Help please When an object is placed at the proper distance

When an object is placed at the proper distance to the left of a converging lens, the image is focused on a screen 46.0 cm to the right of the lens. A diverging lens is now placed 23.0 cm to the right of the lens, and it is found that the screen must be moved 29.6 cm farther to the right to obtain a sharp image. What is the focal length of the diverging lens? cm

Solution

The image from the rst lens becomes the object for the second lens. Since the second lens is placed between the rst lens and the rst image, it will be a virtual object with do < 0.

This distance is d0 = (46 23) = 23 cm

The nal image is 29.6 cm further and is projected onto a screen, so it is a real image with di > 0. The distance is di = 23 +29.6 = +52.6 cm

now i/f=1/d0 + 1/di = 1/-23 + 1/52.6 = -40.87