Three immiscible liquids of densities d_1 d | vedclass

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Question

(A) The apparent depth of an object viewed through multiple layers of different refractive indices is given by the sum of the apparent depths of each

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 refractive indices is given by the sum of the apparent depths of each layer.For a single layer of thickness $t$ and refractive index $\mu$,the apparent depth is $t' = \frac{t}{\mu}$.Here,each liquid has a thickness $t = \frac{h}{3}$.Therefore,the total apparent depth $d_{app}$ is the sum of the apparent depths of the three layers:$d_{app} = \frac{h/3}{\mu_1} + \frac{h/3}{\mu_2} + \frac{h/3}{\mu_3}$$d_{app} = \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 $\frac{h}{3}$. $A$ dot is made at the bottom of the beaker. For near normal vision,find the apparent depth of the dot.

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