Refractive index of the material of a sphere of radius 10 cm is $\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$\sqrt{3}$}\right.$. A point source of light P is placed just inside the surface of the sphere. Then area of the surface of the sphere through which light energy can pass into surrounding air, is

From the ray diagram, $\angle \mathrm{PAC}=\angle \mathrm{PBC}={\mathrm{\theta }}_{\mathrm{C}}={\sin }^{-1}\frac{1}{\left(\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$\sqrt{3}$}\right.\right)}={60}^{\mathrm{o}}$

$\therefore \angle \mathrm{ACB}=4{\mathrm{\theta }}_{\mathrm{C}}={240}^{\mathrm{o}}=\frac{4\mathrm{\pi }}{3}$

Therefore area of surface through area of surface through which rays of light can pass into surrounding area = Area AQB 

Solid angle $\mathrm{ACB}=\mathrm{\varphi }=2\mathrm{\pi }\left(1-\cos \frac{4{\mathrm{\theta }}_{\mathrm{C}}}{2}\right)$

$\Rightarrow \mathrm{\varphi }=2\mathrm{\pi }\left(1-\cos 2\times \raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\right)=2\mathrm{\pi }\left(1+\frac{1}{2}\right)=3\mathrm{\pi }\mathrm{radian}$

$\therefore$ Area $\mathrm{AQB}=3\mathrm{\pi }\times {\mathrm{R}}^2=2\mathrm{\pi }\times {\left(10\right)}^2{\mathrm{cm}}^2=200\mathrm{\pi }{\mathrm{cm}}^2$