It's all in the title. For instance, if I have two lenses , I have been taught to first find the position of the image formed by the first lens, and then use that image to find the final image formed by the 2nd lens, if the first image is formed beynd the 2nd lens. Why does this work?

enter image description here

edit:- image for reference

  • $\begingroup$ could you add a sketch with the configuration that confuses you? $\endgroup$
    – basics
    Nov 3, 2022 at 15:06
  • $\begingroup$ It's most cofigigs in general, but I'll do that $\endgroup$ Nov 3, 2022 at 16:35
  • $\begingroup$ @basics, just did. Sorry for the delay $\endgroup$ Nov 8, 2022 at 16:53
  • $\begingroup$ I don't think so because when we use a barlow lens just before the moon image is formed the image is formed farther away and enlarged. Think that the barlow lens is a concave lens. But if we use the moon image without the barlow lens as the object for the barlow lens the image should form closer to the barlow lens. $\endgroup$ Nov 8, 2022 at 18:50
  • $\begingroup$ what's a Barlow lens $\endgroup$ Nov 9, 2022 at 11:04

2 Answers 2


I addressed this before but will elaborate further. Refer to the diagram here.

Suppose there is an object R on the axis a distance r to the left of a lens with focal length f and r<f. When the rays leave the lens they diverge as if coming from a point P a distance p to the left of the lens. So P is a virtual image. We have

1/r + 1/p = 1/f

1/p = (r-f)/rf

where r>0 and p<0.

Rays are reversible so consider rays from the right heading toward P i.e. P is now a virtual object. They have to converge at R. Your question is basically can we use the same lens equation for this case. Let's see if

1/p = (r-f)/rf

works. Well, f is the same, the absolute values of p and r are the same. In this case r > 0 as it's now a real image and still r<f. So we will end up with the right magnitude for p but it will be negative.

So we conclude that we can use a virtual object if the image distance is negative.

Edit: fixed equation for 1/p. Conclusion still holds.


Because the pattern of rays emerging from the first image is (more or less) the same as the pattern of rays that would emerge from a real object at that point.

  • $\begingroup$ But no rays actually emerge from it. $\endgroup$ Nov 1, 2022 at 14:37
  • $\begingroup$ Is there any way to prove this? $\endgroup$ Nov 1, 2022 at 14:37
  • $\begingroup$ @mathandphysicsforever Get some lenses and make a telescope or microscope. That's how we prove things in physics. But you can make it plausible mathematically by simply extending the rays from the first lens straight through the first image to the second one. $\endgroup$
    – John Doty
    Nov 1, 2022 at 14:44
  • 1
    $\begingroup$ I understood that. But what's troubling me is the situation in which the rays from the first lens would normally converge beyond the second lens. That's what's confusing me. Thanks! $\endgroup$ Nov 1, 2022 at 14:48
  • $\begingroup$ Before the image could form behind the second lens, rays meet the second lens and "deviate". I still can't understand what your doubts are actually, but it's my fault. Try to add a little sketch $\endgroup$
    – basics
    Nov 6, 2022 at 15:28

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