A convex lens of refractive index \(\frac{3}{2}\) has a power 2.5. If it is placed in a liquid of refractive index 2 , the new power of the lens is

Using Lens maker's formula,
\(\frac{1}{\mathrm{f}}=\mathrm{P}=(\mu-1)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right)\)
When placed in liquid the power is given by
\(P^{\prime}=\left(\mu^{\prime}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\)
Where refractive index of lens in the liquid,
\(\begin{aligned}
& \mu^{\prime}=\frac{\mu}{\mu_l}=\frac{3 / 2}{2}=\frac{3}{4} \\
\therefore \quad & \frac{P^{\prime}}{P}=\frac{\mu^{\prime}-1}{\mu-1}=\frac{\frac{3}{4}-1}{\frac{3}{2}-1}=\frac{-\frac{1}{4}}{\frac{1}{2}}=-\frac{1}{2} \\
& P^{\prime}=-\frac{1}{2} \times P=-\frac{1}{2} \times 2.5=-1.25 \mathrm{D}
\end{aligned}\)