Question

If we fill a tank with water to a height of 12.5 cm, the apparent depth of a needle lying at the bottom of the tank is measured to be 9.4 cm. What is the refractive index of water?

Answer 1

1. Actual depth of the needle in water, $h_1=12.5cm$
Apparent depth of the needle in water, $h_2=9.4cm$
Refractive index of water $=\mu$
The value of $=\mu$ 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,
${\mu }^{{}^{\prime }}=1.63$
Hence, the refractive index of water is about 1.33
Water is replaced by a liquid of refractive index,
${\mu }^{{}^{\prime }}=1.63$

2. 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:
${\mu }^{{}^{\prime }}=\frac{h_1}{y}$
$\therefore y=\frac{h_1}{{\mu }^{{}^{\prime }}}$
$=\frac{12.5}{1.63}=7.67cm$
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.
$\therefore$ Distance by which the microscope should be moved up = 9.4 - 7.67
= 1.73 cm