A diverging lens located in the y-z plane at x = 0 forms an image of an arrow at x = x2 = -27.5 cm. The image of the tip of the arrow is located at y = y2 = 5 cm. The magnitude of the focal length of the diverging lens is 49.1 cm.Diagram following:
https://www.smartphysics.com/Content/smartPhysics/Media/Images/EM/26/h26_diverginga1.png

Question

A diverging lens located in the y-z plane at x = 0 forms an image of an arrow at x = x2 = -27.5 cm. The image of the tip of the arrow is located at y = y2 = 5 cm. The magnitude of the focal length of the diverging lens is 49.1 cm.Diagram following:
https://www.smartphysics.com/Content/smartPhysics/Media/Images/EM/26/h26_diverginga1.png

jerry213 · Answer

Question:1)A converging lens of focal length fconverging = 16.09 cm is now inserted at x = x3 = -41.26 cm. In the absence of the diverging lens, at what x co-ordinate, x4, would the image of the arrow form?
2)What is x5, the x co-ordinate of the final image of the combined system?

jerry213 · Answer

https://www.smartphysics.com/Content/smartPhysics/Media/Images/EM/26/h26_diverging_c.png I have no idea about the direction of light here, do not really get this question why is the arrow on the right of the converging lens?Any help will be appreciated!