A spherical mirror forms an erect image three | vedclass

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(C) For an erect image,the magnification $m = +3$. Since $m = -v/u$,we have $v = -3u$. Let the object distance be $u = -x$ (where $x > 0$). Then $v =

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$-30$

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(C) For an erect image,the magnification $m = +3$. Since $m = -v/u$,we have $v = -3u$. Let the object distance be $u = -x$ (where $x > 0$). Then $v = 3x$. The distance between the object and the image is given as $|v - u| = 80 \, cm$. Substituting the values: $|3x - (-x)| = 80 \implies 4x = 80 \implies x = 20 \, cm$. Thus,$u = -20 \, cm$ and $v = 60 \, cm$. Using the mirror formula $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$: $\frac{1}{f} = \frac{1}{60} + \frac{1}{-20} = \frac{1 - 3}{60} = \frac{-2}{60} = \frac{-1}{30}$. Therefore,$f = -30 \, cm$.

$A$ spherical mirror forms an erect image three times the linear size of the object. If the distance between the object and the image is $80 \, cm$,the focal length of the mirror is........ (in $, cm$)

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