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(D) Given magnification $m = -3$. We know that $m = -v/u$,so $-v/u = -3$,which implies $v = 3u$.Since the image is real and formed by a concave mirror

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$15$

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(D) Given magnification $m = -3$. We know that $m = -v/u$,so $-v/u = -3$,which implies $v = 3u$.Since the image is real and formed by a concave mirror,both $u$ and $v$ are negative. The distance between the object and the image is $|v - u| = 20\ cm$.Since the image is formed at a greater distance than the object for $m = -3$,we have $|v| > |u|$. Thus,$v - u = -20\ cm$ (using sign convention where $v$ and $u$ are negative).Substituting $v = 3u$,we get $3u - u = -20$,so $2u = -20$,which gives $u = -10\ cm$.Then $v = 3(-10) = -30\ cm$.Using the mirror formula $1/f = 1/v + 1/u = 1/(-30) + 1/(-10) = (-1 - 3)/30 = -4/30 = -2/15$.So,$f = -7.5\ cm$.The radius of curvature $R = 2|f| = 2 	imes 7.5 = 15\ cm$.

$A$ concave mirror produces an image of an object such that the distance between the object and the image is $20\ cm$. If the magnification of the image is $-3$,then the magnitude of the radius of curvature of the mirror is: (in $cm$)

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