Question

A light beam of diameter $\sqrt{3}$ R is incident symmetrically on a glass homisphere of radius R and of refractive index $n=\sqrt{3}$. Find the radius of the beam at the base of hemisphre

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

Answer 1

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



$\leftarrow x\to \leftarrow r\to$
$\mu ,sini={\mu }_{2}sinr\Rightarrow 1.sin{60}^{\circ }-\sqrt{3}sinr\Rightarrow r={30}^{\circ }tan{30}^{\circ }-\frac{x}{\left(\frac{R}{2}\right)}x=\frac{R}{2\sqrt{3}}r=\frac{\sqrt{3}}{2}r-x=\frac{R}{\sqrt{3}}$