Spherical Aberrations

  1. 1. The problem statement, all variables and given/known data
    A collimated light beam is incident on the plane side of a of index 1.5, diameter 50mm, and radius 40mm. Find the .

    2. Relevant equations
    Refraction in a plane surface:
    Refraction on a spherical surface:
    Spherical Wave Aberrations:
    [tex]a(Q)= \frac{h^4}{8} \left[ \frac{n_1}{s} \left( \frac{1}{s}+\frac{1}{R}\right)^2 + \frac{n_1}{s'} \left( \frac{1}{s'}-\frac{1}{R}\right)^2 \right][/tex]

    R=Radius of curvature
    s=Object distance
    s'=Image distance
    n=index of refraction

    3. The attempt at a solution
    So the light is traveling in a straight beam incident on the plane surface. By equation 1, the image distance is virtual and at infinity. so [itex]s'=-\infty mm[/itex]. I then use this value for the calculation of the image of a spherical surface using equation 2. Plugging everything in, I get a distance of 80mm. So s'=80mm.

    I then used equation 3 to get the spherical wave aberration and used a height of 25mm since the lens is 50mm in diameter. (Also, R<0 since it is a concave surface from left to right.) I got an aberration of -1.29mm. The book says the correct answer is -0.858mm.

    What am I doing wrong? :(?

    EDIT: Figured it out! I used 1.5 for the index in equation 3 instead of 1.
    Last edited: Sep 28, 2011
  2. jcsd
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