(a) A coin lies at the bottom of a beaker. Water is poured into the beaker upto a height of 8 cm. Calculate the shift seen in the position of the coin.
(The refractive index of water is 4/3. The width of the glass wall of the beaker is negligible.)

(b) How will the apparent depth be affected if the temperature of water is increased?

(a) Given,

• Real Depth = 8 cm
• Refractive index of water = $\dfrac{4}{3}$

As we know,

$\text {Shift} = \text {Real depth} \times (1 - \dfrac{1}{$\text a\text μ\text w})

where, _aμ_w is the refractive index of the water with respect to air.

So, substituting the values in the formula we get,

$\text {Shift} = 8 \times (1 - \dfrac{3}{4}) \\[1em] \text {Shift} = 8 \times \dfrac{1}{4} \\[1em] \text {Shift} = \dfrac{8}{4} \\[1em] \Rightarrow \text {Shift} = 2 \text{ cm} \\[1em]$

Hence, the coin appears to be raised by a height of 2 cm when seen from vertically above.

(b) With increase in temperature, the refractive index of the medium decreases and lower the refractive index of the medium, less is the shift so on increasing the temperature of water, apparent depth decreases.