Question

Three immiscible liquids of densities \(d_1>d_2>d_3\) and refractive indices \(μ_1>μ_2>μ_3\) 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.

Answer 1

Let the apparent depth be o₁ for the object seen from μ₁ then

\(o_1=\frac{μ_2h}{μ_13}\) 

If seen from μ₃ the apparent depth is o₂.

\(o_2=\frac{μ_3}{μ2}(\frac{h}{3}+o_1)=\frac{μ_3}{μ_2}(\frac{h}{3}+\frac{μ_2h}{μ_13})=\frac{h}{3}(\frac{μ3}{μ2}+\frac{μ_3}{μ_1})\) 

\(o_3=\frac{1}{μ_3}(\frac{h}{3}+o_2)=\frac{1}{μ_3}[\frac{h}{3}+\frac{h}{3}(\frac{μ_3}{μ_2}+\frac{μ_3}{μ_1})]\) 

\(=\frac{h}{3}(\frac{1}{μ_1}+\frac{1}{μ_2}+\frac{1}{μ_3})\)