For the coin to be visible from E, at least one light ray from O should reach E without being totally internally reflected at the curved interface.

• The extreme ray that might undergo total internal reflection is the one traveling radially outward from O to the hemisphere’s surface at A (the edge of the hemisphere).
• At A, the angle of incidence $i$ is 90° because the ray moves radially outward.

By applying Snell’s law at the interface:

$$\mu \sin i = 1 \cdot \sin r$$

Since TIR occurs if $i > i_c$ (critical angle), the critical angle is given by:

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

For the light to just emerge at E (without total internal reflection):

$$i_c \geq 45^\circ$$

So,

$$\sin 45^\circ = \frac{1}{\sqrt{2}}$$

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

The minimum value of the refractive index so that the coin is visible from E is:

$\mu =\sqrt{2}$