Question

Three immiscible liquids of densities d₁ > d₂ > d₃ and refractive indices µ₁ > µ₂ > µ₃ 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.

Answer 1

Coordinate convention: At the first surface (+ upward and – ve downward)

`mu_2/h^' - mu_1/(-h) = (mu_2 - mu_1)/(oo)` (infinity because the surface is plane),

or `h^' = mu_2/mu_1 h`.

The negative sign shows that it is on the side of the object

`h^'` is the apparent depth of O after refraction from the interface.

The position of image of O after refraction from surface-1. If seen from `mu_2`, the apparent depth is `h_1`

`h_1 = mu_2/mu_1 h/3`

The negative 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-3. If seen from `mu_3`, the apparent depth is `h_2`.

Similarly, the image formed by Medium 2, O₂ acts as an object for MEdium 3.

`h_2 = mu_3/mu_2 (mu_2/mu_1 h/3 + h/3) = - h/3(mu_3/mu_2 + mu_2/mu_1)`

Finally, the image formed by surface-2 will act as an object for surface-2. If seen from the outside, the apparent depth is `h_3`.

`h_3 = - 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)`

Hence apparent depth of dot is `h/3(1/mu_1 + 1/mu_2 + 1/mu_3)`

X_a is apparent depth.