Question

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

• A. $4.00cm$
• B. $3.0cm$
• C. $3.26cm$
• D. $2.52cm$

Answer 1

Answer: (C)

Let $d^{\prime }$ be the diameter of refracted beam Then,
$d=PQ\cos 60^{\circ }$
and $d^{\prime }=PQ\cos r$
i.e. $\frac{d^{\prime }}{d}=\frac{\cos r}{\cos 60^{\circ }}=2\cos r$
or $d=2d\cos r$
$\sin r=\frac{\sin i}{\mu }=\frac{\sqrt{3}/2}{3/2}=\frac{1}{\sqrt{3}}$
$\therefore \cos r=\sqrt{1-{\sin }^{2}r}=\sqrt{\frac{2}{3}}$
$\therefore d=(2)(2)\sqrt{\frac{2}{3}}$
$=4\sqrt{\frac{2}{3}}cm\approx 3.26cm$