Exercise 6.4.4: The medium with \varepsilon_r=\mu_r at oblique incidence

$\sin {\theta }_t=\sin 60^{\circ }/2=0.433$, ${\theta }_t=25.66^{\circ }$, $\cos {\theta }_t=0.901$.

With ${\mu }_1={\epsilon }_1=1$ and ${\mu }_2={\epsilon }_2=2$ (relative values) in Eq. (6.4.23) and Eq. (6.4.30):

The only reflection-free incidence is ${\theta }_i=0$, for both polarisations. $R$ grows with the angle: $0.3\%$ at $30^{\circ }$, $1.9\%$ at $45^{\circ }$, $8.2\%$ at $60^{\circ }$, $29.6\%$ at $75^{\circ }$.

For ${\epsilon }_r=4$, ${\mu }_r=1$ (${\eta }_2={\eta }_0/2$, same ${\theta }_t$): ${\Gamma }_{\perp }=\frac{0.25-0.901}{0.25+0.901}=-0.566$ ($32.0\%$) and ${\Gamma }_{\parallel }=\frac{0.451-0.5}{0.451+0.5}=-0.052$ ($0.27\%$, near its Brewster angle $63.4^{\circ }$).

${\theta }_c=\arcsin (n_2/n_1)=\arcsin (1/2)=30^{\circ }$: total reflection works exactly as in a dielectric of index $2$.

$T_{\perp }=\frac{\cos {\theta }_t}{\cos {\theta }_i}(1+{\Gamma }_{\perp }{)}^2=1.803\times 0.714^2=0.918=1-0.082$.