Question

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

Answer 1

$sin \theta_c=\frac{1}{n}(\theta_c=$critical angle)
$ r'=90^\circ-r \Rightarrow (r')_{min}=90^\circ-(r)_{max}$
and $\, \, \, \, \, \, \, \, n=\frac{sin(i)_{max}}{sin(r)_{max}}=\frac{sin 90^\circ}{sin_{max}}\, \, \, \, \, \, \, \, \, (\because i_{max}=90^\circ$
Then, $\, \, \, \, \, \, \, \, sin(r)_{max}=\frac{1}{n}=sin \theta_c$
$(r)_{max}=\theta_c\, or\, (r')_{min}=(90^\circ-\theta_c)$
Now, if minimum value of $r'\, i.e.\, 90^\circ-\theta_c$ is graeter than $\theta_c$
then obviously all values of r will be greater than $\theta^c$ i. e ., total
internal reflection will take place at face A B in all conditions.
Therefore, the necessary condition is
$ (r')_{min} \ge \theta_c$
$or (90^\circ-\theta_c)\ge \theta_c$
$or\, \, \, \, \, \, \, \, \, \, sin(90^\circ- \theta_c)\ge sin \theta_c$
$or \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, cos \theta_c \ge sin \theta_c$
$or \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, cot \theta_c \ge 1$
$or \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \sqrt{n^2-1} \ge 1$
$or \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, n^2 \ge 2$
$or \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, n \ge \sqrt{2}$
Therefore, minimum value of n is $\sqrt{2}$