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?

Answer 1

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 μ can be obtained as follows:

μ = `"h"_1/"h"_2`

= `12.5/9.4`

= 1.329 ≈ 1.33

Hence, the refractive index of water is about 1.33.

Water is replaced by a liquid of refractive index, μ' = 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:

μ' = `"h"_1/"y"`

∴ y = `"h"_1/(μ"'")`

= `12.5/1.63`

= 7.67 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