Question

Three immiscible liquids of densities $ {d_1} > {d_2} > {d_3} $ and refractive indices $ {\mu _1} > {\mu _2} > {\mu _3} $ are put in a breaker. 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.
(A) $ \dfrac{h}{6}\left( {\dfrac{1}{{{\mu _1}}} + \dfrac{1}{{{\mu _2}}} + \dfrac{1}{{{\mu _3}}}} \right) $
(B) $ \dfrac{h}{6}\left( {\dfrac{1}{{{\mu _1}}} - \dfrac{1}{{{\mu _2}}} - \dfrac{1}{{{\mu _3}}}} \right) $
(C) $ \dfrac{h}{3}\left( {\dfrac{1}{{{\mu _1}}} - \dfrac{1}{{{\mu _2}}} - \dfrac{1}{{{\mu _3}}}} \right) $
(D) $ \dfrac{h}{3}\left( {\dfrac{1}{{{\mu _1}}} + \dfrac{1}{{{\mu _2}}} + \dfrac{1}{{{\mu _3}}}} \right) $

Answer 1

Hint We need to find the apparent depth of the dot of due to each of the layers of the liquid separately. Then on adding the values of the apparent depths of the liquid columns we get the apparent depth of the dot for near normal vision.

Formula Used: In this solution we will be using the following formula,
 $\Rightarrow \mu = \dfrac{{{\text{Real depth}}}}{{{\text{Apparent depth}}}} $
where $ \mu $ is the refractive index of the liquid.

Complete step by step answer
According to the question we have three layers of liquid having real depths $ h/3 $ . Now let us consider the apparent depth of the dot due to the liquid is $ {x_1} $ . We are given that the refractive index of the first layer of the liquid is $ {\mu _1} $ . Now using the formula given as, $ \mu = \dfrac{{{\text{Real depth}}}}{{{\text{Apparent depth}}}} $ we can find the apparent depth due the first level of the liquid. So now substituting the values we get,
 $\Rightarrow {\mu _1} = \dfrac{{h/3}}{{{x_1}}} $
We take the $ {x_1} $ from the RHS to the LHS of the equation and we get,
 $\Rightarrow {x_1} = \dfrac{h}{{3{\mu _1}}} $
Similarly again let us consider the apparent depth of the dot due to the liquid is $ {x_2} $ . We are given that the refractive index of the second layer of the liquid is $ {\mu _2} $ . Now using the same formula given as, $ \mu = \dfrac{{{\text{Real depth}}}}{{{\text{Apparent depth}}}} $ we can find the apparent depth due the second level of the liquid. So now substituting the values we get,
 $\Rightarrow {\mu _2} = \dfrac{{h/3}}{{{x_2}}} $
We take the $ {x_2} $ from the RHS to the LHS of the equation and we get,
 $\Rightarrow {x_2} = \dfrac{h}{{3{\mu _2}}} $
Similarly for the third layer we consider the apparent depth of the dot due to the liquid is $ {x_3} $ . We are given that the refractive index of the second layer of the liquid is $ {\mu _3} $ . So substituting we get,
 $\Rightarrow {\mu _3} = \dfrac{{h/3}}{{{x_3}}} $
We take the $ {x_3} $ from the RHS to the LHS of the equation and we get,
 $\Rightarrow {x_3} = \dfrac{h}{{3{\mu _3}}} $
Now the apparent depth of the dot due to the three layers of the liquid is the sum of the three apparent depths. Hence we have,
 $\Rightarrow x = {x_1} + {x_2} + {x_3} $
So substituting the values we get,
 $\Rightarrow x = \dfrac{h}{{3{\mu _1}}} + \dfrac{h}{{3{\mu _2}}} + \dfrac{h}{{3{\mu _3}}} $
Taking common $ \dfrac{h}{3} $ we get,
 $\Rightarrow x = \dfrac{h}{3}\left( {\dfrac{1}{{{\mu _1}}} + \dfrac{1}{{{\mu _2}}} + \dfrac{1}{{{\mu _3}}}} \right) $
So the correct answer is option D.

Note
The real depth of an object beneath the surface of a liquid is the height of the liquid below which the object is present. The light rays due to passing from a denser to a rarer medium, gets refracted. And as a result the object appears to be raised by a height. This new depth is called the apparent depth of the object.