A convex mirror used on a moving automobile has a radius of curvature of 3.0 m. If a truck is following it at a constant distance of 4.5 m, find (i) the position, (ii) the nature and (iii) the magnification for the image.

Given, a convex mirror.
 
We have, 

Radius of curvature, R = +3.00 m    [R is +ve for a convex mirror]
Object distance, u = - 4.5 m
Image distance, v = ?
Image size,  h' = ? 

Using the relationship between focal length and radius of curvature, we have 

Focal length, $\mathrm{f} = \frac{\mathrm{R}}{2} = +\frac{3.00 \mathrm{m}}{2} = + 1.50 \mathrm{m}$          [ f is +ve for a convex mirror] 

Using the mirror formula, 

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

$\therefore$                     $$\begin{gathered} \frac{1}{\mathrm{v}} = \frac{1}{\mathrm{f}}-\frac{1}{\mathrm{u}} \\ = \frac{1}{1.50} - \frac{1}{-4.5} \\ = \frac{1}{1.50}+\frac{1}{4.5} \\ = \frac{3+1}{4.5} \\ = \frac{4}{4.5} \\ = \frac{8}{9} \end{gathered}$$

Image distance,  $\mathrm{v} = +\frac{9}{8}\mathrm{m}.$ 

The image is formed at a distance of 9/8 m behind the mirror. 

Magnification,
                      $$\begin{gathered} \mathrm{m} = \frac{\mathrm{h}\text{'}}{\mathrm{h}} = -\frac{\mathrm{v}}{\mathrm{u}} \\ = \frac{9/8}{4.5} \\ = + 0.25 \end{gathered}$$ 

The image formed is virtual, erect and smaller in size by a factor of 0.25 (one-fourth) than the object.