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?
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:
According to Snell’s law,
∴ μ = 1.33
Hence, the refractive index of water is about 1.33.
When the water is replaced by a liquid of refractive index,
�’ = 1.63
The actual depth of the needle remains the same, but the apparent depth of the needle changes.
EXPLANATION: This is because the light ray coming from the needle inside liquid suffers refraction at the liquid-air interface, as the refractive index of liquid is greater than air the light ray bends away from the normal at the point of incidence due to which the refracted ray appears to be coming from a point at a depth less than the actual depth of the needle.
Let h2’ be the new apparent depth of the needle.
Hence, we can write the relation:
⇒ 
⇒ h2’ = 7.67 cm
Hence, the new apparent depth of the needle is 7.67 cm. It is less than h2.
∴ Distance by which the microscope should move = 9.4 cm - 7.6 cm
= 1.73 cm upwards.