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What is the proof that we can use the location of virtual image in the lens formula to get the location of the image (convex/concave lens)? The following problem might make my question more clear. The proof for the case of plane mirror is quite easy using ray diagrams but I'm getting contradictory results when I try to use the same for convex/concave lens.

enter image description here

In this image "p" is the location of virtual object and "r" is the location of the image formed after reflection . I would like to restate my question as to Why can we substitute the distance of "P" in the lens formula to get the location of the image formed after refraction by the arriving rays? What is the proof?

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  • $\begingroup$ Typo? Replace virtual image with virtual object in the title $\endgroup$ – Farcher Oct 17 '20 at 7:59
  • $\begingroup$ What is the proof? When the concept of a virtual object is incorporated into the lens/mirror formula the predictions relating to the image are correct. $\endgroup$ – Farcher Oct 17 '20 at 8:01
  • $\begingroup$ In all physics proof is experimentally verifying the prediction to within the limits your model claims to be valid under. I'm not sure if you mean something other than that, but that's what "proof" means to me in physics. Physics is not mathematics in this regard. Even if the math in physics is correct, a model can still be shown to be wrong (or not suited to intended purpose) by experiment. $\endgroup$ – StephenG Oct 17 '20 at 8:15
  • $\begingroup$ "In this image "p" is the location of virtual object and "r" is the location of the image formed after reflection." You mean after refraction. $\endgroup$ – Not_Einstein Oct 18 '20 at 18:19
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Let's look at an example of a virtual image from your diagram by reversing the rays, which is allowable. Let r be the distance from the lens to R and p be the distance from the lens to P. So consider rays leaving R from left to right and diverging from the lens as if coming from the virtual image at P. This is a valid ray trace based on your diagram. For such a case, we would use r>0 and find p<0. Ray diagram

If you're okay with that, let's now let's consider the case in your diagram with rays heading toward P and meeting at R. The focal length is the same and in order to end up with positive r we would have to use negative p in the lens formula. So the use of the distance to P appears to be what works.

I'm not sure this will satisfy you as a proof, but it may make the convention more reasonable.

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