The lateral magnification of the lens with an object located at two different positions $u_{1}$ and $u_{2}$ are $m_{1}$ and $m_{2}$ , respectively. Then the focal length of the lens is :

f = \sqrt{m_{1} m_{2}} (u_{2} - u_{1})

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

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\frac{(u_{2} - u_{1})}{\sqrt{m_{2} m_{1}}}

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\frac{(u_{2} - u_{1})}{(m_{2})^{- 1} - (m_{1})^{- 1}}

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Solution

The correct option is C $\frac{m_{2} u_{2} - m_{1} u_{1}}{m_{2} - m_{1}}$
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
$u = - u$ ; $f = f$
$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$
$v = \frac{f u}{u - f}$
Magnification is: $\frac{f}{u - f}$
$\frac{m_{1}}{m_{2}} = \frac{\frac{f}{u_{1} - f}}{\frac{f}{u_{2} - f}}$
$f = \frac{u_{2} m_{2} - u_{1} m_{1}}{m_{2} - m_{1}}$

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