Home exercise 6.4.3: How a rain sensor sees water

Glass to air: ${\theta }_c=\arcsin (1/1.5)=41.8^{\circ }<45^{\circ }$. The ray is totally reflected, $|\Gamma |=1$ for both polarisations, and all the light returns to the detector.

Glass to water: $\sin {\theta }_c=\frac{4/3}{1.5}=\frac{8}{9}$, ${\theta }_c=62.7^{\circ }>45^{\circ }$, so there is no total reflection. Snell's law gives

With ${\eta }_1=0.6667{\eta }_0$, ${\eta }_2=0.75{\eta }_0$ and $\cos {\theta }_i=0.7071$:

The fractions that leak into the water are $1-0.135^2=98.2\%$ (S) and $1-0.018^2=99.97\%$ (P).

$\tan {\theta }_B=\frac{n_2}{n_1}=\frac{4/3}{1.5}=\frac{8}{9}$: ${\theta }_B=41.6^{\circ }$, close to $45^{\circ }$, which is why P leaks almost completely.

$T_{\perp }=\frac{{\eta }_1\cos {\theta }_t}{{\eta }_2\cos {\theta }_i}{\tau }_{\perp }^2=\frac{0.6667\times 0.6060}{0.75\times 0.7071}\times 1.135^2=0.7618\times 1.288=0.982$.