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 $OA=18cm,$ 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 $A^{{}^{\prime }},$ such that $O{A}^{{}^{\prime }}=27cm,$ to get its inverted real image at $A^{\prime }$ 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,
$P=\frac{2}{18}-\frac{2}{18}\left({\mu }_1-1\right)=\frac{2-2({\mu }_1+2)}{18}$
$\Rightarrow F_m=-\left(\frac{18}{2-{\mu }_1}\right)$
$2=6-3{\mu }_1$
$3{\mu }_1=4$
${\mu }_1=\frac{4}{3}$