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)
$r^{\prime }=90^{\circ }-r\Rightarrow (r^{\prime }{)}_{min}=90^{\circ }-(r{)}_{max}$
and $n=\frac{sin(i{)}_{max}}{sin(r{)}_{max}}=\frac{sin90^{\circ }}{si{n}_{max}}(\because i_{max}=90^{\circ }$
Then, $sin(r{)}_{max}=\frac{1}{n}=sin{\theta }_c$
$(r{)}_{max}={\theta }_{c}or(r^{\prime }{)}_{min}=(90^{\circ }-{\theta }_c)$
Now, if minimum value of $r^{\prime }i.e.90^{\circ }-{\theta }_c$ 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
$(r^{\prime }{)}_{min}\ge {\theta }_c$
$or(90^{\circ }-{\theta }_c)\ge {\theta }_c$
$orsin(90^{\circ }-{\theta }_c)\ge sin{\theta }_c$
$orcos{\theta }_c\ge sin{\theta }_c$
$orcot{\theta }_c\ge 1$
$or\sqrt{n^2-1}\ge 1$
$or{n}^2\ge 2$
$orn\ge \sqrt{2}$
Therefore, minimum value of n is $\sqrt{2}$