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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. |
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Answer» Here, real depth of the dot under liquid of density `d_(1)` is `h//3`. If `x_(1)` is its apparent depth, when seen from air, then from `mu_(1) = (h//3)/(x_(1)) or x_(1) = (h)/(3 mu_(1))` Similarly, apparent depths of the dot when seen from air through two other liquids are `x_(2) = (h)/(3 mu_(2)) and x_(3) = (h)/(3 mu_(3))` `:.` Apparent depth of the dot seen from air through the three liquids is `x = x_(1) + x_(2) + x_(3) = (h)/(3mu_(1)) + (h)/(3 mu_(2)) + (h)/(3 mu_(3)) = (h)/(3)[(1)/(mu_(1)) + (1)/(mu_(2)) + (1)/(mu_(3))]` |
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