Three immiscible liquids of densities `d_1 gt d_2 gt d_3` and refractive indices `mu_1 gt mu_2 gt 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.

Explanation

Let the apparent depth be `O_(1)` for the object seen form `m_(2)` , then
`" " O_(1)=(mu_(2))/(mu_(1))(h)/(3)`
Since, apparent depth =real depth /refractive index `mu`.
Since , the image formed by Medium 1, `O_(2)` act as an obeject for Medium 2
If seen from `mu_(3)`, the apparent depth is `O_(2)`.
Simiarly, the image formed by Mediuam 2, `O_(2),` act as an object for Medium 3
`" " O_(2)=(mu_(3))/(mu_(2))((h)/(3)+O_(1))`
`" " =(mu_(3
))/(mu_(2))((h)/(3)+(mu_(2))/(mu_(1))(h)/(3))=(h)/(3)((mu_(3))/(mu_(2))+(mu_(3))/(mu_(1)))`
Seen from outside , the apparent height is
`" " O_(3)=(1)
/(mu_(3))((h)/(3)+O_(2))=(1)/(mu_(3))[(h)/(3)+(h)/(3)((mu_(3))/(mu_(2))+(mu_(3))/(mu_(1)))]`
`" " =(h)/(3)((1)/(mu_(1))+(1)/(mu_(2))+(1)/(mu_(3)))`

This is the required expression of apparent depth.