Q 12-09-091JEE MainJEE Main 2024 (27 Jan, Shift 1)Easy
Two immiscible liquids of refractive indices $\dfrac85$ and $\dfrac32$ respectively are put in a beaker as shown in the figure. The height of each column is $6\ \text{cm}$. A coin is placed at the bottom of the beaker. For near normal vision, the apparent depth of the coin is $\dfrac{\alpha}{4}\ \text{cm}$. The value of $\alpha$ is ______.
Numerical value type. Enter your answer.
Answer: 31
For near normal viewing, the apparent depth is the sum of $t/\mu$ for each layer:
$$d = \frac{6}{3/2} + \frac{6}{8/5} = 4 + 3.75 = 7.75\ \text{cm} = \frac{31}{4}\ \text{cm}$$
So $\alpha = 31$.
Solution by Sreeraj P, M.Sc Physics