A needle is lying at the bottom of a water tank of height $12 \mathrm{~cm}$. The apparent depth of the…
A needle is lying at the bottom of a water tank of height $12 \mathrm{~cm}$. The apparent depth of the needle measured by a microscope is $9 \mathrm{~cm}$. If the water is replaced by a liquid of refractive index of 1.5 of same height, the distance through which the microscope has to be moved to focus the needle again is
$1.2 \mathrm{~cm}$
$1.1 \mathrm{~cm}$
$1 \mathrm{~cm}$
$1.33 \mathrm{~cm}$
Solution
For water, real depth $=12 \mathrm{~cm}$
Apparent depth $=9 \mathrm{~cm}$
Since, refractive index, $\mu_w=\frac{\text { Real depth }}{\text { Apparent depth }}$
$\Rightarrow \quad \mu_w=\frac{12}{9}$
when water is replaced by liquid of refractive index $\mu=1.5$, we have
New apparent depth $=\frac{\text { real depth }}{\mu_l}=\frac{12}{1.5}=8 \mathrm{~cm}$
Hence, required shifted distance $=9-8=1 \mathrm{~cm}$