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(B) Given: Magnification $m = -3$ (since the image is real,it must be inverted).Object distance $u = -20 \ cm$.Using the magnification formula: $m = -

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$-15$

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(B) Given: Magnification $m = -3$ (since the image is real,it must be inverted).Object distance $u = -20 \ cm$.Using the magnification formula: $m = -\frac{v}{u}$.$-3 = -\frac{v}{-20} \implies v = -60 \ cm$.Using the mirror formula: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$.$\frac{1}{f} = \frac{1}{-60} + \frac{1}{-20}$.$\frac{1}{f} = \frac{-1 - 3}{60} = \frac{-4}{60} = -\frac{1}{15}$.Therefore,$f = -15 \ cm$.

$A$ concave mirror gives an image three times as large as the object placed at a distance of $20 \ cm$ from it. For the image to be real,the focal length should be ...... $cm$.

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