A circular beam of light of diameter d=2 cm falls on a plane surface of glass. The angle of incidence is 60° and refractive index of glass is $\mu$=3/2. The diameter of the refracted beam is

Let d' be the diameter of refracted beam.

Then, $d=PQ\cos {60}^{\circ }$

and d${ }^{\text{'}}$ = PQ cos r

i.e. $\frac{d^{\mathrm{\prime }}}{d}=\frac{\cos r}{\cos {60}^{\circ }}=2\cos r$

or $d^{\mathrm{\prime }}=2d\cos r$

$\begin{array}{l}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\sin r=\frac{\sin i}{\mu }=\frac{\sqrt{3}/2}{3/2}=\frac{1}{\sqrt{3}}\\ \therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\cos r=\sqrt{1-{\sin }^{2}r}=\sqrt{\frac{2}{3}}\\ \therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{d}^{\mathrm{\prime }}=\left(2\right)\left(2\right)\sqrt{\frac{2}{3}}=4\sqrt{\frac{2}{3}}\mathrm{cm}\approx 3.26\mathrm{cm}\end{array}$