Power of a convex lens is + 5D ($μ_g$ = 1.5). When this lens is immersed in a liquid of refractive index $μ$, it acts like a divergent lens of focal length 100 cm. Then refractive index of the liquid will be

$\frac{5}{3}$ $\frac{5}{4}$ $\frac{6}{5}$ none of these

$\frac{5}{3}$

$\frac{P_a}{P_I}=\frac{(\frac{μ_g}{μ_a}-1)}{(\frac{μ_g}{μ_I}-1)}=\frac{5}{-100/100}=-5$ $⇒μ_I=\frac{5}{3}$