Question

A vessel is filled with water to a height of $13 \,cm$. The apparent depth of a screw lying at the bottom of the vessel is measured by a microscope to be $8.5 \, cm$. If water is replaced by a liquid of refractive index $1.70$ upto the same height. Find the distance at which microscope have to be moved to focus on the screw again?

• A. $0.85 \,cm$
• B. $0.52 \,cm$
• C. $0.65 \, cm$
• D. $1.02 \, cm$

Answer 1

Answer: (A)

Actual depth of the screw in water $h_{1}=13\, cm$
and apparent depth is $h_{2}=8.5 \,cm$
$\mu_{\text {water }}=\frac{h_{1}}{h_{2}}=\frac{13}{8.5}=1.53$
When water replaced by a liquid of refractive index $\mu^{\prime}=1.70$
then the actual depth remains same, but its apparent depth changes.
Let $h_{3}$ be the new apparent depth of the needle
$\mu^{\prime}=\frac{h_{1}}{h_{3}} $
or $ h_{3}=\frac{h_{1}}{\mu^{\prime}}=\frac{13}{1.7}=7.65$
here, $h_{3}$ is less than $h_{2}$. Thus for focus the microscope should be moved up.
Distance by which microscope should be moved up
$=8.5-7.65=0.85\, cm$