A light bulb is placed at a distance of 8 cm from the concave mirror having a focal length of 5 cm. Determine the nature of the image.

Object's distance, u = -8 cm

Focal length, f = -5 cm

By, Mirror Formula, we have 

$\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{u}}+\frac{1}{\mathrm{v}}$
$\Rightarrow \frac{1}{\mathrm{v}} = \frac{1}{\mathrm{f}} - \frac{1}{\mathrm{u}}$
 $\Rightarrow \frac{1}{\mathrm{v}} = \frac{1}{(-5)} - \frac{1}{(-8)}$
 $\Rightarrow \frac{1}{\mathrm{v}} = \frac{-1}{5} + \frac{1}{8}$
 $$\begin{gathered} \Rightarrow \frac{1}{\mathrm{v}} = \frac{-8+5}{40} \\ \Rightarrow \frac{1}{\mathrm{v}}=\frac{-3}{40} \end{gathered}$$
 Hence, $\mathrm{v} = -13.33 \mathrm{cm}$

Since the image distance is negative, this indicates that the image is formed on the left side of the pole and as the position is 13.33 cm, hence it is beyond point C, which is at 10 cm. Hence, the image formed by the mirror is real and inverted.