An image formed by a concave mirror of focal | vedclass

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(A) For a concave mirror,the focal length $f = -6 \ cm$. Given magnification $m = \pm 3$. Case $1$: For a real image,$m = -3$. Using the formula $m =

Vedclass · Accepted Answer

$4$

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(A) For a concave mirror,the focal length $f = -6 \ cm$. Given magnification $m = \pm 3$. Case $1$: For a real image,$m = -3$. Using the formula $m = \frac{-v}{u}$,we get $-3 = \frac{-v}{u} \implies v = 3u$. Using the mirror formula $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$: $\frac{1}{-6} = \frac{1}{3u} + \frac{1}{u} = \frac{1+3}{3u} = \frac{4}{3u}$. $3u = -24 \implies u = -8 \ cm$. Case $2$: For a virtual image,$m = +3$. $3 = \frac{-v}{u} \implies v = -3u$. Using the mirror formula: $\frac{1}{-6} = \frac{1}{-3u} + \frac{1}{u} = \frac{-1+3}{3u} = \frac{2}{3u}$. $3u = -12 \implies u = -4 \ cm$. Since the options include $4$ and $8$,and $4$ is a standard solution for virtual magnification,the distance is $4 \ cm$.

An image formed by a concave mirror of focal length $6 \ cm$ is $3$ times the size of the object. What is the distance of the object from the mirror in $cm$?

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