A small bulb is placed at the bottom of a tank containing a transparent liquid (refractive index $n$) to a depth $H$. The radius of the circular area of the surface of liquid, through which light from the bulb can emerge out, is $R$. Then $\left(\frac{R}{H} ight)$ is : (A) $\frac{1}{\sqrt{n^2-1}}$ (B) $\sqrt{n^2-1}$ (C) $\frac{1}{\sqrt{n^2+1}}$ (D) $\sqrt{n^2+1}$