Physics · Geometrical Optics

JEE Main 2025 — 7 April, Morning Shift — Question 67

A container contains a liquid with refractive index of 1.2 up to a height of 60 cm and another liquid having refractive index 1.6 is added to height H above first liquid. If viewed from above, the apparent shift in the position of bottom of container is 40 cm . The value of $H$ is cm (Consider liquids are immiscible).

Answer: 80

Numerical answer — enter this value.

Step-by-step solution

$\Delta t=60\left(1-\frac{1}{1.2}\right)+H\left(1-\frac{1}{1.6}\right)$

\Rightarrow H=80 \mathrm{~cm}

\begin{aligned} & \Rightarrow \quad 40=\left(\frac{1}{6}\right) 60+H\left(\frac{3}{8}\right) & \Rightarrow \quad 30=H \times \frac{3}{8} & 60 \mathrm{~cm} \end{aligned}

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Exam: JEE Main 2025
Paper: 7 April, Morning Shift
Subject: Physics
Chapter: Geometrical Optics
Topic: Apparent Depth, Glass Slab, Prism and Dispersion

Answer: 80

Step-by-step solution

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