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(A) Given,object distance $u = -20\,cm$.Since the image is real,the magnification $m$ must be negative. Given $|m| = 3$,so $m = -3$.Using the magnific

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$15$

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(A) Given,object distance $u = -20\,cm$.Since the image is real,the magnification $m$ must be negative. Given $|m| = 3$,so $m = -3$.Using the magnification formula $m = -v/u$,we have $-3 = -v / (-20)$,which gives $v = -60\,cm$.Using the mirror formula $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$,we substitute the values:$\frac{1}{f} = \frac{1}{-60} + \frac{1}{-20} = \frac{-1 - 3}{60} = \frac{-4}{60} = \frac{-1}{15}$.Therefore,$f = -15\,cm$.The magnitude of the focal length is $15\,cm$.

$A$ 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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