Question

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

• A. $-1.25\text{D}$
• B. $1.25\text{D}$
• C. $1.2\text{D}$
• D. $1.5\text{D}$

Answer 1

The correct option is A $-1.25\text{D}$

Let the radius of curvature of the convex lens is $R_1$ and $R_2$

Focal length of a convex lens having power $2.5\text{D},=\frac{1}{2.5}\text{m}$

Focal length of a lens in air is given by,
$\frac{1}{f}=(\mu -1)(\frac{1}{R_1}-\frac{1}{R_2})$
$\Rightarrow 2.5=\frac{1}{f}=(\frac{3}{2}-1)(\frac{1}{R_1}-\frac{1}{R_2}).....\left(i\right)$

Focal length of the lens in a liquid of refractive index $2$ is
$\Rightarrow \frac{1}{f^{\prime }}=(\frac{\frac{3}{2}}{2}-1)(\frac{1}{R_1}-\frac{1}{R_2})....\left(ii\right)$
Dividing the equation (i) by equation (ii),
$2.5f^{\prime }=\frac{0.5}{-0.25}$
$\Rightarrow \frac{1}{f^{\prime }}=\frac{-2.5\times 0.25}{0.5}=-1.25\text{D}$

Hence, option (a) is correct.