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Two identical convergent lenses

  1. Jul 20, 2008 #1
    1. The problem statement, all variables and given/known data
    the figure shows a combination of two identical lenses. There is a 2cm tall object that is 36 cm away from the first lens (f=9 cm). The second lens is 15 cm away from the first lens. It also has a focal length of 9 cm. So, the first question was "find the position of the final image of the 2 cm tall object". I've solved that one, using 1/f=1/s+1/s' for each lens, making sure to use the "image" from the first lens as the "object" for the second lens. The next question is "find the size of the final image of the 2 cm tall object".

    2. Relevant equations

    I thought I could use h'/h=s'/s, although that didn't seem to work out.
    I already used 1/f=1/s+1/s'

    3. The attempt at a solution

    My attempt was to use h'/h=s'/s. I thought I would solve for h' (height of image).
    I tried to do this in a 2 step process
    a. used original object size & lens #1 h'/2=12/36
    b. used image from 1st lens as "object" for second lens h'/0.67=45/3
    c. got 1.01 cm which is incorrect.

    Any helpful thoughts to move me in the right direction would be hugely appreciated.
    Thanks in advance.
  2. jcsd
  3. Jul 20, 2008 #2


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    OK. You are going about this in the right way, but I think one of your numbers may be off. Are you sure the distances you found are correct?

    I am getting -4.5cm as the image distance from the second lens, not 45cm. Check your work from the first part of the problem.
    Last edited: Jul 20, 2008
  4. Jul 20, 2008 #3
    Sure... for the first part of the problem, my work is:
    (I drew the picture so the object is on the far left of my paper, with the lenses to the right)

    so s' would be 12 cm
    so the image is 12 cm to the right of the first lens

    for the second lens, I am supposed to use the image as the new "object", so
    so I think s' would =-4.5 cm
    meaning that the final image is 4.5 cm to the left of the second lens

    so then I thought h'/h=-s'/s
    I thought I'd use h'/2=-(-4.5)/(-3)
    h'=(-1.5)(2)=-3 cm

  5. Jul 20, 2008 #4


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    You have to use the magnification formula in two steps as you did in your first post.

    Use the magnification formula with the first lens, to find the height of the first image. Then, use that height to find the height of the second image.

    Your numbers look right. (meaning your initial answer of 1.01cm) How do you know your answer is wrong?
  6. Jul 20, 2008 #5
    yes, 3 cm and -3 cm are definitely wrong. I'm stumped.
  7. Jul 20, 2008 #6


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    Do you know the correct answer? If so, post it. Maybe I'll be able to see the mistake we're making if I knew the answer we were looking for.
  8. Jul 20, 2008 #7
    nope, the program just tells me that my answer is incorrect. Oh man.
  9. Jul 21, 2008 #8


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    Well. It could always be that the solution the program has is incorrect. Or, we could both be missing something obvious. I'll keep thinking about it. If I figure anything out, I'll let you know. In the mean time, I suggest you bring this up with your instructor. They would be able to help you if the program was marking you incorrect, even when you were correct.

    Another thing to check is if you are using the correct numbers. Are you sure you copied the numbers from the original problem correctly? (I know this sounds stupid, but always worth double checking something like this.:smile:)

    Also, it could be a rounding error. I was coming up with an answer of 1.00, where you were coming up with an answer of 1.01. Most computer programs for physics problems will allow errors this small, but if your using some obscure program, them maybe this could be it?
  10. Mar 15, 2009 #9
    1.00 is correct with the mastering physics website.
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