Geometry of the hemispherical vessel

Let the radius of the hemispherical vessel be $R$.

• The coin is at the lowest point $O$ of the hemisphere.
• The observer is at point $E$, which is at the level of the rim.
• For the coin to be just visible from $E$, the emergent ray must come out from the liquid surface in such a way that it reaches $E$.

The minimum refractive index corresponds to the limiting case when the ray from the coin emerges from the liquid surface at the point on the rim, i.e. the emergent ray grazes the surface level toward $E$.

Which surface matters?

The vessel is hemispherical and filled completely. Since the observer is outside and looking from the level of the vessel, the relevant refraction occurs at the plane upper surface of the liquid.

Consider the ray from the coin $O$ to the edge point $A$ of the top surface (rim point). This is the extreme ray that can emerge and reach the observer at level $E$.

Find the angle of incidence inside the liquid

In the vertical cross-section:

• Depth of the coin below the top surface = $R$
• Horizontal distance of rim point from the centerline = $R$

So triangle formed has equal vertical and horizontal legs, hence $\tan i = \frac{R}{R} = 1 \implies i = 45^\circ.$

Here $i$ is the angle of incidence at the liquid-air surface, measured from the normal.

Condition for minimum refractive index

For the ray to just emerge and then travel along the surface level toward $E$, the refracted angle must be $r = 90^\circ.$

Therefore the incident angle must be the critical angle $C$: $C = 45^\circ.$

For a liquid-air interface, $\sin C = \frac{1}{\mu}.$

Substituting $C=45^\circ$: $\frac{1}{\mu} = \sin 45^\circ = \frac{1}{\sqrt{2}}.$

Hence, $\mu = \sqrt{2}.$

Check options

• A: $\dfrac{3}{2}$
• B: $\dfrac{\sqrt{3}}{2}$
• C: $\sqrt{3}$
• D: $\sqrt{2}$

Correct option is $\boxed{\sqrt{2}}.$

Comparison with stored answer

Stored correct answer: D

Our derived answer: D ($\sqrt{2}$)

So the derived answer agrees with the stored answer.