Question

Two immiscible liquids of refractive indices $\frac{8}{5}$ and $\frac{3}{2}$ respectively are put in a beaker as shown in the figure. The height of each column is 6 cm. A coin is placed at the bottom of the beaker. For near normal vision, the apparent depth of the coin is $\frac{\alpha}{4}$ cm. The value of $\alpha$ is ______.

Answer 1

For layered media, the apparent depth $d_{\text{app}}$ is given by:

$d_{\text{app}} = \frac{h_1}{\mu_1} + \frac{h_2}{\mu_2}$

where $h_1 = h_2 = 6 \, \text{cm}$, $\mu_1 = \frac{8}{5}$, and $\mu_2 = \frac{3}{2}$.

Calculating:

$d_{\text{app}} = \frac{6}{8/5} + \frac{6}{3/2} = \frac{6 \times 5}{8} + \frac{6 \times 2}{3} = \frac{30}{8} + 4 = \frac{31}{4} \, \text{cm}$

Thus, $\alpha = 31$.