**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, *h*1 = 12.5 cm

Apparent depth of the needle in water, *h*2 = 9.4 cm

Refractive index of water = *μ*

The value of *μ*can 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 *h*2. 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