Exercise 6.4.2: Light trapped in an LED chip

${\theta }_c=\arcsin (1/2.5)=23.6^{\circ }$.

$\frac{2\pi (1-\cos {\theta }_c)}{4\pi }=\frac{1-0.9165}{2}=0.042$: only $4.2\%$ of the emitted light can leave through the top.

$R=(\frac{2.5-1}{2.5+1}{)}^2=0.184$: $81.6\%$ is transmitted at normal incidence.

$\sin {\theta }_t=2.5\sin 15^{\circ }=0.647$, ${\theta }_t=40.3^{\circ }$, $\cos {\theta }_t=0.763$, $\cos {\theta }_i=0.966$. With ${\eta }_1=0.4$, ${\eta }_2=1$:

With $\frac{{\eta }_1\cos {\theta }_t}{{\eta }_2\cos {\theta }_i}=0.4\times 0.789=0.316$: $T_{\perp }=0.316\times 1.520^2=0.730$ and $T_{\parallel }=0.316\times 1.682^2=0.893$, and indeed $1-0.520^2=0.730$, $1-0.327^2=0.893$.

$\tau >1$ is no paradox: the field in air is larger because ${\eta }_2>{\eta }_1$, and the beam in air is narrower by the factor $\cos {\theta }_t/\cos {\theta }_i=0.789$ in the power balance. Averaged over the cone (unpolarised light) about $78\%$ of the light in it escapes, that is about $3.2\%$ of everything emitted.