Q. A vessel of height $2d$ is half-filled with a liquid of refractive index $\sqrt{2}$ and the other half with a liquid of refractive index $n$ (the given liquids are immiscible). Then the apparent depth of the inner surface of the bottom of the vessel (neglecting the thickness of the bottom of the vessel) will be
Explanation
(b) Refractive index $\mu = \frac{Real\ depth\ (d)}{Apparent\ depth\ (x)}$
For 1$^{st}$ liquid, $\sqrt{2} = \frac{d}{x_{1}}$
$\Rightarrow x_{1}\ = \frac{d}{\sqrt{2}}$
Similarly, for 2$^{nd}$ liquid,
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ n = \frac{d}{x_{2}}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x_{2} = \frac{d}{n}$
Total apparent depth $= \ x_{1} + x_{2}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{d}{\sqrt{2}} + \frac{d}{n}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{d(n + \sqrt{2})}{n\sqrt{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.