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 $20 \text{ cm}$. 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 : 5
• D. 5 : 2

Hint:

The power of a lens is directly proportional to $(\mu_{rel} - 1)$. Use this to quickly find the ratio without calculating individual powers.

Answer 1

The Correct Option is A

Step 1: The lens maker's formula for the power $P$ of a lens is given by: \[ P = \left( \frac{\mu_g}{\mu_m} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \] where $\mu_g$ is the refractive index of glass and $\mu_m$ is the refractive index of the surrounding medium.
Step 2: When the lens is in air, $\mu_m = 1$. The power $P_1$ is: \[ P_1 = (1.5 - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) = 0.5 \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Step 3: When the lens is immersed in a liquid of refractive index $1.25$, the power $P_2$ is: \[ P_2 = \left( \frac{1.5}{1.25} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \] \[ P_2 = (1.2 - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) = 0.2 \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Step 4: Calculate the ratio of the power in liquid to the power in air: \[ \frac{P_2}{P_1} = \frac{0.2}{0.5} = \frac{2}{5} \]