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Biconvex Lens

  1. Mar 29, 2005 #1
    An object is located 36 cm to the left of a biconvex lens of index of regraction 1.5. The left surface of the lens has a radius of curvature of 20 cm. The right surface of the lens is to be shaped so that a real image will be formed 72 cm to the right of the lens. What is the required radius of curvature for the second surface?

    I asked earlier on this, but I still got it wrong: this is what I did -

    I used 1/F=(n-1)(1/R1 + 1/R2)

    1/72 = (1.5 - 1) (1/.2 + 1/R2)
    Solved it and got R2 to equal 45 cm. I am pretty sure 1/F shouldnt be 72, but if its unknown that would make it unsolvable unless I am missing something?
     
  2. jcsd
  3. Mar 29, 2005 #2

    Doc Al

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    Staff: Mentor

    First use the thin lens equation to solve for the focal length of the lens:
    [itex]1/o + 1/i = 1/f[/itex]​
     
  4. Mar 29, 2005 #3
    Yeah I thought of that.

    1/f = 1/56 + 1/72

    Is it 56 because there is the 20 cm radius of the left lens and then 36 cm to the object?
     
  5. Mar 29, 2005 #4

    Doc Al

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    No. 20 cm is the radius of curvature of the left surface of the lens, not the thickness of the lens. Assume the lens is thin enough to use the thin lens formula:
    [itex]1/36 + 1/72 = 1/f[/itex]​
     
  6. Mar 29, 2005 #5
    Ah alright, that gives focal length of 4.167 cm, then just plug that into the

    1/f=(n-1)(1/R1 + 1/R2)
     
    Last edited: Mar 29, 2005
  7. Mar 29, 2005 #6

    Doc Al

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    Careful not to mix up f with 1/f.
     
  8. Mar 29, 2005 #7
    Right, thanks for the help
     
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