Question

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

• A. - 1.25 D
• B. - 1.5 D
• C. 1.25 D
• D. 1.5 D

Answer 1

Answer: (A)

Focal length of a convex lens having power $2.5D,=\frac{1}{2.5}m$
Also focal length of a lens in a medium of refractive index $\mu$ is given by $\frac{1}{f}=(\mu -1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$
$\Rightarrow 2.5=\frac{1}{f}=\left(\frac{3}{2}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\dots (i)$ (in air)
$\Rightarrow \frac{1}{f^{\prime }}=\left(\frac{3}{4}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\dots (ii)\left[\because l{\mu }_g=\frac{3}{4}\right]$ (in liquid)
Dividing the two, $2.5f^{\prime }=\frac{0.5}{-0.25}$
$\Rightarrow \frac{1}{f^{\prime }}=\frac{-5}{25\times 0.25}=-1.25D$