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?

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.

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Updated on: 10-Oct-2022

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