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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.