A beaker containing liquid is placed on the 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 focussed on the upper surface of 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 liquid is

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

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

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

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

answer is A.

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Detailed Solution

In first case, the apparent depth of the liquid is $(b - a)$
$\Rightarrow Real depth = μ (b - a)$
In second case, apparent depth $= (d - c)$
$\Rightarrow Real depth = μ (d - c)$
The difference in depth of liquid is $μ (d - c) - μ (b - a)$
Here we use $μ (d - c) > μ (b - a)$
From experimental data, the difference is equal to $(d - b)$

$\Rightarrow μ (d - c) - μ (b - a) = d - b \Rightarrow μ = \frac{d - b}{(a + d - c - b)}$

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