A spherical mirror forms an erect image of an | vedclass

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(C) For an erect image,the magnification $m = +3$.Since $m = -v/u$,we have $3 = -v/u$,which implies $v = -3u$.The distance between the object and the

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$30$

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(C) For an erect image,the magnification $m = +3$.Since $m = -v/u$,we have $3 = -v/u$,which implies $v = -3u$.The distance between the object and the image is given as $|v - u| = 80 \ cm$.Substituting $v = -3u$,we get $|-3u - u| = 80$,so $|-4u| = 80$,which gives $u = -20 \ cm$.Then,$v = -3(-20) = 60 \ cm$.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,$f = -30 \ cm$.

Column-$I$	Column-$II$
$1$. $m = -2$	a. Convex mirror
$2$. $m = -1/2$	b. Concave mirror
$3$. $m = +2$	c. Real image
$4$. $m = +1/2$	d. Virtual image

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

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Draw the ray diagram when an object is placed between the focus $(F)$ and the pole $(P)$ of a concave mirror.

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Match the corresponding entries of Column-$I$ with Column-$II$. [Where $m$ is the magnification produced by the mirror]
Column-$I$ Column-$II$
$1$. $m = -2$ a. Convex mirror
$2$. $m = -1/2$ b. Concave mirror
$3$. $m = +2$ c. Real image
$4$. $m = +1/2$ d. Virtual image

Distance between an object and its three times magnified real image is $40 \ cm$. The focal length of the mirror used is . . . . . . $cm$.

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