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Image Height Question

  1. Nov 17, 2013 #1
    1. The problem statement, all variables and given/known data

    A man holds a double-sided spherical mirror so that he is looking directly into its convex surface, 44.8 cm from his face. The magnification of the image of his face is +0.22. What will be the image distance when he reverses the mirror (looking into its concave surface), maintaining the same distance between the mirror and his face?

    2. Relevant equations

    m = -di/do
    1/f = 1/di + 1/do

    3. The attempt at a solution

    First I solved for the focal length like this:

    di = - (m*do)
    di = - (.22*44.8) = -9.856 cm

    1/f = 1/di + 1/do
    1/f = 1/-9.856 + 1/44.8
    f = -12.64 cm

    Then I tried to solve for the image height of the concave side:

    1/di = 1/f - 1/do
    1/di = 1/-12.64 - 1/40.8
    di = -9.856 cm

    But the answer should have been 17.6, I don't really know what to do. Please help?
     
  2. jcsd
  3. Nov 17, 2013 #2
    Oh, I think I got it! I think the signs in the formula switch to + for concave surfaces. When I do 1/di = 1/-12.64 + 1/44.8 I get -17.6. But I think my signs are wrong altogether, because it's supposed to be positive...
     
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