Question 8.12: For a uniform plane wave normally incident on an interface b......
For a uniform plane wave normally incident on an interface between two media of \eta _{1} and \eta _{2}, derive \left|\Gamma \right|^{2} + (\eta _{1} / \eta _{2}) \left|\tau \right|^{2} = 1 from the law of conservation of energy.
Time-average power densities of the incident, reflected, and transmitted waves are, from Eqs. (8-103)-(8-105),
\pmb{E}^{i} = \pmb{a}_{x} E^{i}_{o} e^{-j\beta _{1} z} (8-103a) \\ \pmb{H}^{i} = \pmb{a}_{y} \frac{E^{i}_{o} }{\eta _{1}} e^{-j\beta _{1} z} (8-103b) \\ \pmb{E}^{t} = \pmb{a}_{x} E^{t}_{o} e^{-j\beta _{2} z} (8-105a) \\ \pmb{H}^{t} = \pmb{a}_{y} \frac{E^{t}_{o} }{\eta _{2}} e^{-j\beta _{2} z} (8-105b) \\ \left\langle \pmb{S} ^{i}\right\rangle = \frac{1}{2} Re \left[\pmb{E}^{i} \times \pmb{H}^{i*}\right] = \pmb{a}_{z} \frac{1}{2 \eta _{1}} \left|E^{i}_{o}\right|^{2} (8-109a)\\ \left\langle \pmb{S} ^{r}\right\rangle = – \pmb{a}_{z}\frac{1}{2\eta _{1}} \left|E^{r}_{o}\right| ^{2} = – \pmb{a}_{z} \frac{1}{2 \eta _{1}} \left|\Gamma \right|^{2} \left|E^{i}_{o}\right|^{2} (8-109b)\\ \left\langle \pmb{S} ^{t}\right\rangle = \pmb{a}_{z}\frac{1}{2\eta _{2}} \left|E^{t}_{o}\right| ^{2} = \pmb{a}_{z} \frac{1}{2 \eta _{2}} \left|\tau \right|^{2} \left|E^{i}_{o}\right|^{2} (8-109c)From the law of conservation of energy, we write
\left|\left\langle \pmb{S} ^{i}\right\rangle\right| = \left|\left\langle \pmb{S} ^{r}\right\rangle \right| + \left|\left\langle \pmb{S} ^{t} \right\rangle \right| (8-110)
Substituting Eq. (8-109) into Eq. (8-110), we have
\boxed{\left|\Gamma \right|^{2} + \frac{\eta _{1}}{\eta _{2}} \left|\tau \right|^{2} = 1 } (8-111)