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.

Options

A. `h/3(1/mu_(1) + 1/mu_(2) + 1/mu_(3))`
B. `3/h(1/mu_(1) + 1/mu_(2) + 1/mu_(3))`
C. `1/h(1/mu_(1) + 1/mu_(2) + 1/mu_(3))`
D. `h(1/mu_(1) + 1/mu_(2) + 1/mu_(3))`

Answer

1

Explanation

The position of imag eof O after refraction from surface-I. if seen from `mu_(2)` , the apparent depth is `h_(1)`
`h_(1)=(mu_(3)h)/(mu_(2)h)`
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 -2. If seen from `mu_(3)` , the apparent depth is `h_(2)`
Similarly, the image formed by medium 2, `O_(2)` acts as an object for Medium 3.
`h_(2) = mu_(3)/mu_(2) ((mu_(2)h)/(mu_(1)3) + h/3) = -h/3 (mu_(3)/mu_(2) + mu_(2)/mu_(1))`
Finally the image formed by sujrface -2 will act as an object for surface -2. If seen from outside, the apparent depth is `h_(3)`
`h_(3) = - 1/mu_(3) [ h/3 + h/3 (mu_(3)/mu_(2) + mu_(3)/mu_(1)) ]`
hence apparent depth of dot is `h/3 (1/mu_(1) + 1/mu_(2) + 1/mu_(3))`