Question

A baker contains water up to a height ${{h}_{1}}$ and kerosene up to a height ${{h}_{2}}$ above water so that the total height of (water $+$ kerosene) is $\left( {{h}_{1}}+{{h}_{2}} \right)$. Refractive index of water is ${{\mu }_{1}}$ and that of kerosene is ${{\mu }_{2}}$. The apparent shift in the position of the bottom of the beaker when viewed from above is:
$\begin{aligned} & \text{A}\text{. }\left( 1-\dfrac{1}{{{\mu }_{1}}} \right){{h}_{1}}+\left( 1-\dfrac{1}{{{\mu }_{2}}} \right){{h}_{2}} \\\ & \text{B}\text{. }\left( 1+\dfrac{1}{{{\mu }_{1}}} \right){{h}_{2}}-\left( 1+\dfrac{1}{{{\mu }_{2}}} \right){{h}_{1}} \\\ & \text{C}\text{. }\left( 1-\dfrac{1}{{{\mu }_{1}}} \right){{h}_{2}}+\left( 1-\dfrac{1}{{{\mu }_{2}}} \right){{h}_{1}} \\\ & \text{D}\text{. }\left( 1+\dfrac{1}{{{\mu }_{1}}} \right){{h}_{1}}-\left( 1+\dfrac{1}{{{\mu }_{2}}} \right){{h}_{2}} \\\ \end{aligned}$

Answer 1

Here we will use the concept of refraction and the apparent shift to find the solution to this question. Obtain the expression for apparent shift and find the required values for both the medium. The total apparent shift will be due to the apparent shift of both the medium.

Complete answer:
Apparent shift or the normal shift can be defined as the shift in the position of an object along the normal during refraction.
Consider a medium A that has a refractive index of ${{n}_{1}}$ and depth or height t is surrounded by another medium B of refractive index ${{n}_{2}}$. Assuming medium A is denser than the medium B, the apparent shift can be given as,
$AS=t\left( 1-\dfrac{{{n}_{2}}}{{{n}_{1}}} \right)$
Given in the question, in a beaker the height of water is ${{h}_{1}}$ and the height of kerosene above the water is ${{h}_{2}}$.
The refractive index of water is ${{\mu }_{1}}$ and the refractive index of kerosene is ${{\mu }_{2}}$.
So, the apparent shift of the bottom of the beaker is the total apparent shift due to the water and the kerosene.
The apparent shift of the bottom of the beaker due to the water will be,
$A{{S}_{w}}={{h}_{1}}\left( 1-\dfrac{1}{{{\mu }_{1}}} \right)$
The apparent shift of the bottom of the beaker due to kerosene is,
$A{{S}_{k}}={{h}_{2}}\left( 1-\dfrac{1}{{{\mu }_{2}}} \right)$
The total apparent shift will be,
$\begin{aligned} & AS=A{{S}_{w}}+A{{S}_{k}} \\\ & AS={{h}_{1}}\left( 1-\dfrac{1}{{{\mu }_{1}}} \right)+{{h}_{2}}\left( 1-\dfrac{1}{{{\mu }_{2}}} \right) \\\ \end{aligned}$
So, the total apparent shift of the bottom of the beaker will be, ${{h}_{1}}\left( 1-\dfrac{1}{{{\mu }_{1}}} \right)+{{h}_{2}}\left( 1-\dfrac{1}{{{\mu }_{2}}} \right)$.

The correct option is (B).

Note:
In the above solution we have found out the apparent shift of the medium with respect to air because we are observing the system from air. The refractive index of air can be expressed as unity or $\mu =1$.