1. (35.1) The numerical aperture of an optical fiber
Figure 35-6 shows a longitudinal section through an optical fiber. A ray emanating from a source $S$ enters the fiber at an angle such that $\theta$ is the critical angle.
(a) Show that $\sin \phi=\left(n_{2}^{2}-n_{1}^{2}\right)^{1 / 2}$. The quantity $\sin \phi$ is the numerical aperture (NA) of the fiber. This expression is valid only if $n_{2}^{2}-n_{1}^{2} \leq 1$. The maximum possible value of $\phi$ is $90^{\circ}$. If the angle $\phi$ increases beyond the value defined by the above equation, then $\theta$ becomes smaller than the critical angle and the ray does not propagate down the guide. This equation therefore defines an upper limit for $\phi$.
(b) Show that, if the source radiates isotropically, then the fraction of the total available light that is collected by the fiber is about $(\mathrm{NA})^{2} / 4$, or about $n_{2} \Delta n / 2$, where $\Delta n=n_{2}-n_{1}$. If $n_{2}=2$ and $n_{1}=1.98$, then $\phi=11.5^{\circ}$ and $F=0.01$. The light collection efficiency is thus very low.