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Optics: Ray matrix for a thin lens

  1. Jan 5, 2010 #1
    Hi guys

    I wonder if you can explain this to me. In http://en.wikipedia.org/wiki/Ray_transfer_matrix_analysis#Table_of_ray_transfer_matrices it says under "Thin lens" that the ray matrix is given by

    \left( {\begin{array}{*{20}c}
    {r_{final} } \\
    {r_{final}^' } \\
    \end{array}} \right) = \left( {\begin{array}{*{20}c}
    1 & 0 \\
    { - 1/f} & 1 \\
    \end{array}} \right)\left( {\begin{array}{*{20}c}
    {r_{initial} } \\
    {r_{initial}^' } \\
    \end{array}} \right)

    Here r is the ray displacement and r' is the slope of the ray. Now the above ray matrix tells us that rfinal = rinitial, but I cannot see why this is the case at all. I mean the displacement of the initial ray is constantly growing, while the displacement of the final ray is constantly decaying. Then how can we equal to the displacements?

    The only argument my book uses is that "Immediately to the right of the lens the ray's displacement is the same as the initial displacement immediately to the left." But still, the displacements are not constant, so I cannot see why this argument works.
  2. jcsd
  3. Jan 5, 2010 #2

    Andy Resnick

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    The thin lens approximation replaces the physical lens with an infinitesimally thin planar object, the only effect of which is to change the direction of the ray. Thus, on either side of the lens, the ray height is a constant.
  4. Jan 5, 2010 #3
    Ok, this I can agree on.

    This I do not agree on. When you say "height", the way I understand it is that we have an axis (e.g. the z-axis), and the ray height at a point z0 is the distance from the axis to the ray at point z0. With this definition, I fail to see why the ray height is constant for both rays.

  5. Jan 5, 2010 #4
    For a thin lens, the ray displacement from the central axis is equal height on both sides of the lens. Only the slope changes.

    For a reasonable thick-lens approximation for a lens with thickness L, insert drifts with 2 x 2 matrix elements R11 and R12 = 1 and L/2, and R21 and R22 = 0 and 1 respectively both before and after the thin lens approximation.
    Bob S
  6. Jan 5, 2010 #5
    Ok, I think my problem is the definition of the ray displacement. When talking about ray displacement, do we mean an expression r(z) showing the displacement from the central axis at some point z, or is the displacement just the total heigh difference from the central axis to the start (or end) or the ray?
  7. Jan 5, 2010 #6
    Yes. for example, in a drift of length L, using the 2 x 2 matrix my previous post
    r(z+L) = R11·r(z) + R12·r'(z) = r(z) + L·r'(z)
    Bob S
    Last edited: Jan 5, 2010
  8. Jan 6, 2010 #7
    But in that case I don't understand why the displacement is the same on both sides of the lens. Lets say we are looking at a distance on the central axis 2L, and the lens is at L.

    The displacement at some point 0<z<L is given by r(z) = r(0) + z·r'(0). The displacement at some point L<z<2L is r(z) = r(L) - z·r'(0), since the slopes just differ in the sign. With this reasoning (which seems pretty straightforward), I cannot see why the displacements are the same.
  9. Jan 6, 2010 #8
    The example is for a drift without a lens, so the only change in displacement is due to z·r'(0).
    r(z) = r(0) + z·r'(0)
    r(L) = r(0) + L·r'(0)

    If z<L then
    r(z) = r(L) + (z-L)·r'(L) = r(L) - (L-z)·r'(L)

    special case: r'(L) = r'(0) for drift
    r(z) = r(L) + (z-L)·r'(L) = r(L) - (L-z)·r'(0)

    Always use 2x2 matrices to go forward or backward.

    You can concatonate 2x2 matrices without permutation: M = M4 x M3 x M2 x M1

    If [r(z),r'(z)] = M[r(0),r'(0)], then
    [r(0),r'(0)] = M-1[r(z),r'(z)]
    where I am using [r(z),r'(z)] to represent a column matrix

    Bob S
    Last edited: Jan 6, 2010
  10. Jan 6, 2010 #9
    Ok, this is clever. I hadn't thought of that.

    But there's no difference between the drift without a lens and drift with a lens other than that the lens changes the slope from r' to -r'. I still cannot see how one can arrive to the statement that "the displacements on both sides of the lens are equal".
  11. Jan 6, 2010 #10
    For a thin lens (using my special notation for a 1 x 2 and a 2 x 2 matrix)
    [r(z),r'(z)]final = [R11,R12,R21,R22]·[r(z),r'(z)]initial = [+1,0,-1/f,+1]·[r(z),r'(z)]initial
    So, using your 2 x 2 matrix for a thin lens:
    r(z)final = r(z)initial
    r'(z)final = -r(z)initial/f +r'(z)initial

    So r(z) on both sides of a thin lens are equal as expected, because the drift length R12 = 0.

    Bob S
  12. Jan 6, 2010 #11


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    Niles, in case it isn't clear yet, Andy and Bob were just talking about the ray height at the lens. To use your example, that height is r(L) just before the lens, and it is r(L) just after the lens. Those two heights are equal; but move away from the lens (z≠L) and we all agree the height will be different.

    The matrix in your OP only applies to a ray just before and just after the lens. Additional matrices, for free-space propagation, are required if you want to know the ray height and angle at some distance from the lens.
  13. Jan 6, 2010 #12
    Thank you very much. This I understand, and it answers my question 100%. That was very kind of you.

    Thanks. My question was a lot more fundemental/simple, and Redbelly hit the nail right on the head. But I still learned something from your posts, so thank you for taking the time to help.

    I hope you all had a happy new year.
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