Exercise 6.2.3: Where does the power go?

$\omega \epsilon =2\pi \times 10^8\times 10\times 10^{-9}/36\pi =0.0556S/m$ and $p=0.1/0.0556=1.80$: neither class. Exactly,

The field falls by $8.686\times 4.82=41.9\mathrm{dB}$ per metre.

Write $\gamma =\alpha +\mathrm{j}\beta$ and ${\eta }_c=|{\eta }_c|e^{\mathrm{j}{\theta }_{\eta }}$. Since $\gamma {\eta }_c=\mathrm{j}\omega \mu$ is imaginary, its real part vanishes: $\alpha \cos {\theta }_{\eta }=\beta \sin {\theta }_{\eta }$. The real part of $\gamma /{\eta }_c$ is then

and therefore

At $z=0$: $2\alpha S_{av}(0)=2\times 4.82\times 0.519=5.00W/{\mathrm{m}}^3$ and $\frac{1}{2}\sigma E_0^2=\frac{1}{2}\times 0.1\times 100=5.00W/{\mathrm{m}}^3$.

$S_{av}(0)(1-e^{-2\alpha \times 0.1})=0.519\times (1-e^{-0.964})=0.321W/{\mathrm{m}}^2$.

The identity in (c) says that the power leaving the beam in a slab $dz$ is exactly the Joule heat $\frac{1}{2}\sigma |E{|}^{2}dz$ in it. With polarisation loss the same holds with $\sigma \to \omega {\epsilon }^{\prime \prime }$, which is how a microwave oven heats food.