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Optical Ray tracing

  1. Feb 11, 2009 #1
    Is anyone familiar with the "y n u" ray trace method for optical systems, using paraxial rays where Snell's law is approximated using:

    n1sinx1 ≈ n1tanx1

    ≈ n1u1 = n2u2

    If anyone is familiar with this, I'd be interested to discuss the finer points of "a-ray" and "b-ray" tracing, as well as aperture stops, as I don't quite understand the methods involved. Thanks.
     
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  3. Feb 12, 2009 #2

    Andy Resnick

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    I'm familiar with y-nu traces (as well as y-y' plots), but I have not heard of a-ray and b-ray traces. Can you giove some more specifics?
     
  4. Feb 12, 2009 #3
    An a-ray is just an axial ray that goes through the aperture stop. Generally, you find the angle in the object plane that corresponds to when the a-ray just kisses the edge of the aperture stop, AKA the Y height at that point is the radius of the aperture stop.

    A b-ray is a ray that goes through the center of the aperture stop. This involves backwards ray tracing, and it helps you to find how much light is coming through the optical system. At the aperture stop surface, the Y height is zero and n*u is determined given whatever parameters you're looking for, such as a certain object half angle or image height.

    Anything sound familiar?
     
  5. Feb 12, 2009 #4

    Andy Resnick

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    Yes, now it does. What you call the 'a-ray' is referred to as the "Marginal ray" and is related to the numerical aperture. the 'b-ray' is more commonly called the "chief ray" and the (y,nu) is written with an overbar. The Chief ray is used to determine the magnification, and relates the the field of view.

    If you like, the chief and marginal rays are the basis set for all paraxial rays that pass through the optical system, and both are defined by the aperture stop, a physical object that limits the passage of light through an optical system. The marginal and chief rays are used in ray-tracing matrices, and are used to define the optical invariant of the system.

    What are you trying to do?
     
  6. Feb 12, 2009 #5
    I have heard those terms briefly. My professor at the institute of optics at the university of rochester probably just wanted to shorten the terms, so he calls them a and b rays.

    Anyways, I'm just trying to get a good grasp of how to evaluate a given optical system using those rays, why you might use them, that sort of thing.

    Say, for example, you've got an optical system with 20 lenses, you're given the curvature, the thickness between surfaces, and indices of refraction, and you want to figure out which lense surface acts as the aperture stop for both a distant object and an object at some finite distance. How would you go about doing this?
     
  7. Feb 13, 2009 #6

    Andy Resnick

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    It's been a while since I had to think about this stuff, but here's my thoughts:

    1) trace two arbitrary rays through the system-these are what I call the 'a-ray' and 'b-ray'.
    2) using superposition, transform these into the chief and marginal rays.
    2a) This is done at two planes- the image plane (y= 0) and the aperture stop plane (y-bar = 0). If you don't know where the aperture stop is, you may start to run into problems.
    2b) choose the a-ray and b-ray carefully- the a-ray starts at the optical axis (for example), and the b-ray starts at the maximum field height (for example).
    3) Now you have the chielf and marginal rays, which you can trace through the whole system and see what's what. For 20 lenses, I hope you take the time and automate the ray trace algorithm.

    I don't know of that many people who use ray tracing for anything other than very simple systems- ray tracing has been computerized now, and optical design in general has gotten much more sophisticated.
     
  8. Feb 14, 2009 #7
    This is simply in the context of the Geometrical Optics course I'm taking. I'm just getting into optics in school, and I am just trying to gain some insight into this method for evaluating optical systems.

    I use Excel to automate the process. The 20 lens system was just hypothetical.

    Moving on, do you happen to know a way to find the effective aperture stop in the system using this method?
     
  9. Feb 16, 2009 #8

    Andy Resnick

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    It's the location where the Chief ray crosses the optical axis, with a diameter set by the height of the marginal ray.
     
  10. Feb 17, 2009 #9

    Andy Resnick

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    I'm not sure my answer was very useful. Let me try again:

    Ok, so you have traced two arbitrary rays through the system, one starting at the optical axis and the other at the maximum field height. Call these (y1, nu1) and (y2, nu2). The next step is to calculate the marginal (y, nu) and chief (y', nu') rays. Becasue paraxial rays obey linear superposition,

    y1 = A y + B y'
    nu1 = A nu + B nu'

    and similarly for y2. At the image plane, y1 = 0 and y = 0 and y' = y2 (image height). At the aperture stop plane, y = Y (radius of stop) and y' = 0. Two sets of equations, solve for the coefficients and then you get the chief and marginal rays.

    Now, looking at the list time I did this, I think I knew where the stops were located and I have some cryptic notes which I can't quite decipher. I have the field stop located at the surface with the smallest r/y' (r is the radius of the surface, and not the radius of curvature), and the aperture stop located at the plane with the smallest r/y. But as I said, it's a bit cryptic.
     
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