A portion of a straight glass rod of diamater 4 cm and refractive index 1.5 is bent into an arc of radius R cm and a parallel beam of light is incident on it as shown in fogure. Find the smallest R, in cm, which permits all the light to pass around the arc.

The necessary and sufficient condition for all the rays to pass around the arc is that the ray with least angle of incidence should get internally reflected i.e., should suffer TIR.
From the figure, it becomes obvious that the ray with least angle of incidence is the one which is incident almost grazing with the inner wall.

For this ray, let α be the angle of incidence, then we observe that

$\sin \mathrm{\alpha }=\frac{\mathrm{R}-\mathrm{d}}{\mathrm{R}}$ , where d is the diameter of tube. 

For TIR, $\mathrm{\alpha }\ge \mathrm{C}\Rightarrow \sin \mathrm{\alpha }\ge \sin \mathrm{C}$ 

$\Rightarrow \frac{\mathrm{R}-\mathrm{d}}{\mathrm{R}}\ge \frac{1}{\mathrm{\mu }}\Rightarrow \mathrm{R}\ge \frac{\mathrm{\mu d}}{\mathrm{\mu }-1}$

Since, $\mathrm{\mu }=1.5\text{ and }\mathrm{d}=4\mathrm{cm}$, 

 $\Rightarrow \mathrm{R}\ge 12 \mathrm{cm}$
Hence, the least radius required is 12 cm