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(C) Given magnification $m = \frac{1}{4}$. Since the image is smaller than the object and formed by a mirror,we assume a concave mirror forming a real

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$10.7$

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(C) Given magnification $m = \frac{1}{4}$. Since the image is smaller than the object and formed by a mirror,we assume a concave mirror forming a real image.For a real image,$m = -\frac{v}{u} = -\frac{1}{4}$,so $u = 4v$.The distance between the object and the image is $|u - v| = 40 \ cm$.Since $u$ and $v$ are on the same side for a real image,$u - v = 40 \ cm$.Substituting $u = 4v$,we get $4v - v = 40 \implies 3v = 40 \implies v = \frac{40}{3} \ cm$.Then $u = 4 	imes \frac{40}{3} = \frac{160}{3} \ cm$.Using the mirror formula $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$ with sign convention ($u = -\frac{160}{3} \ cm$,$v = -\frac{40}{3} \ cm$):$\frac{1}{f} = -\frac{3}{40} - \frac{3}{160} = \frac{-12 - 3}{160} = -\frac{15}{160} = -\frac{3}{32}$.$f = -\frac{32}{3} \approx -10.7 \ cm$.The magnitude of the focal length is $10.7 \ cm$.

$A$ mirror is used to produce an image with magnification of $\frac{1}{4}$. If the distance between the object and its image is $40 \ cm$,then the focal length of the mirror is . . . . . . . (in $cm$)

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