Question

An object is placed 4 cm in front of a concave lens of focal length 3 cm. Using the lens equation, find where the image will form and state whether it is a real or virtual image.

Answer 1

1/s2 = 1/f - 1/s1
we take negative focal length, because it is divergring lens
1/s2 = - 1/3 - 1/4 = - 7/12

s2 = 12/7 cm from the length, on the same side where object is.
The image is virtual