JEE Physics 2025 Question with Solution

Given: A thin plano-convex lens has refractive index $n_{lens} = 1.5$ and is immersed in a liquid of refractive index $n_{medium} = 1.2$. The plane side is silver coated, and the combination behaves like a concave mirror of focal length $F = 0.2 \, \text{m}$.

Find: The radius of curvature $R$ of the curved surface.

For a plano-convex lens in a medium, the lens maker formula is

$$\frac{1}{f} = \left(\frac{n_{lens}}{n_{medium}} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$

Here, $R_1 = R$ and $R_2 = \infty$, so

$$\frac{1}{f} = \left(\frac{1.5}{1.2} - 1\right)\left(\frac{1}{R} - 0\right)$$ $$\frac{1}{f} = \left(1.25 - 1\right)\frac{1}{R} = \frac{0.25}{R} = \frac{1}{4R}$$

Therefore,

$$f = 4R$$

When the plane side is silvered, the effective focal length of the system becomes

$$\frac{1}{F} = \frac{2}{f}$$

Using $F = 0.2 \, \text{m}$,

$$\frac{1}{0.2} = \frac{2}{4R} = \frac{1}{2R}$$ $$5 = \frac{1}{2R}$$ $$2R = 0.2$$ $$R = 0.1 \, \text{m}$$

Therefore, the radius of curvature of the curved surface is $0.10 \, \text{m}$. The correct option is B.

Given: The lens material has refractive index $1.5$ and the surrounding liquid has refractive index $1.2$.

Find: Radius of curvature $R$.

The hint states that for immersion problems, the effective refractive index is the ratio

$$\mu_{rel} = \frac{1.5}{1.2} = 1.25$$

Hence the lens power in the liquid is determined by

$$\frac{1}{f} = (\mu_{rel} - 1)\left(\frac{1}{R} - \frac{1}{\infty}\right)$$ $$\frac{1}{f} = (1.25 - 1)\frac{1}{R} = \frac{0.25}{R}$$ $$f = 4R$$

Because the plane face is silver coated, the light passes through the lens, reflects, and passes through the lens again. So the effective focal length becomes half of the lens focal length:

$$F = \frac{f}{2}$$

Given

$$F = 0.2 \, \text{m}$$

so

$$\frac{f}{2} = 0.2$$ $$f = 0.4 \, \text{m}$$

Now using $f = 4R$,

$$4R = 0.4$$ $$R = 0.1 \, \text{m}$$

Therefore, the required radius of curvature is $0.10 \, \text{m}$.