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

$f=-10cm$

$u=8cm$

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

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

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

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

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

$v=40cm$

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

$\therefore 40cm,virtual$