Saved Bookmarks
| 1. |
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. |
|
Answer» SOLUTION : (Hint : the image FORMED by FIRST medium act as an object for second medium) Let the apparent depth be `O_1` for the object seen from`O_1 =mu_2/mu_1 h/3` image formed by medium 1, O ACTS as an object for medium2. It is seen from `M_3`, the apparent depth is `O_2`.Similarly, the image found by medium 2, `O_2` act as an object formedium 3 `O_2=mu_3/mu_2(h/3+O_1)` `O_3=mu_3(h/3 + O_2)` putting value of `O_2 and O_1` `O_3=h/3(1/mu_1+1/mu_3+1/mu_3)` |
|