A beaker contains water up to a height $ h_{1} $ and kerosene of height $ h_{2} $ above water so that the total height of (water $ + $ kerosene $ ) $ is $ \left(h_{1}+h_{2}\right) . $ Refractive index of water is $ \mu_{1} $ and that of kerosene is $ \mu_{2} . $ The apparent shift in the position of the bottom of the beaker when viewed from above is

Question

A beaker contains water up to a height $ h_{1} $ and kerosene of height $ h_{2} $ above water so that the total height of (water $ + $ kerosene $ ) $ is $ \left(h_{1}+h_{2}\right) . $ Refractive index of water is $ \mu_{1} $ and that of kerosene is $ \mu_{2} . $ The apparent shift in the position of the bottom of the beaker when viewed from above is, $ \left(1-\frac{1}{\mu_{1}}\right) h_{2}+\left(1-\frac{1}{\mu_{2}}\right) h_{1} $, $ \left(1+\frac{1}{\mu_{1}}\right) h_{1}+\left(1+\frac{1}{\mu_{2}}\right) h_{2} $, $ \left(1-\frac{1}{\mu_{1}}\right) h_{1}+\left(1-\frac{1}{\mu_{2}}\right) h_{2} $, $ \left(1+\frac{1}{\mu_{1}}\right) h_{2}-\left(1+\frac{1}{\mu_{2}}\right) h_{1} $

MARKS App · Accepted Answer

Apparent shift due to a transparent material of thickness h is give by, Δ h ⁡ = 1 - 1 μ h ∴ Apparent shift produced by water △ h 1 = 1 - 1 μ 1 h 1 And apparent shift produced by kerosene △ h 2 = 1 - 1 μ 2 h 2 △ h = △ h 1 + △ h 2 = 1 - 1 μ 1 h 1 + 1 - 1 μ 2 h 2

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