Three immiscible liquids of densities d_1 d | vedclass

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(A) The apparent depth of an object viewed through multiple layers of different media is given by the sum of the apparent depths of each layer.For a l

Vedclass · Accepted Answer

$\frac{h}{3} \left( \frac{1}{\mu_1} + \frac{1}{\mu_2} + \frac{1}{\mu_3} \right)$

Vedclass · Answer

(A) The apparent depth of an object viewed through multiple layers of different media is given by the sum of the apparent depths of each layer.For a layer of thickness $t$ and refractive index $\mu$,the apparent depth is $t' = \frac{t}{\mu}$.Here,there are three layers,each with a thickness of $t = \frac{h}{3}$.The apparent depth of the first layer is $d_1' = \frac{h/3}{\mu_1} = \frac{h}{3\mu_1}$.The apparent depth of the second layer is $d_2' = \frac{h/3}{\mu_2} = \frac{h}{3\mu_2}$.The apparent depth of the third layer is $d_3' = \frac{h/3}{\mu_3} = \frac{h}{3\mu_3}$.The total apparent depth $D_{app}$ is the sum of these individual apparent depths:$D_{app} = \frac{h}{3\mu_1} + \frac{h}{3\mu_2} + \frac{h}{3\mu_3} = \frac{h}{3} \left( \frac{1}{\mu_1} + \frac{1}{\mu_2} + \frac{1}{\mu_3} \right)$.

Three immiscible liquids of densities $d_1 > d_2 > d_3$ and refractive indices $\mu_1 > \mu_2 > \mu_3$ are put in a beaker. The height of each liquid column is $h/3$. $A$ dot is made at the bottom of the beaker. For near normal vision,find the apparent depth of the dot.

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