An object of height 6 cm is placed perpendicular to the principal axis of a concave lens of focal length 5 cm. Use lens formula to determine the position, size and nature of the image if the distance of the object from the lens is 10 cm.

Given :    $u=-10$ cm              $f=-5$ cm                $h_o=6$ cm

Using lens formula :       $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

$\therefore$      $\frac{1}{v}-\frac{1}{(-10)}=\frac{1}{-5}$              

$⟹v=-\frac{10}{3}=-3.33$  cm     (- sign shows that image is virtual)

Also   $m=\frac{h_i}{h_o}=\frac{v}{u}$

$\therefore$   $\frac{h_i}{6}=\frac{\frac{-10}{3}}{-10}$                

$⟹h_i=+2$ cm   (+ sign shows that image is erect)