# A tank is filled with water to a height of 12.5 cm.

Question:

A tank is filled with water to a height of 12.5 cm. The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be 9.4 cm. What is the refractive index of water? If water is replaced by a liquid of refractive index 1.63 up to the same height, by what distance would the microscope have to be moved to focus on the needle again?

Solution:

Actual depth of the needle in water, h1 = 12.5 cm

Apparent depth of the needle in water, h2 = 9.4 cm

Refractive index of water = μ

The value of μcan be obtained as follows:

$\mu=\frac{h_{1}}{h_{2}}$

$=\frac{12.5}{9.4} \approx 1.33$

Hence, the refractive index of water is about 1.33.

Water is replaced by a liquid of refractive index, $\mu^{\prime}=1.63$

The actual depth of the needle remains the same, but its apparent depth changes. Let y be the new apparent depth of the needle. Hence, we can write the relation:

$\mu^{\prime}=\frac{h_{1}}{y}$

$\therefore y=\frac{h_{1}}{\mu^{\prime}}$

$=\frac{12.5}{1.63}=7.67 \mathrm{~cm}$

Hence, the new apparent depth of the needle is 7.67 cm. It is less than h2. Therefore, to focus the needle again, the microscope should be moved up.

Distance by which the microscope should be moved up = 9.4 − 7.67

= 1.73 cm