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.

a) Given a convex mirror,

    Focal length of the convex mirror, $\mathrm{f}=+15 \mathrm{cm}$

    Distance of object in front of convex mirror, $\mathrm{u}=-10 \mathrm{cm}$

    Distance of image from convex mirror, $\mathrm{v}=?$

    From mirror formula:

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

   Substituting the values,

                                        $\frac{1}{\mathrm{v}}+\frac{1}{-10}=\frac{1}{15}$

   On simplification,

                                          $\mathrm{v} =6 \mathrm{cm}$

   The positive value of $\mathrm{v}$ indicates that the image is formed behind the mirror and virtual.

b) Magnification, 

                                         $\mathrm{m} =\frac{{\mathrm{h}}_{\mathrm{i}}}{{\mathrm{h}}_{\mathrm{o}}}=\frac{-\mathrm{v}}{\mathrm{u}}$

Substituting the values of $\mathrm{v}$ and $\mathrm{u}$,

                                          $\mathrm{m} =\frac{-6}{-10}=+0.6$

The positive magnification indicates that the image is erect. As magnification is less than 1 it indicates that the image is smaller than the object.

Therefore, the position of image is 6cm behind the mirror. Nature of image is virtual, erect and diminished.