If a concave mirror has a focal length of 10 cm, find the two positions where an object can be placed to give, in each case, an image twice the height of the object.

Focal length, $(f)$ of the concave mirror = $-$10 cm

To find: Distance of the object, $u$ from the mirror.

Solution:

Case-1

The image is real, and its magnification, $m$ is $-$2.

From the magnification formula, we know that-

$m=-\frac{v}{u}$

Substituting the given values in the magnification formula we get-

$-2=-\frac{v}{u}$

$-2u=-v$

$v=2u$

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-

$\frac{1}{(-10)}=\frac{1}{2u}+\frac{1}{u}$

$\frac{1}{-10}=\frac{1+2}{2u}$

$\frac{1}{-10}=\frac{3}{2u}$

$2u=-10\times {3}$

$u=\frac {-10\times 3}{2}$

$u=-15cm$

Thus, the object should be placed at a distance of 15 cm in front of the concave mirror.

Case-2

The image is virtual and its magnification, $m$ is +2.

From the magnification formula, we know that-

$m=-\frac{v}{u}$

Substituting the given values in the magnification formula we get-

$-2=-\frac{v}{u}$

$-2u=-v$

$v=2u$

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-

$\frac{1}{(-10)}=\frac{1}{(-2u)}+\frac{1}{u}$

$\frac{1}{-10}=-\frac{1}{2u}+\frac{1}{u}$

$\frac{1}{-10}=\frac{-1+2}{2u}$

$\frac{1}{-10}=\frac{1}{2u}$

$2u=-10$

$u=\frac{-10}{2}$

$u=-5cm$

Thus, the object should be placed at a distance of 5 cm in front of the concave mirror.