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Homework Help: Serveral Lens generate image

  1. Jun 18, 2009 #1
    Good afternoon,

    I want to check another problem about optical physics:
    There are taken N equal converging lens, ordered one by one in a distance of f, where f is the focal distance.
    There is an object standing in a distance of 0,5*f of the first lens.

    A Sketch:
    http://img35.imageshack.us/img35/8288/aufzeichnenc.png [Broken]

    For which N, a real image is generatet behind the last lens? What is the factor of incresement of the image compared with the object?

    My solution is that a real image is generated
    if 3|N (image size = object size) or 3|(N+1) (image size = 2* object size)


    My approach:

    in general: 1/f = 1/g + 1/b

    after first lens: g=f/2
    -> b=-f
    -> a virtual image with size A=b/g=-2

    after second lens: g=2*f
    -> a real image with size A=1, compared with the size of the virtual image
    -> size of real image = size of object * 2

    after third lens: g=-f
    -> a real image with size A=0,5, compared with the size of the real image (2)
    -> size of real image = size of object

    after fourth lens - nth lens:
    images seems to repeat -> virtual image if 3|(N+2), real image if 3|N or 3|(N+1)

    Is this calculation right?

    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jun 20, 2009 #2
    No opinions?
  4. Jun 21, 2009 #3


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    Have you tried the geometrical approach ... somehow I think that 2 lenses will be enough.
  5. Jun 21, 2009 #4


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    Either that or infinity.
  6. Jun 22, 2009 #5
    Why do you think that two lenses will be enough?
    Which geometrical approch do you mean?
    If two lenses will be enough, where is the mistake in my consideration?

  7. Jun 22, 2009 #6


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    http://img34.imageshack.us/img34/5571/clipboard1n.jpg [Broken]

    Something like this if I remember correctly.

    So the first object which is at f/2 distance get's magnified like with a usual magnifiing glass so that a virtual image appears on the same side but twice as big.

    The virtual image then passes through the second lens using the same geometric method and a real but inverted image if it appears. the final real image is twice the originals size.

    Again if I remember it right.
    Last edited by a moderator: May 4, 2017
  8. Jun 22, 2009 #7
    It seems you used the same way of solution:
    Obviously, there does not arise a real image after the first lens, but after the second.
    But I think that there is an image after the third lens, too.
    After the fourth lens only a virtual image arises, and so on.

  9. Jun 22, 2009 #8


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    Yes a third lens would create a real image. And after the cycle begins.

    About my infinity ... I was trying to see if the first lens does converge the light so that the second one would be useless. But it doesn't.
  10. Jun 23, 2009 #9


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    Staff Emeritus
    Science Advisor
    Homework Helper

    You're solution looks right.
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