Question

A beaker contains water up to a height $h_1$ and kerosene up to height $h_2$ above water. 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 the top is

• A. $(1-\frac{1}{{\mu }_1})h_1+(1-\frac{1}{{\mu }_2})h_2$
• B. $(1+\frac{1}{{\mu }_1})h_1-(1+\frac{1}{{\mu }_2})h_2$
  
• C. $(1-\frac{1}{{\mu }_1})h_2+(1-\frac{1}{{\mu }_2})h_1$
• D. $(1+\frac{1}{{\mu }_1})h_2-(1+\frac{1}{{\mu }_2})h_1$

Answer 1

The correct option is A $(1-\frac{1}{{\mu }_1})h_1+(1-\frac{1}{{\mu }_2})h_2$

When the observer is viewing from top, he is observing from a rarer medium (air). Thus, the bottom of the beaker will appear shifted above as a result of refraction at two interfaces.
Shift $1:$ Due to water,
$\Delta d_1=h_1-\frac{h_1}{{\mu }_1}$
Here, $\frac{h_1}{{\mu }_1}$ is the apparent depth of bottom below water surface.
Shift $2:$ Due to kerosene,
$\Delta d_2=h_2-\frac{h_2}{{\mu }_2}$
Thus, total shift $=\Delta d_1+\Delta d_2$
$\Rightarrow \Delta d_{total}=(1-\frac{1}{{\mu }_1})h_1+(1-\frac{1}{{\mu }_2})h_2$

Why this question? Tips: Use the principle of superposition to add the shift produced in position of object due to individual medium or liquid layer.