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Refraction (magnification) - very basic

  1. Dec 10, 2006 #1

    A lens produces a real image that is twice as large as the object and is located 15 cm from the lens. Find (a) the object distance and (b) the focal length of the

    [tex] m = \frac{-i}{p} [/tex]

    So the image is twice as large so obviously [tex] m = 2 [/tex]. The lens produces a real image that is 15cm from the lens. Thus, [tex] i > 0 [/tex] RIGHT??? So we are simply left with,

    [tex] p = \frac{-15cm}{2} [/tex]. How can [tex] p [/tex] (the objects distance) be negative? This doesn't make any sense to me. I can resolve this physically but not in the math... I would like to know why.

    This problem should be trivial.
  2. jcsd
  3. Dec 10, 2006 #2
    The issue here is that there are bunch of easy-to-forget sign conventions that go along with the optics formulas. In particular: we are dealing here with a REAL image. Real images are ALWAYS inverted. Because the real image is inverted, it's magnification is technically -2, not +2. This will solve your sign problem and make the object distance positive.
  4. Dec 11, 2006 #3
    Thanks :)

    I figured it would be something simple like that. Much appreciated.
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