Question

A circular beam of light of diameter $d=2\, cm$ 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. $4.00\, cm$
• B. $3.0\, cm$
• C. $3.26 \, cm$
• D. $2.52 \, cm$

Answer 1

Answer: (C)

Let $d^{\prime}$ be the diameter of refracted beam Then,
$d=P Q \cos 60^{\circ}$
and $ d^{\prime}=P Q \cos r$
i.e. $ \frac{d^{\prime}}{d}=\frac{\cos r}{\cos 60^{\circ}}=2 \cos r $
or $ d=2 d \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}}$
$=4 \sqrt{\frac{2}{3}} cm \approx 3.26 \,cm$