JEE Physics 2025 Question with Solution

Given: A thin convex lens of refractive index $\mu_2$ is kept in a liquid of refractive index $\mu_1$ with $\mu_1 < \mu_2$. The magnitudes of radii of curvature are $|R_1|$ and $|R_2|$, and the second surface is silver polished.

Find: The object distance on the optic axis for which a real and inverted image is formed at the same place.

The solution uses the lens-maker relation for a lens in a medium:

$$\frac{1}{f} = \left(\frac{\mu_2}{\mu_1} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$

Equivalently,

$$\frac{1}{f'} = \left(\frac{\mu_2 - \mu_1}{\mu_1}\right)\left(\frac{1}{|R_1|} - \frac{1}{|R_2|}\right)$$

After the second surface is silvered, the arrangement behaves as a lens-mirror system. Using the relation quoted in the solution,

$$\frac{1}{F} = \frac{2}{f'} + \frac{1}{r}$$

where for the silvered second surface,

$$r = -|R_2|$$

Substituting,

$$\frac{1}{F} = \frac{2 \mu_1}{(\mu_2 - \mu_1)}\left(\frac{1}{|R_1|} - \frac{1}{|R_2|}\right) - \frac{1}{|R_2|}$$

The extracted solution then simplifies this to the required object distance for the autoconjugate condition, stated there as $u = F$:

$$u = \frac{\mu_1 |R_1| |R_2|}{\mu_2 (|R_1| + |R_2|) - \mu_1 |R_2|}$$

Therefore, the correct option is B.

Given: The image must coincide with the object after refraction, reflection from the silvered second surface, and refraction again.

Find: The required object position.

The solution states that for coincidence of object and image in this lens-mirror system, the relevant condition is expressed as

$$v = u$$

and for the system the placement corresponds to the effective focal distance obtained after combining the lens action with the mirror action.

From the solution, the final simplified result is

$$u = \frac{\mu_1 |R_1| |R_2|}{\mu_2 (|R_1| + |R_2|) - \mu_1 |R_2|}$$

This matches option B exactly.

Hence the object should be placed at

$$\frac{\mu_1 |R_1| |R_2|}{\mu_2 (|R_1| + |R_2|) - \mu_1 |R_2|}$$

from the lens on the optic axis.