Exercise 6.3.2: Leaving the glass

$E^i=8$ V/m, $E^r=1.6$ V/m, $E^t=9.6$ V/m. With ${\eta }_1=251.3\Omega$: $H^i=8/251.3=31.8$ mA/m, $H^r=0.2\times 31.8=6.37$ mA/m (directed against $H^i$, since $H^r=-\frac{E_0^r}{{\eta }_1}{\mathrm{e}}^{\mathrm{j}{\beta }_{1}x}{a}_z$), and $H^t=9.6/377=25.5$ mA/m.

${\mathcal{P}}^i=\frac{8^2}{2\times 251.3}=0.1273$ W/m${}^2$, ${\mathcal{P}}^r=0.04{\mathcal{P}}^i=0.0051$ W/m${}^2$, ${\mathcal{P}}^t=\frac{9.6^2}{2\times 377}=0.1222$ W/m${}^2$, and $0.1273-0.0051=0.1222$ $✓$. The larger electric field goes with a smaller magnetic field ($25.5<31.8$ mA/m): in power terms $T=\frac{{\eta }_1}{{\eta }_2}{\tau }^2=\frac{2}{3}\times 1.44=0.96$.

$S=1.2/0.8=1.5$. With $\Gamma >0$ the incident and reflected fields add at the face: $|E_1{|}_{\max }=8\times 1.2=9.6$ V/m at $x=0$, and maxima repeat every ${\lambda }_1/2={\lambda }_0/(2n)=200$ nm. The minima, $8\times 0.8=6.4$ V/m, lie ${\lambda }_1/4=100$ nm from the maxima.

Continuity at $x=0$: $8+1.6=9.6$ V/m $✓$ and $31.8-6.37=25.5$ mA/m $✓$. The reflectance $4\%$ is the same as for light entering the glass from air: $R$ depends only on $|{\eta }_2-{\eta }_1|/({\eta }_2+{\eta }_1)$.