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A spherical mirror forms an image of magnification 3. The object distance, if focal length of mirror is $24 \mathrm{~cm}$, may be
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$32 \mathrm{~cm}, 16 \mathrm{~cm}$
$m=\frac{f}{f-u}$
If $m=+3$, then $3=\frac{-24}{-24-u}$ or $-24-u=-8$ or $u+24=8$ or $u=8-24 \mathrm{~cm}=-16 \mathrm{~cm}$ If $m=-3$, then $-3=\frac{-24}{-24-u}$ $u+24=-8$ or $u=-32 \mathrm{~cm}$
If $m=+3$, then $3=\frac{-24}{-24-u}$ or $-24-u=-8$ or $u+24=8$ or $u=8-24 \mathrm{~cm}=-16 \mathrm{~cm}$ If $m=-3$, then $-3=\frac{-24}{-24-u}$ $u+24=-8$ or $u=-32 \mathrm{~cm}$
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