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(D) From the mirror formula,$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.Differentiating both sides with respect to $u$,we get $-\frac{1}{v^2} dv - \frac

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

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(D) From the mirror formula,$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.Differentiating both sides with respect to $u$,we get $-\frac{1}{v^2} dv - \frac{1}{u^2} du = 0$.This implies $dv = -\frac{v^2}{u^2} du$. The magnitude of the image length $|dv|$ is given by $|dv| = \left( \frac{v}{u} \right)^2 |du|$.Since the object is small,$|du| = b$,so the image length is $L = \left( \frac{v}{u} \right)^2 b$.From the mirror formula,$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{u - f}{fu}$,which gives $\frac{v}{u} = \frac{f}{u - f}$.Substituting this into the expression for $L$,we get $L = b \left( \frac{f}{u - f} \right)^2$.

$A$ small linear object of length $b$ is placed on the axis of a concave mirror of focal length $f$ at a distance $u$ from its pole. What is the approximate length of the image?

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