Question

A quarter cylinder of radius $R$ and refractive index $1.5$ is placed on a table. A point object $P$ is kept at a distance of $mR$ from it as shown in figure. For what value of $m$ for which a ray from $P$ will emerge parallel to the table?

• A. $\frac{2}{3}$
• B. $\frac{3}{2}$
• C. $\frac{4}{3}$
• D. $\frac{3}{4}$

Answer 1

The correct option is C $\frac{4}{3}$

Refraction at plane surface (1),

$n_1=1$, $n_2=1.5=\frac{3}{2}$, $u=-mR$, $v=$?

Using formula, $\frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R}$

$\because R=\infty$ for plane surface.

$\because \frac{3/2}{v}-\frac{1}{(-mR)}=0$

$\therefore v=-\frac{3}{2}mR$

$u=-(\frac{3}{2}mR+R)$ and $R=-R$, $v=\infty$, $n_1=\frac{3}{2}$, $n_2=1$

$\frac{1}{\infty }-\frac{3/2}{[-(\frac{3}{2}mR+R)]}=\frac{1-3/2}{-R}$

$\Rightarrow \frac{3/2}{\frac{R(3m+2)}{2}}=\frac{1/2}{R}$

$\Rightarrow \frac{6}{3m+2}=1\Rightarrow 3m+2=6$

$m=\frac{4}{3}$