An object kept $ 60 \mathrm{~cm} $ from a lens gives a virtual image $ 20 \mathrm{~cm} $ in front of the lens. What is the focal length of the lens? Is it a converging lens or diverging lens?


Given,
Object distance, u = -60cm 
Image distance, v = -20cm
(According to the sign convention of the lens, if the image and the object are on the same side of the lens, the distance of the image will be negative and the image will be virtual.)

To find = focal length, f

Solution:
Here we use the Lens formula, which is given as: 
$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$
$\frac{1}{f}=\frac{1}{(-20)}-\frac{1}{(-60)}$
 $\frac{1}{f}=\frac{1}{60}-\frac{1}{20}$

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

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

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

$f=-30cm$

So, the focal length, f = -30cm.

Since the focal length is negative, it is a diverging or concave lens.

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Updated on: 10-Oct-2022

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