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At what distance from a concave mirror focal length 10 cm should an object 2 cm long be placed in order to get an erect image 6 cm tall?
Given:
Height of the object, $h_{1}$ = 2 cm
Focal length of the mirror, $f$ = $-$10 cm
Height of the image, $h_{2}$ = 6 cm
To find: Distance of the object from the mirror, $u$.
Solution:
From the magnification formula, we know that-
$m=\frac{{h}_{2}}{{h}_{1}}=-\frac{v}{u}$
Substituting the given values in the magnification formula we get-
$\frac{6}{2}=-\frac{v}{u}$
$3=-\frac{v}{u}$
$v=-3u$
Now, from the mirror formula, we know that-
$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$
Substituting the given values in the mirror formula we get-
$\frac{1}{(-10)}=\frac{1}{(-3u)}+\frac{1}{u}$
$\frac{1}{3u}-\frac{1}{u}=\frac{1}{10}$
$\frac{1-3}{3u}=\frac{1}{10}$
$\frac{-2}{3u}=\frac{1}{10}$
$3u=-20$
$u=\frac{-20}{3}$
$u=-6.67cm$
Thus, the distance of the object from the mirror $u$ is -6.67 cm.