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.

Solution to 2008 Problem 76