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Image Distance and Magnification

  1. Nov 18, 2006 #1
    Question

    An object is a distance of 23.4 cm from the center of a silvered spherical glass Christmas tree ornament which has a diameter of 5.70 cm.

    A What is the position of its image?

    B What is the magnification of its image?

    I think the object can be treated as a convex mirror.

    s (object distance) = 23.4 cm
    R = -(5.70)/2 = -2.85 cm

    [tex]\frac{1}{s} + \frac{1}{s'} = \frac{2}{R}[/tex]

    Using that equation, I get [tex]\frac{1}{s'} = \frac{2}{-2.85} - \frac{1}{23.4}[/tex].

    Then, s' = -1.12 cm.

    And, [tex]m = -\frac{s'}{s}[/tex], so m = .048.

    I know for sure that -1.12 cm is not the correct answer, so most likely my second answer is also incorrect.

    What did I do wrong?
     
  2. jcsd
  3. Nov 19, 2006 #2
    Did I not provide enough work?
     
  4. Nov 19, 2006 #3

    Doc Al

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    Staff: Mentor

    Your method is fine but check your arithmetic.
     
  5. Nov 19, 2006 #4
    Sorry, I still get the same answer.
     
  6. Nov 20, 2006 #5

    Doc Al

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    Staff: Mentor

    Please redo this calculation.
     
  7. Nov 24, 2006 #6
    Doc Al,

    Indeed, it was incorrect. Thanks!
     
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