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Magnification by Thin Lenses

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Question

The magnification produced by a convex lens for two different positions of an object are $m_{1}$ and $m_{2}$ respectively $(m_{1} > m_{2})$ . If 'd ' is the separation between the two positions of the object then the focal length of the lens is

A

\sqrt{m_{1} m_{2}}

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B

\frac{1}{\sqrt{m_{1} - m_{2}}}

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C

\frac{d m_{1} m_{2}}{m_{1} - m_{2}}

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D

\frac{d}{m_{1} - m_{2}}

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Solution

The correct option is C $\frac{d m_{1} m_{2}}{m_{1} - m_{2}}$
$\begin{matrix} for convex lens m = \frac{f}{f - u} where m = magnification \begin{matrix} f & = focal length u & = object distance. \end{matrix} \end{matrix}$

$m_{2} = \frac{f}{f - x} - (1)$
$m_{1} = \frac{f}{f - (x + d)}$
$f - x - d = \frac{f}{m_{1}}$
$f - d - \frac{f}{m_{1}} = x - (2)$
substitute (2) in (1)
$m_{2} = \frac{f}{f - (f - d - \frac{f}{m})}$
$m_{2} = \frac{f}{d + \frac{f}{m_{1}}}$
$m_{2} d + \frac{m_{2}}{m_{1}} f = f$
$m_{2} d = f [1 - \frac{m_{2}}{m}]$
$m_{2} d = f (\frac{m_{1} - m_{2}}{m_{1}})$
$f_{i} = \frac{m_{1} m_{2} d}{m_{1} - m_{2}}$
Hence option c is correct

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