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(D) For the near end of the rod,the mirror formula is $\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$.Using the sign convention for a concave mirror,$u$ and

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$\frac{f^2}{u - f}$

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(D) For the near end of the rod,the mirror formula is $\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$.Using the sign convention for a concave mirror,$u$ and $f$ are negative.Thus,$\frac{1}{-f} = \frac{1}{-u} + \frac{1}{v}$,which gives $\frac{1}{v} = \frac{1}{u} - \frac{1}{f} = \frac{f - u}{uf}$.So,$v = \frac{uf}{f - u} = -\frac{uf}{u - f}$. The distance of the image from the mirror is $|v| = \frac{uf}{u - f}$.The far end of the rod is at infinity $(u = \infty)$. According to the mirror formula,the image of the far end is formed at the focus $(v = f)$.The length of the image is the difference between the image positions of the two ends: $L = |v| - f$.$L = \frac{uf}{u - f} - f = \frac{uf - f(u - f)}{u - f} = \frac{uf - uf + f^2}{u - f} = \frac{f^2}{u - f}$.

An infinitely long rod lies along the axis of a concave mirror of focal length $f$. The near end of the rod is at a distance $u > f$ from the mirror. Its image will have a length

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