Solution to 2008 Problem 76

 We first apply Snell's Law to the upper surface of the fiber and find the condition for total internal internal reflection.Total internal reflection occurs when Snell's Law\begin{align*}n_1 \sin \theta_1 = n_2 \sin \theta_2\end{align*}fails because of the fact that $\sin$ cannot be larger than $1$. If we take medium $1$ to be the optical fiber and medium $2$ to be free space, then $n_1 = n$ and $n_2 = 1$. So, Snell's Law will fail when \begin{align*}n \sin \theta_1 > 1\end{align*}or\begin{align*}\theta_1 > \sin^{-1} \frac{1}{n}\end{align*}Now, from the right triangle in the diagram, we see that an application of Snell's Law to the leftmost surface of the fiber gives\begin{align*}\sin \theta = n \sin \left(\frac{\pi}{2} - \theta_1 \right) = n \cos \theta_1\end{align*}\begin{align*}\cos \left(\sin^{-1} \frac{1}{n}\right) = \frac{\sqrt{n^2 - 1}}{n}\end{align*}So, light will stay in the fiber when\begin{align*}\theta_1 < \sin^{-1} \frac{1}{n} \Longrightarrow \frac{\sqrt{n^2 - 1}}{n} > \cos \theta_1\end{align*}So, the condition for total internal reflection is\begin{align*} \sin \theta < \sqrt{n^2 - 1}\end{align*}or \begin{align*}\boxed{\theta < \sin^{-1} \sqrt{n^2 - 1}}\end{align*}Therefore, answer (B) is correct.