An object 5.0 cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position of the image, its nature and size.

a) Given,  here we are using sign convention rules 

     Object distance,$\mathrm{u}=-20 \mathrm{cm}$

     Radius of curvature, $\mathrm{R} = 30 \mathrm{cm}$

     Focal length,

                                            $\mathrm{f}=\frac{\mathrm{R}}{2}=15 \mathrm{cm}$

     Size of image, $\mathrm{h}=5 \mathrm{cm}$

     Let size of image be $\mathrm{h}\text{'}$

     From mirror formula,

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

     Substituting the values,

                                             $\frac{1}{\mathrm{v}}+\frac{1}{-20}=\frac{1}{15}$

     On simplification,

                                           $\Rightarrow \mathrm{v}= 8.57 \mathrm{cm}$

The positive value of v indicates that the image is formed behind the mirror.

b) Equation of magnification is,

                                            $\mathrm{m}=\frac{ {\mathrm{h}}^{\text{'}}}{\mathrm{h}}=\frac{-\mathrm{v}}{\mathrm{u}}$

    The height of image,

                                           $\Rightarrow {\mathrm{h}}^{\text{'}}=\mathrm{h}\times \frac{-\mathrm{v}}{\mathrm{u}}$

    Substituting values,

                                          $\Rightarrow {\mathrm{h}}^{\text{'}}=5\times \frac{-8.57}{-20}$

    On simplification,

                                         $\Rightarrow {\mathrm{h}}^{\text{'}}=2.14 \mathrm{cm}$

Since the height of image is positive and less than the value of height of object, it will be erect and diminished.

Thus, the image is formed at a distance of $8.57 \mathrm{cm}$ behind the mirror. Nature of image is virtual, erect and diminished. And the size of the image is ${\mathrm{h}}^{\text{'}}=2.14 \mathrm{cm}$