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Lens Maker's Formula

The magnifica...

Question

The magnifications produced by a convex lens for two different of an object are $m_{1}$ and $m_{2}$ respectively $(m_{1} > m_{2})$ . If 'd' is the distance of 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{d}{\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 D $\frac{d}{m_{1} - m_{2}}$
Let us consider $m_{1} > m_{2}$
$\frac{v}{u} = m_{1}$
For case ii, $m_{2} = \frac{u}{v}$
$m_{1} m_{2} = 1$
$\frac{1}{u} + \frac{1}{v} = \frac{1}{f}$
$\frac{1}{u} + \frac{1}{m_{1} u} = \frac{1}{f}$
$u = \frac{f (1 + m_{1})}{m_{1}}$
$u = m_{2} v$
$v = \frac{f (1 + m_{2})}{m_{2}}$
$d = v - u = f (\frac{1 + m_{2}}{m_{2}} - \frac{1 + m_{1}}{m_{1}})$
$d = f (\frac{m_{1} + m_{1} m_{2} - m_{2} - m_{1} m_{2}}{m_{1} m_{2}})$
$f = \frac{m_{1} m_{2}}{m_{1} - m_{2}}$
$m_{1} m_{2} = 1$
$f = \frac{d}{m_{1} - m_{2}}$

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