A hemispherical vessel is completely filled with a liquid of refractive index μ. A small coin is kept at the lowest point (O) of the vessel as shown in figure. The minimum value of the refractive index of the liquid so that a person can see the coin from point E (at the level of the vessel) is _________.

Solution

Physics sir · 4 steps

To see the coin from E, a ray must leave the surface grazing it, so look for the ray that hits at the edge B.

1. Grazing emergence. E is level with the liquid surface, so a ray from O reaches E only if it leaves the surface at 90° to the normal. That means it must hit the surface at exactly the critical angle C, where μ sin C = 1.
2. Largest possible angle. O is at depth R below the centre C, and the edge B is a distance R from C. So the ray OB makes 45° with the vertical normal at B. No ray from O can hit the surface at a larger angle than this.
3. Smallest μ. A smaller μ needs a larger critical angle. The largest angle available is 45°, so the minimum μ comes from setting C = 45°.

   μₘᵢₙ = 1/(sin 45°) = √2

4. So. For any μ below √2, even the ray to B emerges tilted upward and misses E. The minimum refractive index is √2, option (d).

The idea to keep. Seeing an object from eye level at the surface means grazing emergence: the incidence angle must equal the critical angle. The minimum μ comes from the largest incidence angle the geometry allows.

Answer from the NTA final answer key. Solution by PrepWiser.