JEE Physics 2025 Question with Solution

Given: A spherical surface separates media of refractive indices $n_1 = 1$ and $n_2 = 1.5$. From the solution working, the object distance is $u = -0.2 \, \text{m}$ and the radius of curvature is $R = +0.4 \, \text{m}$. Light travels from left to right.

Find: The image distance $v$ and the correct option.

For refraction at a spherical surface,

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

Substituting the given values,

$$\frac{1.5}{v} - \frac{1}{-0.2} = \frac{1.5 - 1}{0.4}$$

So,

$$\frac{1.5}{v} + 5 = 1.25$$

Hence,

$$\frac{1.5}{v} = 1.25 - 5 = -3.75$$

Therefore,

$$v = \frac{1.5}{-3.75} = -0.4 \, \text{m}$$

The negative sign shows that the image is formed to the left of the spherical surface, on the same side as the object.

Therefore, the image is at $0.4 \, \text{m}$ left to the spherical surface.

The solution marks option B, but the extracted working gives $v = -0.4 \, \text{m}$, which does not match the listed option values. Since the solution explicitly declares The Correct Option is B, the answer is recorded as B despite this discrepancy.

Given: Incident light is taken from left to right. Distances measured in the direction of incident light are positive, and those opposite are negative.

Find: The sign and magnitude of the image distance.

• The object is on the left, so $u = -0.2 \, \text{m}$.
• The center of curvature is on the right, so $R = +0.4 \, \text{m}$.
• Refractive indices are $n_1 = 1$ and $n_2 = 1.5$.

Using

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

we get

$$\frac{1.5}{v} - \frac{1}{-0.2} = \frac{0.5}{0.4}$$

Now evaluate the numerical terms:

$$\frac{1}{0.2} = 5, \qquad \frac{0.5}{0.4} = 1.25$$

Thus,

$$\frac{1.5}{v} + 5 = 1.25$$ $$\frac{1.5}{v} = -3.75$$ $$v = -0.4 \, \text{m}$$

Since $v$ is negative, the image lies to the left of the spherical surface and is virtual.

So the computed result is $0.4 \, \text{m}$ left to the spherical surface.