Question

A convex lens of refractive index $\frac{3}{2}$ has a power of 2.5 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 D
• B. - 1.5 D
• C. 1.25 D
• D. 1.5 D

Answer 1

The correct option is A - 1.25 D

According to Lens Maker's Formula,

Power of lens$=P=({\mu }_r-1)(\frac{1}{R_1}-\frac{1}{R_2})$

where ${\mu }_r$ is the refractive index of material of lens relative to the surrounding medium$=\frac{{\mu }_{material}}{{\mu }_{medium}}$

For convex lens in air, ${\mu }_r=\frac{3/2}{1}=\frac{3}{2}$

For convex lens in the liquid, ${\mu }_r=\frac{3/2}{2}=\frac{3}{4}$

Thus $\frac{P_2}{P_1}=\frac{{\mu }_{r_2}-1}{{\mu }_{r_1}-1}$

$=\frac{\frac{3}{4}-1}{\frac{3}{2}-1}=-\frac{1}{2}$

$⟹P_2=-\frac{1}{2}\times 2.5D=-1.25D$