Worked example 6.3.2: Wi-Fi through a window pane

${\eta }_2={\eta }_0/\sqrt{6}=153.9\Omega$. In units of ${\eta }_0$, ${\eta }_1=1$ and ${\eta }_2=0.408$ (${\eta }_2^2=1/6$).

${\lambda }_0=c/f=122.4$ mm and ${\lambda }_2={\lambda }_0/\sqrt{6}=50.0$ mm, so

${\sin }^2{\beta }_{2}d=0.469$, ${\cos }^2{\beta }_{2}d=0.531$:

so $32.8\%$ is reflected, $67.2\%$ transmitted, and the loss is $10{\log }_{10}(1/0.672)=1.7$ dB.

At $5.8$ GHz: ${\lambda }_0=51.7$ mm, ${\lambda }_2=21.1$ mm, ${\beta }_{2}d=1.785$ rad $=102.3^{\circ }$, ${\sin }^2{\beta }_{2}d=0.955$, $|\Gamma |=0.706$, $49.9\%$ reflected, loss $3.0$ dB.

No reflection for ${\beta }_{2}d=\pi$, $d={\lambda }_2/2=25.0$ mm. Worst for ${\beta }_{2}d=\pi /2$, $d={\lambda }_2/4=12.5$ mm, where ${\eta }_{\mathrm{in}}={\eta }_2^2/{\eta }_0$ and ${\Gamma }_d=\frac{{\eta }_2^2-{\eta }_1^2}{{\eta }_2^2+{\eta }_1^2}=\frac{1/6-1}{1/6+1}=-\frac{5}{7}$, $|\Gamma {|}^2=51\%$ ($3.1$ dB).

No thickness can reflect more than the quarter-wave value $51\%$, and both answers stay below it $✓$ (Figure 6.3.22).