Question

An object is placed at a distance of 50 cm from a concave lens of focal length 20 cm. Find the nature and position of the image.

Answer 1

Focal length of concave lens is negative.

$f=-20cm$

$u=-50cm$

We know that :

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

$\frac{1}{-20}=\frac{1}{v}-\frac{1}{-50}$

$v=-14.3cm$.

The image is virtual, erect.