A boy stands 30 ,cm in front of a mirror. The | vedclass

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(B) Given: Object distance $u = -30 \,cm$. Magnification $m = +1/5$ (since the image is erect).For a spherical mirror,magnification $m = -v/u$.Substit

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(B) Given: Object distance $u = -30 \,cm$. Magnification $m = +1/5$ (since the image is erect).For a spherical mirror,magnification $m = -v/u$.Substituting the values: $1/5 = -v / (-30)$.$1/5 = v / 30 \implies v = 6 \,cm$.Since the image distance $v$ is positive,the image is formed behind the mirror.Using the mirror formula: $1/f = 1/v + 1/u$.$1/f = 1/6 + 1/(-30) = 5/30 - 1/30 = 4/30 = 2/15$.$f = 15/2 = +7.5 \,cm$.Since the focal length $f$ is positive,the mirror is a convex mirror.

$A$ boy stands $30 \,cm$ in front of a mirror. The height of his erect image is $1/5 \,th$ of his actual height. The mirror used is ..... .

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