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Object-to-image distance

  1. Sep 6, 2013 #1
    1. The problem statement, all variables and given/known data
    A 50 mm focal length thin lens in air is used to image a 250 mm x 250 mm object onto a 10 mm x 10 mm detector. What is the overall object-to-image distance (throw)? Besides the exact solution, can you think of a reasonable approximation?


    2. Relevant equations
    Throw, T = l' - l
    For a thin lens: (1/l') = (1/F) + (1/l)
    For a thin lens in air: f = -F

    3. The attempt at a solution

    Since f = 50 mm was given, F = -50 mm. I'm thinking that if I could find either l' or l, I could then use the second equation to find the remaining unknown, and then plug into the equation for T. I'm not sure how to use the given dimensions for the object and detector. I know of no equations that incorporate the area, and I can't determine what else they could be used to describe.
     
  2. jcsd
  3. Sep 6, 2013 #2

    ehild

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    You get the magnification from the given sizes of object and image. The image is on the detector, so it is real. How are l and l' related ?

    ehild
     
  4. Sep 6, 2013 #3
    M = l' / l

    Is it correct to think that the lateral magnification is the ratio between the horizontal length of the image and that of the object? If so, then M = 10 mm / 250 mm = .04

    l' = l*M so 1/F = 1/(l*M) - 1/l = 1/(l*M) - M/(l*M) = (1 - M)/(l*M)
    => l = F*(1 - M)/M
     
  5. Sep 6, 2013 #4

    ehild

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    take care of the sign of M.

    ehild
     
  6. Sep 6, 2013 #5
    Why would either 10 mm or 250 mm be negative? I understand l would be negative since it is measured from the lens to the object, so I agree that M should be negative, but I didn't think the dimensions of the object and image were given relative to anything. Is taking the ratio between horizontal lengths "one of those" formulas where I calculate the magnitude and have to come up with the sign independently?
     
  7. Sep 6, 2013 #6

    ehild

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    We take the real image inverted, so its height negative.

    ehild
     
  8. Sep 6, 2013 #7
    Oh, it's the ratio of heights, not widths. Thank you very much for your help. :)
     
  9. Sep 7, 2013 #8

    ehild

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    You are welcome. :smile:

    ehild
     
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