Question

A tank is filled with water to a height of 12.5cm. The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be 9.4cm. 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=12.5cm
The apparent depth of the needle in water,h₂=9.4cm
The refractive index of water = \(\mu\)
The value of μ can be obtained as follows:
\(\mu=\frac{h_1}{h_2}=\frac{12.5}{9.4}\approx1.33\)
Hence,the refractive index of water is about 1.33.
Water is replaced by a liquid of refractive index,\(\mu'\)=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 relationship \(\frac{h_1}{y}\)
∴y=\(\frac{h_1}{\mu'}\)=\(\frac{12.5}{1.63}\)=7.67cm
Hence, the new apparent depth of the needle is 7.67cm.It is less than h₂. Therefore,to focus the needle again, the microscope should be moved on.
Distance by which the microscope should be moved up=9.4-7.67=1.73cm.