Physics - Geometrical Optics Question with Solution | TestHub

From the diagram, diameter of incident beam is$d_1=PQ \mathrm{c}\mathrm{o}\mathrm{s}{60}^{\mathrm{o}}$and diameter of refracted beam is$d_2=PQ\mathrm{c}\mathrm{o}\mathrm{s}r$.
$\therefore$$\frac{d_1}{d_2}=\frac{1}{2\mathrm{c}\mathrm{o}\mathrm{s}r}$
$\Rightarrow d_2=2d_1\mathrm{c}\mathrm{o}\mathrm{s}r\mathrm{ }$

From Snell's law:
$1\times \mathrm{s}\mathrm{i}\mathrm{n}i=\mu \mathrm{s}\mathrm{i}\mathrm{n}r$
$\mathrm{s}\mathrm{i}\mathrm{n}r=\frac{\mathrm{s}\mathrm{i}\mathrm{n}60°}{\mu }=\frac{1}{\sqrt{3}}$
$\Rightarrow \mathrm{c}\mathrm{o}\mathrm{s}r=\sqrt{\frac{2}{3}}$
$\therefore$$d_2=2\times 2\times \sqrt{\frac{2}{3}}=3.26\mathrm{ }\mathrm{c}\mathrm{m}$