Question

A light beam of diameter $\sqrt{5}R$ is incident symmetrically on a glass hemisphere of radius II and of refractive index $n=\sqrt{3}$. The radius of the beam at the base of hemisphere is:

• A. $\frac{R}{\sqrt{3}}$
• B. $\frac{R}{2\sqrt{3}}$
• C. $\frac{\sqrt{3}}{2}R$
• D. $\frac{R}{2}$

Answer 1

The correct option is A $\frac{R}{\sqrt{3}}$

From the given ray diagram shown in figure below

${\mu }_{1}sini={\mu }_{2}sinr$

$\Rightarrow 1\times sin{60}^o=\sqrt{3}\times sinr$

$\Rightarrow r={30}^o$

In $\Delta MNP$

$tan{30}^o=\frac{X}{\left(R/2\right)}$

$\Rightarrow X=\frac{R}{2\sqrt{3}}$

From geometry of the figure

$\Rightarrow r=\frac{\sqrt{3}}{2}R-X=\frac{\sqrt{3}}{2}R-\frac{R}{2\sqrt{3}}$

$=\left(\frac{3-1}{2\sqrt{3}}\right)R=\frac{R}{\sqrt{3}}$