A thin convex lens $\mathrm{L}$ (refractive index $=1.5$ ) is placed on a plane mirror $\mathrm{M}$. When a pin is placed at $\mathrm{A}$, such that $\mathrm{O}\mathrm{A}=18 \mathrm{c}\mathrm{m},$ its real inverted image is formed at $\mathrm{A}$ 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 ${\mathrm{A}}^{\text{'}},$ such that $\mathrm{O}{\mathrm{A}}^{\mathrm{\text{'}}}=27\mathrm{c}\mathrm{m},$ to get its inverted real image at $\mathrm{A}\text{'}$ itself. The value of ${\mu }_l$ will be

• Correct answer
  A

  $\frac{4}{3}$
• B

  $\sqrt{3}$
• C

  $\frac{3}{2}$
• D

  $\sqrt{2}$

Hence, the correct answer is option: 'A'.