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Finding radius of curvature of an eyeball

  1. Nov 1, 2014 #1
    1. The problem statement, all variables and given/known data
    Consider a simplified model of the human eye, in which all internal elements have the
    same refractive index of n = 1.40. Furthermore, assume that all refraction occurs at the
    cornea, whose vertex is 2.50 cm from the retina. Calculate the radius of curvature of the
    cornea such that the image of an object 40.0 cm from the vertex of the cornea is focussed
    on the back of the eye (the retina).

    2. Relevant equations
    1/s+1/s'=1/f

    1/f=(n-1)(1/R_1-1/r_2)

    3. The attempt at a solution

    I attempted to find the focal point with

    1 / -40 + 1 / 2.5 = 0.375cm
    With that I figured I should use the lensmaker equation but I've never seen a problem where you have to solve the radius of curvature and couldn't find any examples like this online.

    I changed the equation to 1/f=(n-1)R and solved for R but I'm not sure if that would be the right thing to do.

    Thanks
     
  2. jcsd
  3. Nov 6, 2014 #2
    Thanks for the post! Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
  4. Nov 7, 2014 #3

    rude man

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    Gold Member

    Why is "40" prefixed by a minus sign?
    Make it 1/f = (n-1)/R and you got a deal. It assumes the eye lens is plano-convex, i.e. only one surface is curved. That seems to be the intent of the problem (it mentions only one radius of curvature). In reality the cornea is convex-convex, though.
     
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