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A breaker 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 $(h_{1} + h_{2})$ . Refractive index of water is $μ_{1}$ and that of kerosene is $μ_{2}$ . The apparent shift in the position of the bottom of the beaker when viewed from above is

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$(1 + \frac{1}{μ_{1}}) h_{1} - (1 + \frac{1}{μ_{2}}) h_{2}$

$(1 + \frac{1}{μ_{1}}) h_{2} - (1 + \frac{1}{μ_{2}}) h_{1}$

$(1 - \frac{1}{μ_{1}}) h_{2} + (1 - \frac{1}{μ_{2}}) h_{1}$

$(1 - \frac{1}{μ_{1}}) h_{1} + (1 - \frac{1}{μ_{2}}) h_{2}$

answer is D.

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Detailed Solution

$S = t (1 - \frac{1}{μ})$

$= (1 - \frac{1}{μ_{1}}) h_{1} + (1 - \frac{1}{μ_{2}}) h_{2}$

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