Question

A beaker contains water upto a height $h_1$ and kerosene of height $h_2$ above water so that the total height of (water $+$ kersene) is $(h_1+h_2)$. 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 of the beaker when viewed from above is:

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

Answer 1

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

We Rnow, apparent shift

$\Delta h=(1-\frac{1}{\mu })h$

Apparent shift produced by water,

$\Delta h_1={(1-\frac{1}{{\mu }_1})}^{h_1}$

and apparent shift produced by Kerosene

$\begin{array}{c}\Delta h_2=(1-\frac{1}{{\mu }_2})\widehat{h_2}\\ \Delta h=\Delta h_1+\Delta h_2=(1-\frac{1}{u_1})h_1+(1-\frac{1}{u_2})h_2\end{array}$

Hence obtion 'C' is correct