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Homework Help: Human eye optics

  1. Jan 19, 2010 #1
    1. The problem statement, all variables and given/known data

    The far point of a nearsighted person is 6.0 m from her eyes, and she wears contacts that enable her to see distant objects clearly. A tree is 18.0 m away and 2.0 m high. (a) when she looks through the contacts at the tree, what is its image distance? (b) How high is the image formed by the contacts?

    2. Relevant equations

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

    3. The attempt at a solution
    I'm not sure how to start the problem. Can someone point me in the right direction?
     
  2. jcsd
  3. Jan 19, 2010 #2

    tiny-tim

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    Hi runfast220! :wink:
    The contact is a lens that makes an object at ∞ have an image at 6.0 m.

    So calculate its f, and then apply that to an object at 18.0 m. :smile:
     
  4. Jan 19, 2010 #3
    ok so...

    1/f = 1/infin + 1/6
    so f=6.0

    1/di = 1/6 - 1/18 = .11
    di = 9.0 m

    hi/ho = - di/do
    hi/2 = -9/18

    hi = -1.0 m

    does that look right?
     
  5. Jan 19, 2010 #4

    tiny-tim

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    Hi runfast220! :smile:
    Nooo :redface: … you've got the tree focussing at 9.0 m, which is too far away for the nearsighted person to see it!

    Hint: is the contact a converging or a diverging lens? :wink:
     
  6. Jan 19, 2010 #5
    Its a converging lens so does that make the di a virtual image thus making it -9.0m?
     
  7. Jan 19, 2010 #6

    tiny-tim

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    No, it's a diverging lens …

    a converging lens wold make parallel lines (from ∞) come closer, but these contacts make them diverge, so that they appear to come from 6.0 m in front of the person.

    And (from the PF Library on lens …)
     
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