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` upto the same height, by what distance would the microscope have to be moved to focus on the needle again ?

Explanation

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 `mucan` be obtained as follows:
`mu=h_(1)/h_(2)`
`=12.5/1.63= 7.67cm`
Hence, the 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 relation.
`~~ 13.3`
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 relation.
`mu^(')=h_(1)/y`
`therefore y=h_(1)/mu^(')`
`=12.5/1.63=7.67 cm`
Hence, the new apparenet 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 whichthe microscope be moved up =`9.4-7.67=1.73 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