A portion of a straight glass rod of diameter 4 cm and refractive index 1.5 is bent into an arc of radius $R\text{ cm}$  and a parallel beam of light is incident on it as shown in figure. The smallest R,  in cm, which permits all the light to pass around the arc is

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 grazingly with the inner wall.

For this ray, let  $\alpha$ be the angle of incidence, then we observe that
$\sin \alpha =\frac{R-d}{R}$  where  d is the diameter of tube
For  $TIR,\alpha \ge C$
 $\Rightarrow \sin \alpha \ge \sin C\Rightarrow \frac{R-d}{R}\ge \frac{1}{\mu }\Rightarrow R\ge \frac{\mu d}{\mu -1}$
Since,  $\mu =1.5\text{ and }d=4\text{ cm,}$ so we get  $R=12\text{ cm}$
Hence, the least required is 12 cm.