An object 60 cm from a lens gives a virtual image at a distance of 20 cm in front of the lens. What is the focal length of the lens? Is the lens converging or perging? Give reasons for your answer.

​Object distance, $u$ = $-$60 cm                           (object distance is always taken negative, as it is placed on the left side of the lens)

​Image distance, $v$ = $-$20 cm                          (image distance is taken negative as the image is virtual)

To find: Focal length of the lens, $f$.

Solution:

From the lens formula, we know that-

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

Substituting the given values in the formula, we get-

$\frac {1}{(-20)}-\frac {1}{(-60)}=\frac {1}{f}$

$-\frac {1}{20}+\frac {1}{60}=\frac {1}{f}$

$\frac {1}{f}=\frac {1}{60}-\frac {1}{20}$

$\frac {1}{f}=\frac {1-3}{60}$

$\frac {1}{f}=-\frac {2}{60}$

$\frac {1}{f}=-\frac {1}{30}$

$f=-30cm$

Thus, the focal length of the lens is 30 cm, and the negative sign implies that the lens is diverging in nature because a negative focal length indicates a concave lens.