Question

The radius of curvature of each surface of a convex lens having refractive index 1.8 is 20 cm. The lens is now immersed in a liquid of refractive index 1.5. The ratio of power of lens in air to its power in the liquid will be:

Hint:

When a lens is immersed in a medium with a different refractive index, the power of the lens changes based on the ratio of the refractive indices.

Answer 1

Correct Answer: 4

The formula for the power of a lens is given by: \[ P = \frac{1}{f} \] Where \( f \) is the focal length of the lens. The focal length is related to the radius of curvature (\( R \)) and refractive index \( n \) by the lens-maker's formula: \[ \frac{1}{f} = (n - 1)\left( \frac{1}{R_1} - \frac{1}{R_2} \right) \] For a convex lens, \( R_1 = R \) and \( R_2 = -R \), so the focal length in air is: \[ \frac{1}{f_{\text{air}}} = (1.8 - 1) \left( \frac{1}{R} + \frac{1}{R} \right) = 0.8 \times \frac{2}{R} \] Thus: \[ f_{\text{air}} = \frac{R}{1.6} \] Now, when the lens is immersed in a liquid of refractive index 1.5, the power changes. The new focal length is given by: \[ \frac{1}{f_{\text{liquid}}} = (1.8 - 1.5) \left( \frac{1}{R} + \frac{1}{R} \right) = 0.3 \times \frac{2}{R} \] Thus: \[ f_{\text{liquid}} = \frac{R}{0.6} \] The ratio of the power in air to the power in liquid is: \[ \text{Ratio} = \frac{P_{\text{air}}}{P_{\text{liquid}}} = \frac{1/f_{\text{air}}}{1/f_{\text{liquid}}} = \frac{R/1.6}{R/0.6} = 2 \] Thus, the ratio of the power in air to the power in liquid is 4.