Q. A beaker containing liquid is placed on a table, underneath a microscope which can be moved along a vertical scale. The microscope is focussed, through the liquid onto 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) $\frac{d - b}{a + b - c - d}$
Explanation
(a) The real depth $= \mu$ [apparent depth]
$\Rightarrow$ In first case, the real depth $h_{1} = \mu(b - a)$
Similarly in the second case, the real depth $h_{2} = \mu(d - c)$
Since $h_{2} > h_{1},$ the difference of real depths
$= h_{2} - h_{1} = \mu(d - c - b + a)$
Since the liquid is added in second case,
$h_{2} - h_{1} = (d - b) \Rightarrow \mu = \frac{(d - b)}{(d - c - b + a)}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.