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Physics 11

  1. Jul 12, 2007 #1
    Use the mirror equation to find the image location and the height of an object placed at the centre of curvature of a concave mirror. Also find the magnification. Hint: What is the relation between the focal length and the object distance, do, for this situation?

    I'm really confused about this question... Can someone please help me????
     
  2. jcsd
  3. Jul 12, 2007 #2

    Chi Meson

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    What's the mirror equation? (It's the same as the thin-lens equation)

    What's the relationship between the center of curvature and the focal point?
     
  4. Jul 12, 2007 #3
    centre of curvature is half the focal point and the mirror equation is 1/di + 1/do = 1/f
     
  5. Jul 12, 2007 #4
    but i still don't understand how that answers the question..
     
  6. Jul 12, 2007 #5

    Chi Meson

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    I was answering your other thread.

    Center of curvature is not half the focal length, it's the other way 'round.

    Center of curvature is twice the focal length. So focal length is "f" and the object distance is "2f." Do a little algebra and find the image.
     
  7. Jul 12, 2007 #6
    Would the focal length be equal to the object distance?
     
  8. Jul 12, 2007 #7

    Chi Meson

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    No, try again. You're close.

    But I have to go now, keep at it.
     
  9. Jul 12, 2007 #8
    Okay, thnx for your help
     
  10. Jul 12, 2007 #9

    Kurdt

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    All you have to do is substitute 2f = do into the equation you listed above and solve for di.
     
    Last edited: Jul 12, 2007
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