Search any question & find its solution
Question:
Answered & Verified by Expert
A vessel of height $2 d$ 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
Options:
Solution:
1567 Upvotes
Verified Answer
The correct answer is:
$\frac{d(n+\sqrt{2})}{n \sqrt{2}}$
Refractive index $\mu=\frac{\text { Real depth }(d)}{\text { Apparent depth }(x)}$
For 1st liquid, $\sqrt{2}=\frac{d}{x_1}$
$\Rightarrow \quad x_1=\frac{d}{\sqrt{2}}$
Similarly, for 2nd liquid,
$n=\frac{d}{x_2}$
$\Rightarrow \quad x_2=\frac{d}{n}$
Total apparent depth $=x_1+x_2$
$=\frac{d}{\sqrt{2}}+\frac{d}{n}=\frac{d(n+\sqrt{2})}{n \sqrt{2}}$
For 1st liquid, $\sqrt{2}=\frac{d}{x_1}$
$\Rightarrow \quad x_1=\frac{d}{\sqrt{2}}$
Similarly, for 2nd liquid,
$n=\frac{d}{x_2}$
$\Rightarrow \quad x_2=\frac{d}{n}$
Total apparent depth $=x_1+x_2$
$=\frac{d}{\sqrt{2}}+\frac{d}{n}=\frac{d(n+\sqrt{2})}{n \sqrt{2}}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.