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(A) Given: Object distance $u = -20 \ cm$ (by sign convention).Magnification $m = -3$ (since the image is real and magnified).Using the magnification

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$15$

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(A) Given: Object distance $u = -20 \ cm$ (by sign convention).Magnification $m = -3$ (since the image is real and magnified).Using the magnification formula $m = -\frac{v}{u}$,we have:$-3 = -\frac{v}{-20} \Rightarrow v = -60 \ cm$.Using the mirror formula $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$,we get:$\frac{1}{f} = \frac{1}{-60} + \frac{1}{-20} = \frac{-1 - 3}{60} = \frac{-4}{60} = -\frac{1}{15}$.Therefore,the focal length $f = -15 \ cm$.The magnitude of the focal length is $15 \ cm$.

An object is placed at $20 \ cm$ in front of a concave mirror,which produces a three times magnified real image. What is the focal length of the concave mirror (in $cm$)?

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