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

As we know,

$\text {Shift} = \text {Real depth} \times (1 – \dfrac{1}{$aμm})

Given,

$\text {Refractive index of the water} = \dfrac{4}{3} \\[0.5em]$

and

Real depth = 12 cm

So, substituting the values in the formula we get,

$\text {Shift} = 12 \times (1 – \dfrac{3}{4}) \\[0.5em] \text {Shift} = 12 \times \dfrac{1}{4} \\[0.5em] \text {Shift} = \dfrac{12}{4} \\[0.5em] \Rightarrow \text {Shift} = 3 \\[0.5em]$

Hence, the coin appears to be raised by a height of 3 cm when seen from vertically above.