An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position and nature of the image.

We are given a convex mirror. 
Here, 

Object distance, u = -10 cm 
Focal length, f = + 15 cm [f is +ve for a convex mirror]
Image distance,  v = ? 

Using the mirror formula, 

                     $\frac{1}{\mathrm{u}}+\frac{1}{\mathrm{v}} = \frac{1}{\mathrm{f}}$ 
we have, 

$\therefore$                  $$\begin{gathered} \frac{1}{\mathrm{v}} = \frac{1}{\mathrm{f}}-\frac{1}{\mathrm{u}} \\ = \frac{1}{+15}-\frac{1}{-10} \\ = \frac{2+3}{30} \\ =\frac{1}{6} \end{gathered}$$ 

Thus, image distance,  v = + 6 cm.
 
As image distance is +ve, so a virtual, erect image is formed at a distance 6 cm behind the mirror.