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Combinations of Lenses

  1. Nov 9, 2008 #1
    The statement is that the image formed by the 1st lens acts as the object for the 2nd lens.
    For 2 thin lenses aligned along a common axis, would this statement be true for when the image formed by the 1st lens is real and past the 2nd lens (while ignoring the effect of the 2nd lens in this step) so that the image does not form between the 2 lenses? If it is...then how?
  2. jcsd
  3. Nov 9, 2008 #2
  4. Nov 9, 2008 #3


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    Yes, it is still true. In that case you would say that you have a virtual object for the second lens, and the object distance for the second will be negative.
  5. Nov 9, 2008 #4
    I see...problem is is that I don't see it matching; I drew out a ray diagram using biconvex lenses...the 2nd image did come out virtual in respect to the 1st image as the object, but the rays didn't head in the same direction or intersect at the same point as they would if the rays from the original object were simply continued from the 1st lens to the 2nd. I guess I'll have to look over my drawings again.
  6. Nov 15, 2008 #5
    Can anyone provide a visual representation of this (a drawing)? That is sort of where I am having trouble...visualizing it.
  7. Nov 15, 2008 #6


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    Can you post your ray diagram?
  8. Nov 15, 2008 #7
    Here it is...the diagram is from text...it is actually a single thick lens, but the idea still applies (in this case, it is stated that the image formed by the 1st surface, R1, acts as the object for the second, R2...but when I start drawing it out, it doesn't seem to work).

    Attached Files:

  9. Nov 15, 2008 #8


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    Your attachment has not been approved yet, but have you tried it with two thin lenses? Just off the top of my head I think it might be easier.
  10. Nov 15, 2008 #9
    I have done so...same problem occurs...the drawings don't seem to follow the statement.
  11. Nov 15, 2008 #10


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    I was hoping you could post your work with the two lenses, because I wasn't sure what you meant when you said a few posts back:

    The reason I'm confused is because I think you should only be using rays that are coming from the first lens. I had some trouble getting some understandable pictures, so I've created several (you might have to view them full size for them to look okay). Here is the procedure:

    To get the first image from the first lens, do the standard thing with the three special rays: the ray that passes through the focal point emerges parallel, the ray the comes in parallel passes through the focal point on the other side, and the ray that passes through the center of the lens goes straight through undeflected. So you get something like:


    which is just the standard thing. Now put in the second lens along with its focal points:


    Now the idea is that we need to find the three special rays again: the one coming in parallel, the one passing through the focal point, and the one going through the center. We already have the parallel one, so that gives the red ray in the following:


    To get the other two rays takes a bit of thought. The idea is that if the second lens was not there, all the rays that leave the first lens passes through the image of the first lens (where the blue lines cross). So use that to find the other two special rays (I have not yet had them affected by the second lens):


    Finally, let the one that passes through the second lens focal point come out parallel to get the final answer:


    Where the blue lines cross is the image from the first lens; where the red lines cross is the real final image.

    (I guess the pictures make more sense if you open each full-sized image in an individual tab and just go from tab to tab.)
  12. Nov 16, 2008 #11
    I see...and the final image is a virtual image of the 1st image...right?
  13. Nov 16, 2008 #12


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    No, the final image is a real image, because the light rays actually are crossing where the final image is located (at least in the example I gave).
  14. Nov 16, 2008 #13
    If an object, O2 was placed where the 1st image is (where the blue lines meet) while the original object remains where it is as O1, then would it be possible that while disregarding the 1st lens, the image formed by O2 in respect to the second lens would be able to match the final image from O1 (where the red lines meet)? It doesn't seem likely considering that the image would have to be a virtual image on the same side as O2 but at the same time, would have to be in front of O2 (closer to the 2nd lens).
  15. Nov 16, 2008 #14


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    Do you mean if you view it from the other side of lens 2? (Because you only see the real image if you are viewing from the right of both lenses; if you wanted to see a virtual image of an object that is to the right of the lens, you would need to be looking from the left.)

    Let me rephrase, so you can check to see if I understand what you are asking: In my final diagram, we have an object to the left of both lenses, lens 1, lens 2, and a position where the blue lines cross, and a final image position where the red lines cross.

    If we remove lens 1 and the original object, and put an object at the blue line crossing, will it form a final virtual image where the red lines cross? (Is that the question?)

    If so, then for the situation I have drawn it will not be true. The reasons is that the converging lens 2 will always produce the final image closer to the lens than its virtual object. (So the final image is between the lens and the virtual object.)

    However, if you use the converging lens to create a virtual image of a real object, the virtual image will always be farther from the lens (the object will be between the lens and the image).
  16. Nov 16, 2008 #15
    Yes, that was my question.

    Doesn't this disprove the statement that the image formed by the 1st lens acts as the object for the 2nd lens in this particular case, or am I simply not understanding the statement?
  17. Nov 16, 2008 #16


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    No, because one is a real object and one is a virtual object; one has a positive object distance and one has a negative image distance; etc.

    If the second lens was farther out, so that the blue lines crossed before reaching the second lens, then whether the real object was a physical object or just an image from the first lens, the final image behavior would be the same if they were at the same spot.
  18. Nov 16, 2008 #17
    So...what you're saying is the statement is true for one case but not the other or is it that the statement is true for both cases?
  19. Nov 16, 2008 #18


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    The statement "the image of the first lens is the object for the second lens" is a true statement.

    However, if you compare the image from a virtual object 10cm to the right of a lens, and the image of a real object 10cm to the right of a lens, those image will bes different. For one thing, the passage of the light is opposite (from left to right in the first case, from right to left in the second), so you have to be looking on differents sides of the lens to see the different images; why would they be the same?

    So the statement "the image of the first lens is the object for the second lens" is true (and is very useful for determining object distances, for example), but if that object is virtual then there is no reason for it to "behave" like a real object.
  20. Nov 16, 2008 #19
    I understand now...I was misunderstanding what was a virtual object and a real object. I'll have a better look at the mathematics to see how this works out...thanks.
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