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Optics Matrix Methods

  1. Nov 26, 2006 #1
    So I've gotten through half of this very long problem involving thick lens matrices and using the calculated system matrix for image locations. I need help at part d!!!! I'm going mad with it.

    [tex]S = \begin{array}{cc}\frac{11}{8}&\frac{5}{2}\\\frac{3}{20}&1\end{array}[/tex]

    So I ended up with a system matrix with elements a11= 11/8 a12= 5/2 a21= 3/20 a22= 1. Calculations were S1=

    r2 x t1 x r1. Looks good so far.

    I did this correct I think, I'm trying to get my scanner to work now.
    I got an answer of -60/11 cm, that's to the left of the right side of the lens.

    I got a system matrix of elements a11= 11/8 + 6x/40, a12= 11s/8 + 6sx/40 + x + 5/2, a21= 6/40, a22= 1 +

    6s/40. I took the original three refraction and transfer matrices and added two more transfer matrices, one

    for s and x. Calculations were S1= t3 x r2 x t2 x r1 x t1.

    I have no idea how to do this. I figured out the focal lengths, and they came out to be the same as in part

    B, but the matrix in part D has the first two elements with respect to x when s= 30cm. Do I use matrix

    algebra to solve for x? What about magnification. I think I missed a lecture.

    The rest of the problem:
    It acts like a thin lens doesn't it? p and q will be zero, focal points will be on their respective input and

    output planes. That's all I have figured out.

    Simple; you multiply the system matrix by an object matrix o11, o21 to solve an image matrix i11, i21 with

    height and angle as elements.
    Last edited: Nov 27, 2006
  2. jcsd
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