Question

A parallel beam of light is incident on a glass prism in the shape of a quarter cylinder of radius $R \,= \,0.05\, m$ and refractive index n = $1.5$ placed on a horizontal table as shown in the figure. Beyond the cylinder, a patch of light is found whose nearest distance $x$ from the cylinder is

• A. $\left(3\sqrt{3}-4\right)\times10^{-2}m$
• B. $\left(2\sqrt{3}-2\right)\times10^{-2}m$
• C. $\left(3\sqrt{5}-5\right)\times10^{-2}m$
• D. $\left(3\sqrt{2}-3\right)\times10^{-2}m$

Answer 1

Answer: (C)

Given, radius $R = 0 .05 \,m$
Refractive index $n = 1.5$

The critical angle, $\sin c=\frac{1}{n}=\frac{1}{1 \cdot 5}$
$=\frac{10}{15}=\frac{2}{3}$
From above figure.
$\cos c =\frac{R}{R+X} $
$ \sqrt{1-\sin ^{2} c} =\frac{R}{R+X}$
$ \sqrt{1-\frac{4}{9}} =\frac{R}{R+X} $
$ \frac{\sqrt{5}}{3} =\frac{R}{R+X} $
$ \sqrt{5}(R+X) =3 R $
$\sqrt{5} R+\sqrt{5} X =3 R$
$\therefore X = (3\sqrt{5} - 5 ) \times 10^{-2} m$