An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed, so that a sharp focused image can be obtained? Find the size and the nature of the image.

a) Given that, here we are using sign convention rules 

    Size of object, ${\mathrm{h}}_{\mathrm{o} }=+7 \mathrm{cm}$

    Focal length of concave mirror, $\mathrm{f}=-18 \mathrm{cm}$

   Object distance, $\mathrm{u}=-27 \mathrm{cm}$

   Using mirror’s formula:

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

   Substituting the values,

                                          $\frac{1}{\mathrm{v}}+\frac{1}{-27}=\frac{1}{-18}$

   On simplification,

                                        $\Rightarrow \mathrm{v}= -54 \mathrm{cm}$

The negative value of $\mathrm{v}$ indicates that the screen should be placed at a distance of $54 \mathrm{cm}$ in front of the mirror and the image is real.

b)Equation of magnification is,

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

    The size of image,

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

    Substituting values,

                                          $\Rightarrow {\mathrm{h}}_{\mathrm{i}}=7\times \frac{-(-54)}{-27}$

    On simplification,

                                         $\Rightarrow {\mathrm{h}}_{\mathrm{i}}= -14 \mathrm{cm}$

Since the height of image is negative and greater than the value of size of object, it will be inverted and enlarged.

c)Nature of image is real, inverted and enlarged.