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A beaker containing liquid is placed on a table underneath a microscope with can be moved along a vertical scale. The microscope is focussed, through the liquid into a mark on the table when the reading on the scale is a. It is next focussed on the upper surface of the liquid and the reading be \mathrm{b}. More the liquid is added and the observations are repeated, the corresponding readings are \mathrm{C} and \mathrm{d}.The refractive index of the liquid is   

Option: 1

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


Option: 2

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


Option: 3

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


Option: 4

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


Answers (1)

best_answer

The real depth = ( R.I. ) apparent depth

\Rightarrow In first case, the real depth \mathrm{h}_{1}=\mathrm{n}(\mathrm{b}-\mathrm{a})

Similarly in the second case, the real depth \mathrm{h}_{2}=\mathrm{n}(\mathrm{d}-\mathrm{c})

Since, \mathrm{h_{2}>h_{1}} the difference of real depths \mathrm{=h_{2}-h_{1}=n(d-c-b+a)}.

Since the liquid is added in second case, \mathrm{h_{2}-h_{1}=d-b}

\mathrm{\Rightarrow \quad \mathrm{n}=\frac{\mathrm{d}-\mathrm{b}}{\mathrm{d}-\mathrm{c}-\mathrm{b}+\mathrm{a}}}

\mathrm{\therefore \quad A }

 

Posted by

Rishabh

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