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Lens question

  1. Mar 24, 2006 #1

    Hi... Do you know how to prove R=2f where R is the radius of curvature and f is the focal length?
  2. jcsd
  3. Mar 24, 2006 #2


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    The parabola [itex]y= \frac{1}{4f}x^2[/itex] has focus at (0, f) so focal length f. The curvature of a function y= y(x) is given by
    [tex] \frac {|y"|} {(1+(y')^2)^{ \frac {3}{2}}}[/tex]

    In this case, y'= (1/(2f)x and y"= 1/(2f). At x= 0, the curvature is
    1/(2f) and so the radius of curvature is 2f.
    Last edited by a moderator: Mar 24, 2006
  4. Mar 24, 2006 #3
    Sorry, I can't see the image of the graph. Can you plz send it again? thank you very much!!
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