Optics - What is the magnification of the lens?

[rajiv_putcha]

Object O1 is 15.0 cm to the left of a converging lens of 10.0-cm focal length. A second lens is positioned 10.0 cm to the right of the first lens and is observed to form a final image at the position of the original object O1. (a) What is the focal length of the second lens? (b) What is the overall magnification of this system? (c) What is the nature (i.e., real or virtual, upright or inverted) of the final image?

Answers:
a) -11.1 cm
b) M = positive 2.50
c) virtual, upright

Please Explain how to get slove! (I tried to get the picture and the ray diagrams but they are not working out)

Thanks in Advance!

 

[member 553083]

Solution: a) The focal length of the second lens can be calculated using the lens equation: 1/f2 = 1/f1 + (1/do) - (1/di), where f1 is the focal length of the first lens, do is the object distance, and di is the image distance. In this case, f1 = 10 cm, do = 15 cm, and di = 0 cm, so the focal length of the second lens is given by: f2 = 1/(1/10 + 1/15) = -11.1 cm.b) The magnification of this system can be calculated using the magnification equation: M = hi/ho, where hi is the height of the final image and ho is the height of the original object. In this case, the final image is at the position of the original object, so hi = ho, and thus M = 1. c) The final image is a virtual, upright image since it is formed at the position of the original object, which is to the left of the converging lens.