Question

Refractive index of a glass convex lens is 1.5. The radius of curvature of each of the two surfaces of the lens is 40 cm. The ratio of the power of the lens when immersed in a liquid of refractive index 1.25 to that when placed in air is

• A. 2:3
• B. 3:2
• C. 2:5
• D. 5:2

Hint:

Power of a lens decreases when immersed in a medium with refractive index greater than 1.

Answer 1

The Correct Option is C

Step 1: Formula
Power $P = \frac{1}{f} = (\mu_{rel} - 1)(\frac{1}{R_1} - \frac{1}{R_2})$.

Step 2: Analysis
- In air: $P_a = (1.5 - 1)K = 0.5K$ (where $K$ is the curvature constant).
- In liquid: $P_l = (\frac{1.5}{1.25} - 1)K = (1.2 - 1)K = 0.2K$.

Step 3: Calculation
Ratio $\frac{P_l}{P_a} = \frac{0.2K}{0.5K} = \frac{2}{5}$.

Step 4: Conclusion
Hence, the ratio is 2:5. Final Answer: (C)