Question

A lens with refractive index \( \frac{3}{2} \) has a power of +5 diopters in air. If it is completely immersed in water, its power is (in diopters).
The refractive index of water is \( \frac{4}{3} \) 
 

• A. 1.25
• B. 1.3
• C. 1.35
• D. 1.20

Hint:

The power of a lens changes when it is placed in different media due to the change in refractive index.

Answer 1

The Correct Option is A

The lens maker's formula relates the focal length $f$ of a lens to the refractive index $n$ of the lens relative to the surrounding medium, and the radii of curvature $R_1$ and $R_2$ of the lens surfaces: \[\frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right).\]The power $P$ of a lens is the reciprocal of its focal length: \[P = \frac{1}{f}.\]Let $n_l = \frac{3}{2}$ be the refractive index of the lens, and let $n_w = \frac{4}{3}$ be the refractive index of water. In air, the power of the lens is \[P_a = (n_l - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) = 5,\]so \[\left( \frac{3}{2} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) = 5,\]which simplifies to \[\frac{1}{R_1} - \frac{1}{R_2} = 10.\]In water, the power of the lens is \[P_w = \left( \frac{n_l}{n_w} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) = \left( \frac{3/2}{4/3} - 1 \right) \cdot 10 = \left( \frac{9}{8} - 1 \right) \cdot 10 = \frac{1}{8} \cdot 10 = \frac{5}{4} = \boxed{1.25}.\] Final Answer: 1.25.