Question 8.21: With reference to the three waves expressed by Eqs. (8-153a)......
With reference to the three waves expressed by Eqs. (8-153a), (8-153b), and (8- 153c) in Example 8-18,
\pmb{E}^{i} = \frac{4}{\sqrt{13} } (3 \pmb{a}_{x} – 2 \pmb{a}_{z} ) e^{-j (2x + 3z)} (8-153a)
\pmb{E}^{r} = (\pmb{a}_{x} \cos \theta _{r} + \pmb{a}_{z} \sin \theta _{r}) \Gamma _{\parallel } E^{i}_{o} e^{ – j[(k_{r} \sin \theta _{r})x -(k_{r} \cos \theta _{r})z]} \\ \quad = (0.83 \pmb{a}_{x} + 0.55 \pmb{a}_{z})( – 0.15 \times 4) e^{- j [2x – 3z]} \\ \quad = – (0.50 \pmb{a}_{x} + 0.33 \ \pmb{a}_{z}) e^{-j [2x – 3z]} (8-153b)
\pmb{E}^{t} = (\pmb{a}_{x} \cos \theta _{t} – \pmb{a}_{z} \sin \theta _{t}) \tau _{\parallel } E^{i}_{o} e^{ – j[(k_{t} \sin \theta _{t})x +(k_{t} \cos \theta _{t})z]} \\ \quad = (0.93 \pmb{a}_{x} – 0.37 \pmb{a}_{z}) 0.76 \times 4 e^{- j [2.00x + 5.03z]} \\ \quad = (2.84 \pmb{a}_{x} – 1.13 \pmb{a}_{z}) e^{-j [2.00x + 5.03z]} (8-153c)
(a) find the power of each wave per unit area of the interface, and
(b) verify the law of conservation of energy at the interface.
(a) Incident power per unit area of the interface is
I_{i}\cos \theta _{i} = \frac{1}{2} \sqrt{ \frac{\varepsilon_{o}}{ \mu_{o}}}(E^{i}_{o})^{2}\cos \theta _{i} = \frac{377}{2} \left[\left\lgroup\frac{12}{\sqrt{13} } \right\rgroup^{2} + \left\lgroup\frac{8}{\sqrt{13} } \right\rgroup^{2} \right] \cos 33.7^{\circ } \\ \quad \quad \quad \quad = 2.509 [W]Reflected power per unit area of the interface is
I_{r}\cos \theta _{r} = \frac{1}{2} \sqrt{ \frac{\varepsilon_{o}}{ \mu_{o}}}(E^{r}_{o})^{2}\cos \theta _{r} = \frac{377}{2} \left[\left\lgroup 0.50 \right\rgroup^{2} + \left\lgroup 0.33 \right\rgroup^{2} \right] \cos 33.7^{\circ } \\ \quad \quad \quad \quad = 56 [W]Transmitted power per unit area of the interface is
I_{t}\cos \theta _{t} = \frac{1}{2} \sqrt{2.25} \sqrt{ \frac{\varepsilon_{o}}{ \mu_{o}}}(E^{t}_{o})^{2}\cos \theta _{t} = 1.5 \frac{377}{2} \left[\left\lgroup 2.84 \right\rgroup^{2} + \left\lgroup 1.13 \right\rgroup^{2} \right] \cos 21.7^{\circ } \\ \quad \quad \quad \quad = 2.454 [W](b) The incident power is 2,509[W] , whereas the sum of the reflected and transmitted powers is calculated as 2,510[W] . Ignoring floating point errors, the law of conservation of energy is satisfied.