An object is placed on the principal axis of a concave mirror of focal length 10 cm at a distance of 8.0 cm from the pole. Then the position and nature of the image is

$\text{u}=\text{8cm}$

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

$-\frac{1}{10}=\frac{-1}{8}+\frac{1}{v}$

$\frac{1}{v}=\frac{-1}{10}+\frac{1}{8}$

$\frac{1}{v}=\frac{-4+5}{40}$   (or) $v=\frac{8\times 10}{-8+10}$

$\frac{1}{v}=\frac{1}{40}$                                                                        

$v=\frac{+80}{2}$

$\text{v}=\text{4}0\text{cm}$             $\text{v}=\text{4}0\text{cm}$

Image distance is +ve. So image is formed behind the mirror. So it is virtual image

$\therefore \text{4}0\text{cm}$ ,virtual