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Homework Help: Finding length and longitudinal magnification

  1. Apr 13, 2007 #1
    1. The problem statement, all variables and given/known data
    A small object of length dL is placed on the axis of a converging mirror of focal length f. The center of the object is a distance o from the mirror. Determine the length of the image and thus effective determine a longitudinal m' = dL'/dL for the lens.


    2. Relevant equations




    3. The attempt at a solution

    I know that the distance from the center to the object to the mirror is o so one end of the mirror is o+dL/2 and the other end is o - dL/2. I will call x = o + dL/2 and y = o -dL/2

    I know 1/f = 1/o + 1/i

    1/f = 1/x + 1/i
    1/f - 1/x = 1/i
    i = (x - f)/xf
    x' = (x-f)/xf

    For the other one

    i = (y-f)/yf
    y' (y-f)/(yf)

    I know dL =|x-y|
    which is just dL

    dL'=|x'-y'|
    =[(x-f)*yf - (y-f)*xf]/xyf^2
    [xyf - yf^2 - xyf -xf^2] / xyf^2
    [-yf^2 - xf^2] / (xyf^2)
    (-y - x) /xy

    [-(o - dL/2) - (o+dL/2) ]/ [(o-dL/2)(o+dL/2)]
    dL'=-2o / (o-dL^2/4)

    m' = -2o/(o-dL^2/4)/dL
    m' = -2o/(dL*o - dL^3/4)

    The correct answer is m'=-m^2. By definition, m = -i/o. I don't see how to get this. Did I approach the problem incorrectly or am I overlooking some kind of algebra manipulations? How do I get to -m^2?
     
  2. jcsd
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