Question

A parallel beam of light is incident on a glass prism in the shape of a quarter cylinder of radius $R=0.05m$ 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. $\left(3\sqrt{3}-4\right)\times 10^{-2}m$
• B. $\left(2\sqrt{3}-2\right)\times 10^{-2}m$
• C. $\left(3\sqrt{5}-5\right)\times 10^{-2}m$
• D. $\left(3\sqrt{2}-3\right)\times 10^{-2}m$

Answer 1

Answer: (C)

Given, radius $R=0.05m$
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.
$\cos c=\frac{R}{R+X}$
$\sqrt{1-{\sin }^{2}c}=\frac{R}{R+X}$
$\sqrt{1-\frac{4}{9}}=\frac{R}{R+X}$
$\frac{\sqrt{5}}{3}=\frac{R}{R+X}$
$\sqrt{5}(R+X)=3R$
$\sqrt{5}R+\sqrt{5}X=3R$
$\therefore X=(3\sqrt{5}-5)\times 10^{-2}m$