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 a liquid of refractive index ${\mu }_1$ 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 ${\mu }_1$ will be :

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

Answer 1

Answer: (B)

$\frac{1}{f_1}=\frac{1}{2}\times \frac{2}{18}=\frac{1}{18}$
$\frac{1}{f_2}=\frac{\left({\mu }_1-1\right)}{-18}$
when ${\mu }_1$ is filled between lens and mirror
$P=\frac{2}{18}-\frac{2}{18}\left({\mu }_1-1\right)=\frac{2-2{\mu }_1+2}{18}$
$=F_m=-\left(\frac{18}{2-{\mu }_1}\right)$
$2=6-3{\mu }_1$
$3{\mu }_1=4$
${\mu }_1=4/3$