Question

A coin is placed at the bottom of a beaker containing water (refractive index = $\frac{4}{3}$) to a depth of 12 cm. By what height the coin appears to be raised when seen from vertically above?

Answer 1

Refractive index of the water, ${\mu }_w$ = $\frac{4}{3}$

Real depth at which the coin is placed = 12 cm

Shift in the image = ?

Shift = Real depth $\times$ (1 - $\frac{1}{\mu }$)

Shift = 12 $\times$ (1 - $\frac{3}{4}$)

R = $\frac{12}{4}$ = 3 cm