Q. A beaker contains water up to a height $h₁$ and kerosene of height $h₂$ above water so that the total height of (water +kerosene) is $\left( h_{1} + h_{2} \right).$ Refractive index of water is $u_{1}$ and that of kerosene is $u_{2}.$ The apparent shift in the position of the bottom of the beaker when viewed from above is
A) $\left( 1 - \frac{1}{u_{1}} \right)h_{2} + \left( 1 - \frac{1}{u_{2}} \right)h_{1}$
B) $\left( 1 + \frac{1}{u_{1}} \right)h_{1} + \left( 1 + \frac{1}{u_{2}} \right)h_{2}$
C) $\left( 1 - \frac{1}{u_{1}} \right)h_{2} + \left( 1 - \frac{1}{u_{2}} \right)h_{2}$
D) $\left( 1 + \frac{1}{u_{1}} \right)h_{2} - \left( 1 + \frac{1}{u_{2}} \right)h_{1}$
Explanation
(c) Apparent shift $h = \left( 1 - \frac{1}{\mu} \right)h$
$\therefore$ Apparent shift produced by water
$\bigtriangleup h_{1} = \left( 1 - \frac{1}{\mu_{1}} \right)h_{1}$
And apparent shift produced by kerosene
$\bigtriangleup h_{2} = \left( 1 - \frac{1}{\mu_{2}} \right)h_{2}$
$\bigtriangleup h = \ \bigtriangleup h₁\ + \bigtriangleup h₂ = \left( 1 - \frac{1}{\mu ₁} \right)h₁ + \left( 1 - \frac{1}{\mu ₂} \right)h₂$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.