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 $\frac{\text{h}}{3}$. 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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Key concept: Coordinate convention: At the first surfece (+upward and -ve downward) $\frac{\mu_2}{\text{h}'}-\frac{\mu_1}{(-\text{h})}=\frac{\mu_2-\mu_1}{\infty}$ (infinity because the surface is plane), or $\text{h}'=-\frac{\mu_2}{\mu_1}\text{h}$. The negative sign shown that it is on the side of the object h' is the apparent depth of O after refraction from interface. 
The position of image if O after refraction from surface-1. If seen from $\mu_2$ the apparent depth is $ h_1. \text{h}_1=-\frac{\mu_2}{\mu_1}\frac{\text{h}}{3}$ The nagative sign shows that it is on the side of the object.
Since, the image formed by surface-1 will act as an object for surface-2. IF seen form $\mu_3$, the apparent depth is $h_2$. Similarly, the image formed by Medium 2, $O_2$ acts as an object for Medium 3. $\text{h}_2=-\frac{\mu_3}{\mu_2}\Big(\frac{\mu_2}{\mu_1}\frac{\text{h}}{3}+\frac{\text{h}}{3}\Big)=-\frac{\text{h}}{3}\Big(\frac{\mu_3}{\mu_1}+\frac{\mu_2}{\mu_1}\Big)$ Finally the oimage formed by surface-2 will act as an object for surface-2. If seen from outside, the apparent depth is $h_3. \text{h}_3=-\frac{1}{\mu_3}\bigg[\frac{\text{h}}{3}+\frac{\text{h}}{3}\Big(\frac{\mu_3}{\mu_2}+\frac{\mu_3}{\mu_1}\Big)\bigg]=-\frac{\text{h}}{3}\Big(\frac{1}{\mu_1}+\frac{1}{\mu_2}+\frac{1}{\mu_3}\Big)$ Hence apparent depth of dot is $\frac{\text{h}}{3}\Big(\frac{1}{\mu_1}+\frac{1}{\mu_2}+\frac{1}{\mu_3}\Big)$.Important point:
Apparent depth (distance of final image from final surface) $=\frac{\text{t}_1}{\text{n}_{1\text{real}}}+\frac{\text{t}_2}{\text{n}_{2\text{real}}}+\frac{\text{t}_3}{\text{n}_{3\text{real}}}+\ .....\ +\frac{\text{t}_\text{n}}{\text{n}_\text{nreal}}$
The position of image if O after refraction from surface-1. If seen from $\mu_2$ the apparent depth is $ h_1. \text{h}_1=-\frac{\mu_2}{\mu_1}\frac{\text{h}}{3}$ The nagative sign shows that it is on the side of the object.Since, the image formed by surface-1 will act as an object for surface-2. IF seen form $\mu_3$, the apparent depth is $h_2$. Similarly, the image formed by Medium 2, $O_2$ acts as an object for Medium 3. $\text{h}_2=-\frac{\mu_3}{\mu_2}\Big(\frac{\mu_2}{\mu_1}\frac{\text{h}}{3}+\frac{\text{h}}{3}\Big)=-\frac{\text{h}}{3}\Big(\frac{\mu_3}{\mu_1}+\frac{\mu_2}{\mu_1}\Big)$ Finally the oimage formed by surface-2 will act as an object for surface-2. If seen from outside, the apparent depth is $h_3. \text{h}_3=-\frac{1}{\mu_3}\bigg[\frac{\text{h}}{3}+\frac{\text{h}}{3}\Big(\frac{\mu_3}{\mu_2}+\frac{\mu_3}{\mu_1}\Big)\bigg]=-\frac{\text{h}}{3}\Big(\frac{1}{\mu_1}+\frac{1}{\mu_2}+\frac{1}{\mu_3}\Big)$ Hence apparent depth of dot is $\frac{\text{h}}{3}\Big(\frac{1}{\mu_1}+\frac{1}{\mu_2}+\frac{1}{\mu_3}\Big)$.Important point:
Apparent depth (distance of final image from final surface) $=\frac{\text{t}_1}{\text{n}_{1\text{real}}}+\frac{\text{t}_2}{\text{n}_{2\text{real}}}+\frac{\text{t}_3}{\text{n}_{3\text{real}}}+\ .....\ +\frac{\text{t}_\text{n}}{\text{n}_\text{nreal}}$
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