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(D) For a divergent (concave) lens,the focal length $f$ is negative,so we take focal length as $-f$.Given the object distance $u = -mf$.Using the lens

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$1/(m + 1)$ times

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(D) For a divergent (concave) lens,the focal length $f$ is negative,so we take focal length as $-f$.Given the object distance $u = -mf$.Using the lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.Substituting the values: $\frac{1}{v} - \frac{1}{-mf} = \frac{1}{-f}$.$\frac{1}{v} + \frac{1}{mf} = -\frac{1}{f}$.$\frac{1}{v} = -\frac{1}{f} - \frac{1}{mf} = -\left(\frac{m+1}{mf}\right)$.Therefore,$v = -\frac{mf}{m+1}$.The magnification $M$ is given by $M = \frac{v}{u}$.$M = \frac{-mf/(m+1)}{-mf} = \frac{1}{m+1}$.Thus,the size of the image is $\frac{1}{m+1}$ times the size of the object.

An object is placed at a distance $m$ times the focal length of a divergent lens. The size of the image is shorter than that of the object by what factor?

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