Light is incident at an angle $\alpha$ on one planar end of a transparent cylindrical rod of refractive index n. The least value of n so that the light entering the rod does not emerge from the curved surface of the rod for any value of $\alpha$ is

Ray OA is incident at an angle $\alpha$ at the planar face of the cylindrical rod. Let $\theta$ be the angle of refraction. From Snell's law, we have

$n=\frac{\sin \alpha }{\sin \theta } or \sin \theta =\frac{\sin \alpha }{n} .......... \left(1\right)$

The ray AB is incident at point B of the curved surface of the cylinder at an angle (90° - $\theta$). This ray is travelling in a denser medium of refractive index n and is incident at the cylinder-air interface at point B. The ray will not emerge from the curved surface if it suffers total internal reflection at B. For this to happen (90° - $\theta$) $\ge$ i_c, the critical angle.

$$\begin{gathered} \Rightarrow \sin (90° - \theta ) \ge \sin i_c or \cos \theta \ge \sin i_c \\ \Rightarrow {\left(1-{\sin }^2\theta \right)}^{1/2}\ge \sin i_c \\ \Rightarrow 1-{\sin }^2\theta \ge {\sin }^2\left(i_c\right) ...................... \left(2\right) \\ The critical angle is given by \\ \sin i_c=\frac{1}{n} .............. \left(3\right) \\ U\sin g Eqs. \left(1\right) and \left(3\right) in Eq. \left(2\right) we get \\ 1-\frac{ {\sin }^2\alpha }{n^2}\ge \frac{1}{n^2} or n^2-{\sin }^2\alpha \ge 1 \\ or n^2 \ge \left(1+{\sin }^2\alpha \right) \\ Since the maximum value of {\sin }^2\alpha = +1, it follows that \\ { n}_{min}^2 \ge 2 or n_{min}\ge \sqrt{2} \end{gathered}$$

This is the minimum value of refractive index of the cylindrical rod for the ray AB to suffer total internal reflection at point B. By symmetry, ray BC will be totally reflected along CD suffering another total internal reflection at D and so on until the ray finally emerges from the opposite planar face of the rod.