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The lower half of a vessel of depth 2dcm is filled with a liquid of refractive index $μ_{1}$ and the upper half with a liquid of refractive index $μ_{2}$ . The apparent depth of a vessel seen perpendicularly is

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$2 d [\frac{1}{μ_{1} μ_{2}}]$

$d [\frac{1}{μ_{1}} + \frac{1}{μ_{2}}]$

$d [\frac{μ_{1} μ_{2}}{μ_{1} + μ_{2}}]$

$2 d [\frac{1}{μ_{1}} + \frac{1}{μ_{2}}]$

answer is B.

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Detailed Solution

${(d_{1})}_{a p p} = \frac{d}{μ_{1}}; {(d_{2})}_{a p p} = \frac{d}{μ_{2}}$

So; $d_{a p p} = \frac{d}{μ_{1}} + \frac{d}{μ_{2}} = d [\frac{1}{μ_{1}} + \frac{1}{μ_{2}}]$

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