An object 3cm high is placed 12cm from a converging lens. A real
image is found to form 36cm from the lens.
(a)what is the height of the image?
You can easily use E maths' similarity here
If you draw only the ray that passes through the optical centre
of the lens, you will see similar triangles. We can ignore the
other ray that cuts through the focal length, because the distances
of the images from the lens are already given.
So using similarity, height of image / 36 = 3 / 12
height of image = 9 cm