A container contains a liquid with refractive index of 1.2 up to a height of 60 cm and another liquid having…
(Consider liquids are immisible)
Solution

$y=$ apparent depth of bottom
$\frac{\mathrm{y}}{1}=\frac{\mathrm{H}}{1.6}+\frac{60}{1.2}$
Shift $=40$
$H+60-y=40$
$\mathrm{H}+60-\frac{\mathrm{H}}{1.6}-\frac{60}{1.2}=40$
$\frac{6}{16} \mathrm{H}=30$
$\mathrm{H}=80 \mathrm{~cm}$
Asked in: JEE Main 2025 (07 Apr Shift 1)