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(D) For a real image, magnification $m = -3$. Since $m = -v/u$, we have $-3 = -v/u$, which implies $v = 3u$.Given that the distance between the object

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$-15$

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(D) For a real image, magnification $m = -3$. Since $m = -v/u$, we have $-3 = -v/u$, which implies $v = 3u$.Given that the distance between the object and the image is $40 \ cm$, and for a real image formed by a concave mirror, both object and image are on the same side, the distance is $|v - u| = 40 \ cm$.Substituting $v = 3u$, we get $|3u - u| = 40$, so $|2u| = 40$, which gives $u = -20 \ cm$ (using sign convention).Then $v = 3(-20) = -60 \ cm$.Using the mirror formula $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$:$\frac{1}{f} = \frac{1}{-60} + \frac{1}{-20} = \frac{-1 - 3}{60} = \frac{-4}{60} = \frac{-1}{15}$.Therefore, $f = -15 \ cm$.

Distance between an object and its three times magnified real image is $40 \ cm$. The focal length of the mirror used is . . . . . . $cm$.

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