Question

A beaker container water up to a height ${h_1}$ and kerosene of height ${h_2}$ above water so that 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:
A. $\left( {1 + \dfrac{1}{{{\mu _1}}}} \right){h_1} - \left( {1 + \dfrac{1}{{{\mu _2}}}} \right){h_2}$
B. $\left( {1 - \dfrac{1}{{{\mu _1}}}} \right){h_1} + \left( {1 - \dfrac{1}{{{\mu _2}}}} \right){h_2}$
C. $\left( {1 + \dfrac{1}{{{\mu _1}}}} \right){h_2} - \left( {1 + \dfrac{1}{{{\mu _2}}}} \right){h_1}$
D. $\left( {1 - \dfrac{1}{{{\mu _1}}}} \right){h_2} + \left( {1 - \dfrac{1}{{{\mu _2}}}} \right){h_1}$

Answer 1

Use the formula for the apparent shift in the position of an object due to a liquid. This formula gives the relation between the apparent shift in position of the object, real position of the object and refractive index of the liquid. Rewrite this formula for apparent shift due to water and kerosene. Then determine the total shift in position of bottom of the beaker.

Formula used:
The apparent shift $d$ in position of an object from the surface of a liquid is given by
$d = h\left( {1 - \dfrac{1}{\mu }} \right)$ …… (1)
Here, $h$ is the real depth of the object from the surface of the liquid and $\mu$ is the refractive index of the liquid.

Complete step by step answer:
We have given that in a beaker the height of water from the bottom of the beaker is ${h_1}$ and the height of kerosene from the surface of the water is ${h_2}$.We have also given that the refractive index of water is ${\mu _1}$ and the refractive index of kerosene is ${\mu _2}$.We have asked to calculate the total apparent shift in the position of bottom of the beaker due to water and kerosene.

Let us first determine the apparent shift in the position of bottom of the beaker due to water.Rewrite equation (1) for the apparent shift ${d_1}$ in the position of the bottom of the beaker due to water.
${d_1} = {h_1}\left( {1 - \dfrac{1}{{{\mu _1}}}} \right)$
Now let us determine the apparent shift in the position of the bottom of the beaker due to kerosene.Rewrite equation (1) for the apparent shift ${d_2}$ in the position of the bottom of the beaker due to kerosene.
${d_2} = {h_2}\left( {1 - \dfrac{1}{{{\mu _2}}}} \right)$

The total shift $d$ in the position of the bottom of beaker is the sum of the apparent shift in the position of the bottom of the beaker due to water and kerosene.
$d = {d_1} + {d_2}$
Substitute ${h_1}\left( {1 - \dfrac{1}{{{\mu _1}}}} \right)$ for ${d_1}$ and ${h_2}\left( {1 - \dfrac{1}{{{\mu _2}}}} \right)$ for ${d_2}$ in the above equation.
$\therefore d = {h_1}\left( {1 - \dfrac{1}{{{\mu _1}}}} \right) + {h_2}\left( {1 - \dfrac{1}{{{\mu _2}}}} \right)$
Therefore, the total shift in the position of the bottom of the beaker due to water and kerosene is ${h_1}\left( {1 - \dfrac{1}{{{\mu _1}}}} \right) + {h_2}\left( {1 - \dfrac{1}{{{\mu _2}}}} \right)$.

Hence, the correct option is B.

Note: The students can also solve the same question by using another formula. The students can use the formula for the refractive index in terms of real depth and apparent depth and then convert this formula to determine the apparent shift in the position of the bottom of the beaker. The obtained formula will also give the same value of the total apparent shift in position of the bottom of the beaker.