A convex mirror used for rear-view on an automobile has a radius of curvature of 3.00 m. If a bus is located at 5.00 m from this mirror, find the position, nature and size of the image.  

Given, a convex mirror.

We have, 

Radius of curvature, R = + 3.00 m          [R is +ve for a convex mirro]
Object distance, u = - 5.00 m 
Image distance, v = ?
Height of the image, h'  = ? 

Therefore,
Focal length,  $\mathrm{f} = \frac{\mathrm{R}}{2} = +\frac{3.00 \mathrm{m}}{2} = + 1.50 \mathrm{m}$ 

Using the mirror formula, 

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

$\therefore$                    $$\begin{gathered} \frac{1}{\mathrm{v}} = \frac{1}{\mathrm{f}}-\frac{1}{\mathrm{u}} \\ = +\frac{1}{1.50}-\frac{1}{(-5.00)} \\ = \frac{1}{1.50}+\frac{1}{5.00} \end{gathered}$$

                            $= \frac{5.00+1.50}{7.50}$ 

                          $$\begin{gathered} \mathrm{v} = \frac{+7.50}{6.50} \\ = + 1.15 \mathrm{m} \end{gathered}$$ 

The image is formed at a distance of 1.15 m behind the mirror. 

Now, 
Magnification,  $\mathrm{m} = \frac{\mathrm{h}\text{'}}{\mathrm{h}} = -\frac{\mathrm{v}}{\mathrm{u}}$ 

                          $$\begin{gathered} = -\frac{1.15 \mathrm{m}}{-5.00 \mathrm{m}} \\ = + 0.23 \end{gathered}$$ 

The image is virtual, erect and smaller in size by a factor of 0.23.