JEE Main 2025 Apr 7 Shift 1, Physics Q23: Refraction at Plane and Spherical surfaces

JEE Main2025Apr 7, Shift 1Physics

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)

Your answer

Solution

Apparent depth of a layer of real depth $h$ and refractive index $μ$ viewed from a rarer medium (air) above is $h / μ$ . The apparent shift for that layer is $h - h / μ = h (1 - \frac{1}{μ})$ .

Total apparent shift (two layers, both contribute):

$shift = H (1 - \frac{1}{1.6}) + 60 (1 - \frac{1}{1.2})$

Compute each parenthetical:

$1 - \frac{1}{1.6} = 1 - \frac{5}{8} = \frac{3}{8}$

$1 - \frac{1}{1.2} = 1 - \frac{5}{6} = \frac{1}{6}$

Set the total equal to 40 cm:

$\frac{3 H}{8} + \frac{60}{6} = 40$

$\frac{3 H}{8} + 10 = 40 \Rightarrow \frac{3 H}{8} = 30 \Rightarrow H = 80$ cm.

$∴$ Answer = 80.

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JEE Main 2025 Apr 7 Shift 1, Physics Q23: Refraction at Plane and Spherical surfaces

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