JEE Physics 2025 Question with Solution

Given: $\mu_1 = 1$ for air, $\mu_2 = 1.5$ for glass, $v = +200 \, \text{cm}$, and $R = +50 \, \text{cm}$.

Find: The magnitude of the object distance from the spherical refracting surface.

For refraction at a spherical surface, use

$$\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$$

Substituting the given values,

$$\frac{1.5}{200} - \frac{1}{u} = \frac{1.5 - 1}{50}$$ $$\frac{1.5}{200} - \frac{1}{u} = \frac{0.5}{50} = \frac{1}{100}$$ $$\frac{3}{400} - \frac{1}{u} = \frac{1}{100}$$

Rearranging,

$$\frac{1}{u} = \frac{3}{400} - \frac{1}{100} = \frac{3}{400} - \frac{4}{400} = -\frac{1}{400}$$

Hence,

$$u = -400 \, \text{cm}$$

The negative sign shows that the object is on the incident side. Therefore, the magnitude is

$$|u| = 400 \, \text{cm} = 4 \, \text{m}$$

Therefore, the required distance is $4$.

Given: Light goes from air to glass. Take the Cartesian sign convention where distances measured in the direction of incident light are positive.

Find: The magnitude of the distance of the source from the glass surface.

Because the image is formed inside the glass, $v$ is positive:

$$v = +200 \, \text{cm}$$

The spherical surface is convex towards the object, so

$$R = +50 \, \text{cm}$$

Also,

$$\mu_1 = 1, \qquad \mu_2 = 1.5$$

Now apply

$$\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$$

Thus,

$$\frac{1.5}{200} - \frac{1}{u} = \frac{0.5}{50}$$ $$\frac{1.5}{200} - \frac{1}{u} = \frac{1}{100}$$

Move terms carefully:

$$\frac{1}{u} = \frac{1.5}{200} - \frac{1}{100}$$ $$\frac{1}{u} = \frac{1.5}{200} - \frac{2}{200} = -\frac{0.5}{200} = -\frac{1}{400}$$

So,

$$u = -400 \, \text{cm}$$

Hence the required magnitude is

$$400 \, \text{cm} = 4 \, \text{m}$$

So the numerical value of the answer is $4$.