Question

A tank is filled with water to a height of $12.5\text{cm}$. The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be $9.4\text{cm}$. 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

When tank is filled with water $\left(n_w\right)$
Given,
Actual depth of needle/object $\left(h\right)=12.5\text{cm}$
Apparent depth of needle/object $\left(h_1\right)=9.4\text{cm}$
Refractive index of liquid $=\frac{\text{Real depth}}{\text{appartent depth}}$
$n_w=\frac{h}{h_1}=\frac{12.5}{9.4}$
$n_w=1.33$
When water is replaced by a liquid of refractive index 1.63 up to the same height
Actual depth of needle/object $\left(h\right)=12.5\text{cm}$
Apparent depth of needle/object = $h_2$
$n_L=\frac{\text{Real depth}}{\text{apparent depth}}$
$1.63=\frac{12.5}{h_2}$
$h_2=\frac{12.5}{1.63}=7.67\text{cm}$
Displacement of microscope to see the needle at position $h_1$:
$\Delta h=h_1-h_2$
$\Delta h=9.4-7.67=1.73\text{cm}$
Final Answer : $1.33;1.7\text{cm}$