Question

A beaker containing liquid is placed on a table, underneath a microscope with can be moved along a vertical scale. The microscope is focused, through the liquid into a mark on the table when the reading on the scale is a. it is next focused on the upper surface of the liquid and the reading is $b$. More liquid is added and the observations are repeated. The corresponding readings are $c$ and $d$. The refractive index of the liquid is

• A. $\frac{d-b}{d-c-b+a}$
• B. $\frac{b-d}{d-c-b+a}$
• C. $\frac{d-c-b+a}{d-b}$
• D. $\frac{d-b}{a+b-c-d}$

Answer 1

Answer: (A)

$\frac{d-b}{d-c-b+a}$

The real depth in the first case is given as,

$h_1=\mu \left(b-a\right)$

The real depth in the second case is given as,

$h_2=\mu \left(d-c\right)$

The liquid is added in second case so the difference in real depths is given as,

$h_2-h_1=\left(d-b\right)$

$\mu \left(d-c\right)-\mu \left(b-a\right)=\left(d-b\right)$

$\mu =\frac{\left(d-b\right)}{\left(d-c-b+a\right)}$

Thus, the refractive index of the liquid is $\frac{\left(d-b\right)}{\left(d-c-b+a\right)}$.