A spherical mirror forms an erect image three | vedclass

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(C) For an erect image,the magnification $m = +3$. Since the image is erect,the mirror must be a concave mirror forming a virtual image.Given $m = -v/

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$30$

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(C) For an erect image,the magnification $m = +3$. Since the image is erect,the mirror must be a concave mirror forming a virtual image.Given $m = -v/u = 3$,so $v = -3u$.Let the object distance be $u = -x$ (where $x > 0$). Then the image distance $v = 3x$.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 $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$:$\frac{1}{60} + \frac{1}{-20} = \frac{1}{f}$$\frac{1 - 3}{60} = \frac{1}{f} \implies \frac{-2}{60} = \frac{1}{f}$$f = -30 \, cm$.The magnitude of the focal length is $30 \, cm$.

$A$ 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........$cm$.

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