Light enters at an angle of incidence in a transparent rod of refractive index $$\mu .$$ For what value of the refractive index of the material of the rod the light once entered into it will not leave it through its lateral face whatsoever be the value of angle of incidence ?

Solution :
Suppose a light ray enters at $$A$$ and refracted beam is $$AB.$$  At the lateral face, the angle of incidence is $$\theta .$$ For no refraction at this face, $$\theta > C$$
$$\sin \theta > \sin C$$

$$\eqalign{ & {\text{but}}\,\,\theta + r = {90^ \circ }\,\,\left( {{\text{According to geometry of figure}}} \right) \cr & \Rightarrow \theta = \left( {{{90}^ \circ } - r} \right) \cr & \therefore \sin \left( {{{90}^ \circ } - r} \right) > \sin C \cr & {\text{or}}\,\,\cos r > \sin C\,......\left( {\text{i}} \right) \cr} $$
Now from Snell’s law,
$$\eqalign{ & \mu = \frac{{\sin i}}{{\sin r}} \cr & \Rightarrow \sin r = \frac{{\sin i}}{\mu } \cr} $$
As from trigonometry,
$$\eqalign{ & \cos r = \sqrt {1 - {{\sin }^2}r} \cr & \therefore \cos r = \sqrt {\left( {1 - \frac{{{{\sin }^2}i}}{{{\mu ^2}}}} \right)} \cr} $$
$$\therefore $$ Eq. (i) gives,
$$\eqalign{ & \sqrt {1 - \frac{{{{\sin }^2}i}}{{{\mu ^2}}}} > \sin C \cr & \Rightarrow 1 - \frac{{{{\sin }^2}i}}{{{\mu ^2}}} > {\sin ^2}C \cr & {\text{Also}}\,\,\sin C = \frac{1}{\mu } \cr & \therefore 1 - \frac{{{{\sin }^2}i}}{{{\mu ^2}}} > \frac{1}{{{\mu ^2}}} \cr} $$
$$\eqalign{ & {\text{or}}\,\,1 > \frac{1}{{{\mu ^2}}} + \frac{{{{\sin }^2}i}}{{{\mu ^2}}} \cr & {\text{or}}\,\frac{1}{{{\mu ^2}}}\left( {{{\sin }^2}i + 1} \right) < 1 \cr & {\text{or}}\,\,{\mu ^2} > {\sin ^2}i + 1 \cr} $$
The maximum value of $$\sin i$$  is 1. So,
$$\eqalign{ & \therefore {\mu ^2} > 2 \cr & {\text{or}}\,\,\mu > \sqrt 2 \cr} $$