JEE Physics 2025 Question with Solution

Given: A hemispherical vessel is completely filled with a liquid of refractive index $\mu$. A coin is at the lowest point $O$, and the observer is at point $E$ at the level of the vessel.

Find: The minimum value of $\mu$ for which the coin is visible from $E$.

For the coin to be just visible from $E$, the emergent ray at the liquid surface must travel horizontally. Hence the angle of refraction is $90^\circ$, and the angle of incidence at the surface is the critical angle.

Using Snell's law at the liquid-air interface:

$$\mu \sin \theta_i = 1 \cdot \sin 90^\circ$$

So,

$$\mu \sin \theta_i = 1$$ $$\sin \theta_i = \frac{1}{\mu}$$

Now use the geometry of the hemisphere. The limiting ray goes from $O$ to the rim point and forms a right triangle in which the relevant ratio is

$$\sin \theta_i = \frac{R}{R\sqrt{2}} = \frac{1}{\sqrt{2}}$$

Equating the two expressions for $\sin \theta_i$:

$$\frac{1}{\mu} = \frac{1}{\sqrt{2}}$$

Therefore,

$$\mu = \sqrt{2}$$

Thus, the minimum refractive index of the liquid for the coin to be visible from $E$ is $\sqrt{2}$. Therefore, the correct option is A.

Given: The observer is at the same level as the top surface of the hemispherical vessel.

Find: The least refractive index for visibility.

The minimum refractive index corresponds to the limiting condition of visibility. In that case, the refracted ray grazes the surface, so the refracted angle becomes $90^\circ$.

Hence the incident angle inside the liquid equals the critical angle $\theta_c$, where

$$\sin \theta_c = \frac{1}{\mu}$$

From the geometry shown, the limiting ray makes an angle such that

$$\sin \theta_c = \frac{1}{\sqrt{2}}$$

Therefore,

$$\frac{1}{\mu} = \frac{1}{\sqrt{2}}$$ $$\mu = \sqrt{2}$$

So the least value of the refractive index is $\sqrt{2}$.