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Homework Help: Optics question

  1. Nov 2, 2003 #1
    Hi,

    I'm hoping I can get some help. (I'm a first time poster, so if there's something that isn't quite clear, please let me know).

    A converging lens (n=1.355) in air has diameter 4.0cm, thickness 0.50cm and zero thickness at its edges.

    A point object at 'So' = 18cm has an image at 'Si'= x cm. (where 'So' and 'Si' are distances from the center of the lens)

    (a) Write an expression for the Optical path length (OPL) of the ray passing through the edge of the lens and (b) an expression for the OPL of the ray passing through the centre of the lens, and (c) determine the distance x.

    (sqrt= square root, sq = squared) For (a) I used pythagorus. OPL = sqrt(18 sq + 2 sq) + sqrt(x sq + 2 sq)

    (b) OPL = 17.75 +(x-0.25) + 0.5n
    =17.5+x+.6775
    =18.1775+x
    =18.18+x

    (c) Now I have both in term of x. Since OPL1=OPL2,
    sqrt(18 sq + 2 sq) + sqrt(x sq + 2 sq)=18.18+x
    (square both sides) 18 sq + 4 + x sq + 4 = 18.18 sq + x sq
    332 + x sq = 330.5 + x sq

    I can't think of any other way to do it, yet something must be wrong as I have to calculate x. Is there a reason why I'm 1.5 off other than rounding? Wouldn't the x sq both cancel out leaving me with nothing?

    Thanks
    goodo
     
    Last edited by a moderator: Nov 2, 2003
  2. jcsd
  3. Nov 2, 2003 #2
    Welcome goodo,
    I think everything's OK up to here:
    The square of a sum is not the sum of squares.
    Remember: (a + b)2 = a2 + 2ab + b2.

    BTW, the 1st root has no x, so you can calculate its value. Makes things easier...
     
    Last edited: Nov 2, 2003
  4. Nov 2, 2003 #3
    Thansk for your response arcnets.
     
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