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. \(2:5\)
• C. \(3:4\)
• D. \(5:2\)

Hint:

Power is proportional to (n_lens / n_medium - 1). Compare 1.5/1.25 - 1 with 1.5 - 1.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The power of a thin lens depends on the refractive index of the lens relative to the medium around it, and on the radii of curvature.

Step 2: Key Formula or Approach:
\[ P = \frac1f = \left(\frac{n_{lens}}{n_{medium}} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \]
The radii do not change when the lens goes into the liquid, so the second bracket is the same.

Step 3: Detailed Explanation:
In air: \(n_{rel} = 1.5\), so the first bracket is \(1.5 - 1 = 0.5\).
In the liquid: \(n_{rel} = \dfrac{1.5}{1.25} = 1.2\), so the first bracket is \(1.2 - 1 = 0.2\).
Ratio of power in liquid to power in air:
\[ \frac{P_{liquid}}{P_{air}} = \frac{0.2}{0.5} = \frac25 \]
Option (D) 5:2 is the inverse ratio. Options (A) and (C) do not follow from 0.2 and 0.5.

Final Answer:
The ratio is 2:5, option (B). \[ \boxed{2:5 \text{ (B)}} \]