An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm.

solution:

Given, object distance, $\mathrm{u}=-10 \mathrm{~cm}$;

focal length, $f=+15 \mathrm{~cm}$;

image distance, $\mathrm{v}=? ;$ magnification, $\mathrm{m}=?$

Mirror formula,

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f} \quad$ or $\quad \frac{1}{v}+\frac{1}{(-10)}=\frac{1}{(+15)}$

or $\frac{1}{v}=\frac{1}{10}+\frac{1}{15}=\frac{3+2}{30}=\frac{5}{30}=\frac{1}{6}$ or $v=+6 \mathrm{~cm}$

Now, magnification, $m=-\frac{v}{u}=-\frac{(+6)}{(-10)}=+0.6$

The image is located 6 cm behind the mirror ; it is a virtual, erect and diminished image.