Question

what is the nature and focal length of the lens that will form a virtual image having one-quarter the dimensions of an object spaced 40 cm from the lens?

Answer 1

The formula for finding Focal length ia;

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

$v=40$

$u=10$

$\frac{1}{10}+\frac{1}{40}=\frac{1}{f}$

Reducing the sum you find $\frac{1}{8}$

As a result the focal length is 8 cm while the nature of the lens is convex since the power is positive.