Question

A thin convex lens $L$ (refractive index $=1.5$ ) is placed on a plane mirror $M$ . When a pin is placed at $A$ , such that $OA=18 \, cm,$ its real inverted image is formed at $A$ itself, as shown in figure. When liquid of refractive index $\mu _{l}$ is put between the lens and the mirror, the pin has to be moved to $A^{'},$ such that $OA^{'}=27 \, cm,$ to get its inverted real image at $A'$ itself. The value of $\mu _{l}$ will be

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

Answer 1

Answer: (A)

$\frac{1}{f_{1}}=\frac{1}{2}\times \frac{2}{18}$
$\frac{1}{f_{2}}=\frac{\left(\mu_{1}-1\right)}{-18}$
When $\mu _{1}$ is filled between lens and mirror,
$P=\frac{2}{18}-\frac{2}{18}\left(\mu _{1} - 1\right)=\frac{2 - 2( \mu _{1} + 2)}{18}$
$\Rightarrow F_{m}=-\left(\frac{18}{2 - \mu _{1}}\right)$
$2=6-3\mu _{1}$
$3\mu _{1}=4$
$\mu _{1}=\frac{4}{3}$