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. |