Question

Light is incident at angle $\theta$ on one plane end of a transparent cylindrical rod of $R.l.n$. Determine the least value of $n$ so that the light entering the rod does not come out of the curved surface of the rod irrespective of the value of $\theta$.

Answer 1

The ray of light is incident at $A$ & it just gets reflected totally at $B$. Therefore incident angle at $B$ is equal to critical angle given as ${\theta }_c={\sin }^{-1}\frac{1}{n}$
Snells law for refreatrion at $A$ yields
$\frac{\sin \theta }{\sin r}=n\Rightarrow \sin r+\frac{\sin \theta }{n}$ $\left(1\right)$
Since $r+{\theta }_c={90}^o\Rightarrow \sin ({90}^o-{\theta }_c)=\cos {\theta }_c$. For a ray not to come through the curved surface,
$r\le {90}^o-{\theta }_c$
$\Rightarrow \sin r\le \sqrt{1-{\sin }^2{\theta }_o}\le \sqrt{1-\frac{1}{n^2}}$ ($2$)
Eliminating $\sin r$ from $\left(i\right)$ & $\left(ii\right)$ we obtain
$\frac{\sin \theta }{n}\le \sqrt{1-\frac{1}{n^2}}\Rightarrow \sin \theta \le \sqrt{n^2-1}$
$n^2-1\ge {\sin }^2\theta$; $n^2\ge 1+\sin \theta$
$n\ge \sqrt{1+{\sin }^2\theta }$
$\Rightarrow n_{min}=\surd 2$
Since $\sin \theta \le 1$;
$\Rightarrow n_{min}=\sqrt{2}$