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Physics Question on Ray optics and optical instruments

A beaker contains water up to a height h1h_1 and kerosene of height h2h_2 above water so that the total height of (water + kerosene) is (h1+h2)(h_1 + h_2). Refractive index of water is μ1\mu_{1} and that of kerosene is μ2\mu_{2}. The apparent shift in the position of the bottom of the beaker when viewed from above is :

A

(1+1μ1)h1−(1+1μ2)h2\left(1+\frac{1}{\mu_{1}}\right)h_{1}-\left(1+\frac{1}{\mu_{2}}\right)h_{2}

B

(1−1μ1)h1+(1−1μ2)h2\left(1-\frac{1}{\mu_{1}}\right)h_{1}+\left(1-\frac{1}{\mu_{2}}\right)h_{2}

C

(1+1μ1)h2−(1+1μ2)h1\left(1+\frac{1}{\mu_{1}}\right)h_{2}-\left(1+\frac{1}{\mu_{2}}\right)h_{1}

D

(1−1μ1)h2+(1−1μ2)h1\left(1-\frac{1}{\mu_{1}}\right)h_{2}+\left(1-\frac{1}{\mu_{2}}\right)h_{1}

Answer

(1−1μ1)h1+(1−1μ2)h2\left(1-\frac{1}{\mu_{1}}\right)h_{1}+\left(1-\frac{1}{\mu_{2}}\right)h_{2}

Explanation

Solution

Apparent shift : =h1(1−1μ1)+h2(1−1μ2).= h_{1}\left(1-\frac{1}{\mu_{1}}\right)+h_{2}\left(1-\frac{1}{\mu_{2}}\right).