The fraction of the incident power that is transmitted is
2
1 −| s | = .951

InExample12.7,reflectioncoefficientvaluesforthetwopolarizationswerefound
to be negative. The meaning of a negative reflection coefficient is that the component
of the reflected electric field that is parallel to the interface will be directed opposite
the incident field component when both are evaluated at the boundary.
This effect isalsoobserved when the second medium is a perfect conductor. Inthis
case, we know that the electric field inside the conductor must be zero. Consequently,
η 2 = E 20 /H 20 = 0, and the reflection coefficients will be p = s =−1. Total
reflection occurs, regardless of the incident angle or polarization.