JEE Physics 2025 Question with Solution

Given: Two plano-convex lenses are made of the same glass of refractive index $1.5$. For the first lens, radius of curvature of the first surface is $2 \, \text{cm}$ and it is in air. For the second lens, radius of curvature of the first surface is $3 \, \text{cm}$ and it is immersed in a liquid of refractive index $1.2$.

Find: The ratio $f_1 : f_2$.

For a plano-convex lens, the lens maker's formula is

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

Since for a plano-convex lens $R_2 = \infty$, it becomes

$$\frac{1}{f} = (n - 1)\frac{1}{R_1}$$

For the first lens in air:

$$\frac{1}{f_1} = (1.5 - 1)\frac{1}{2}$$ $$\frac{1}{f_1} = 0.5 \times \frac{1}{2} = 0.25 \, \text{cm}^{-1}$$ $$f_1 = \frac{1}{0.25} = 4 \, \text{cm}$$

For the second lens in liquid, the effective refractive index is

$$n_{\text{eff}} = \frac{1.5}{1.2}$$

Hence,

$$\frac{1}{f_2} = \left(\frac{1.5}{1.2} - 1\right)\frac{1}{3}$$ $$\frac{1}{f_2} = \left(\frac{0.3}{1.2}\right) \times \frac{1}{3} = \frac{1}{12}$$ $$f_2 = 12 \, \text{cm}$$

Therefore,

$$\frac{f_1}{f_2} = \frac{4}{12} = \frac{1}{3}$$

So, the ratio is $1 : 3$. The correct option is B.

Given: A plano-convex lens has focal length inversely proportional to $\left(\mu_{\text{relative}} - 1\right)\frac{1}{R}$.

Find: The ratio $f_1 : f_2$.

For the first lens in air,

$$\frac{1}{f_1} = \left(\frac{1.5}{1} - 1\right)\frac{1}{2} = \frac{0.5}{2} = \frac{1}{4}$$

So,

$$f_1 = 4 \, \text{cm}$$

For the second lens in liquid,

$$\frac{1}{f_2} = \left(\frac{1.5}{1.2} - 1\right)\frac{1}{3} = \frac{0.25}{3} = \frac{1}{12}$$

So,

$$f_2 = 12 \, \text{cm}$$

Thus,

$$f_1 : f_2 = 4 : 12 = 1 : 3$$

Therefore, the correct option is B.