A thin glass (refractive index $1.5$) lens has optical power of $-5\text{D}$ in air. Its optical power in a liquid medium with refractive index $1.6$ will be:

Using the Lens Maker's Formula: $P=\frac{1}{f}=\left(\frac{{\mu }_l}{{\mu }_m}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$.Let $K=\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$.In air (${\mu }_m=1$):$P_{air}=(1.5-1)K=0.5K=-5⟹K=-10$.In liquid (${\mu }_m=1.6$):$P_{liq}=\left(\frac{1.5}{1.6}-1\right)K=\left(\frac{1.5-1.6}{1.6}\right)(-10)$$P_{liq}=\left(\frac{-0.1}{1.6}\right)(-10)=\frac{1}{1.6}=0.625\text{ D}$.None of the options perfectly match $0.625\text{ D}$, but the calculation follows the standard method. Based on the provided source key (5/8 D), $0.625$ is correct.