A vessel of depth 2h is half filled with a liquid of refractive index 2√(2) and the upper half with another liquid of refractive index √(2). The liquids are immiscible. The apparent depth of the inner surface of the bottom of vessel will be: (a) h/√(2) (b) (h)/(3√(2)) (c) 3/4h√(2) (d) (h)/(2(√(2)+1))

Answer: (c). For layered liquids, apparent depth = Σ (ti)/(μi). dapp = (h)/(2√(2)) + h/√(2) = (h + 2h)/(2√(2)) = (3h)/(2√(2)) (3h)/(2√(2)) = (3√(2))/(4)h So the answer is (c) 3/4h√(2).

A vessel of depth 2h is half filled with a liquid of refractive index… | JEE Main 2020 (9 Jan, Shift 1)

Q64JEE Main 2020 · 9 JAN · SHIFT 1Medium

Refraction, Glass Slab & Apparent Depth

A vessel of depth 2h is half filled with a liquid of refractive index $2\sqrt{2}$ and the upper half with another liquid of refractive index $\sqrt{2}$ . The liquids are immiscible. The apparent depth of the inner surface of the bottom of vessel will be:

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Answer: (c)

For layered liquids, apparent depth $= \sum \frac{t_i}{\mu_i}$ .
$d_{app} = \frac{h}{2\sqrt{2}} + \frac{h}{\sqrt{2}} = \frac{h + 2h}{2\sqrt{2}} = \frac{3h}{2\sqrt{2}}$
$\frac{3h}{2\sqrt{2}} = \frac{3\sqrt{2}}{4}h$
So the answer is (c) $\frac{3}{4}h\sqrt{2}$ .

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