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(D) From the mirror formula,$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$ .....$(i)$Differentiating equation $(i)$ with respect to $u$,we get:$0 = - \frac

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$l{\left( {\frac{f}{{u - f}}} \right)^2}$

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(D) From the mirror formula,$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$ .....$(i)$Differentiating equation $(i)$ with respect to $u$,we get:$0 = - \frac{1}{v^2} \frac{dv}{du} - \frac{1}{u^2}$$\frac{dv}{du} = - \left( \frac{v}{u} \right)^2$Since the object length $l = du$ is small,the image length $dv$ is given by $dv = - \left( \frac{v}{u} \right)^2 du$ .....$(ii)$From the mirror formula,we can express the magnification $m = \frac{v}{u}$ as:$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{u - f}{fu}$$\frac{v}{u} = \frac{f}{u - f}$ .....$(iii)$Substituting equation $(iii)$ into equation $(ii)$,we get:$dv = - \left( \frac{f}{u - f} \right)^2 l$The magnitude of the size of the image is $l' = |dv| = l \left( \frac{f}{u - f} \right)^2$.

$A$ short linear object of length $l$ lies along the axis of a concave mirror of focal length $f$ at a distance $u$ from the pole of the mirror. The size of the image is approximately equal to

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