JEE Physics 2025 Question with Solution

Given: A convex lens of glass has refractive index $\mu_g = 1.5$ and focal length in air $f_1 = 24 \, \text{cm}$. The refractive index of water is $\mu_w = 1.33$.

Find: The new focal length of the lens when immersed in water.

For a lens in a medium, lens maker's formula is

$$\frac{1}{f} = \left( \frac{\mu_l}{\mu_m} - 1 \right)\left( \frac{1}{R_1} - \frac{1}{R_2} \right)$$

For the same lens, $\left( \frac{1}{R_1} - \frac{1}{R_2} \right)$ remains constant.

In air,

$$\frac{1}{f_1} = \left( \frac{\mu_g}{\mu_a} - 1 \right)\left( \frac{1}{R_1} - \frac{1}{R_2} \right)$$

with $\mu_a = 1$.

In water,

$$\frac{1}{f_2} = \left( \frac{\mu_g}{\mu_w} - 1 \right)\left( \frac{1}{R_1} - \frac{1}{R_2} \right)$$

Taking ratio,

$$\frac{f_2}{f_1} = \frac{\left( \frac{\mu_g}{\mu_a} - 1 \right)}{\left( \frac{\mu_g}{\mu_w} - 1 \right)}$$

Substitute the values,

$$\frac{f_2}{24} = \frac{\left( \frac{1.5}{1.0} - 1 \right)}{\left( \frac{1.5}{1.33} - 1 \right)}$$ $$\frac{f_2}{24} = \frac{0.5}{1.1278 - 1} = \frac{0.5}{0.1278} \approx 3.91$$ $$f_2 = 24 \times 3.91 = 93.84 \approx 96 \, \text{cm}$$

Therefore, the focal length in water is approximately $96 \, \text{cm}$. The correct option is B.

Given: The focal length in air is $24 \, \text{cm}$, refractive index in air is taken as $1$, and the lens is placed in water of refractive index $1.33$.

Find: The changed focal length in water.

Using the relation shown in the solution,

$$\frac{1}{f'} = \left( \frac{\mu_{\text{water}} - 1}{\mu_{\text{air}} - 1} \right) \cdot \frac{1}{f}$$

Substitute the values,

$$\frac{1}{f'} = \left( \frac{1.33 - 1}{1.5 - 1} \right) \cdot \frac{1}{24}$$ $$\frac{1}{f'} = \frac{0.33}{0.5} \cdot \frac{1}{24}$$

This gives a focal length close to $96 \, \text{cm}$ as concluded in the provided working.

Therefore, the focal length changes to $96 \, \text{cm}$, so the correct option is B.