(a) A coin kept inside water [µ=$\frac{4}{3}$] when viewed from air in a vertical direction appears to be raised by 3.0 mm. Find the depth of the coin in water.

(a) Given, µ=$\frac{4}{3}$

Shift = 3.0 mm

We know,

refractive index = $\frac{\text{real depth}}{\text{apparent depth}}$

$\frac{4}{3}$ = $\frac{\text{x}}{\text{apparent depth}}$

apparent depth = $\frac{\text{3x}}{4}$

Shift = real depth - apparent depth

$$\begin{aligned}3 = \text{x} - \frac{\text{3x}}{4} \\[1em] \Rightarrow 3 = \frac{\text{4x - 3x}}{4} \\[1em] \Rightarrow 3 = \frac{\text{x}}{4} \\[1em] \Rightarrow \text{x} = 4 \times 3 = 12\end{aligned}$$

∴ Depth of coin in water = 12 mm

(b) Critical angle [C] = sin^-1 $\frac{1}{μ}$