Two immiscible liquids of refractive indices 8/5 and 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 (α)/(4) cm. The value of α is _________. Numerical answer type. Work it out, then open the answer.

Answer: 31. Apparent depth = (t₁)/(μ₁) + (t₂)/(μ₂). d = 6/8/5 + 6/3/2 = 3.75 + 4 = 7.75 cm 7.75 = 31/4 ⇒ α = 31 So α = 31.

Two immiscible liquids of refractive indices 8/5 and 3/2 respectively… | JEE Main 2024 (27 Jan, Shift 1)

Q29JEE Main 2024 · 27 JAN · SHIFT 1Medium

Refraction, Glass Slab & Apparent Depth

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 $6\text{ cm}$ . A coin is placed at the bottom of the beaker. For near normal vision, the apparent depth of the coin is $\frac{\alpha}{4}\text{ cm}$ . The value of $\alpha$ is _________.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 31

Apparent depth $= \frac{t_1}{\mu_1} + \frac{t_2}{\mu_2}$ .
$d = \frac{6}{8/5} + \frac{6}{3/2} = 3.75 + 4 = 7.75\text{ cm}$
$7.75 = \frac{31}{4} \Rightarrow \alpha = 31$
So $\alpha = 31$ .

Report a mistake

Refraction, Glass Slab & Apparent Depth question from JEE Main 2024 (27 Jan, Shift 1)

More questions on Refraction, Glass Slab & Apparent Depth