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Parabola incoming ray not parallel to axis

  1. Dec 25, 2014 #1
    If incoming light to a parabola is parallel to the central axis the light is reflected through the focal point and the back to its source.

    What about light coming in not parallel to the central axis. What path do these rays take?
     
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  3. Dec 25, 2014 #2

    phinds

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    No, it is not. It is reflected to the focal point of the parabola and if there isn't anything there to stop it, it spreads out in a wide cone, not reflected back to its source
    That will be a function of the exact angle but it will be some kind of cone as it leaves the parabolic reflector, just not as symmetrical a cone as you get with rays parallel to the axis.


    EDIT: Ah ... I see that I am assuming a short reflector like a car headlamp or a photographers reflector and you are likely assuming a much longer one, in which case you are right although your statement leave out the significant factor of its hitting the reflector wall again and THEN being reflected back to the source. My short reflector isn't long enough for it to hit the reflector again, which is why I said what I said.
     
    Last edited: Dec 25, 2014
  4. Dec 25, 2014 #3
    Yeah sorry I should have been more clear.

    The incoming ray reflects of the parabola wall through the focal point then reflects of the parabola wall on the opposite side of the focal point back to the light source.

    In this case what happens to rays shining into the parabola not parallel to the central access?
     
  5. Dec 25, 2014 #4

    phinds

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    Why don't you draw one and see?
     
  6. Dec 25, 2014 #5
    Would I do it by having the angle of incidence equal to the angle of reflection at each point of the parabola wall?
     
  7. Dec 25, 2014 #6

    phinds

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    Uh ... isn't that how reflection WORKS ?
     
  8. Dec 25, 2014 #7
    Yeah, I was hoping the focal point would provide a short cut as finding the normal the parabola wall isn't as quick as the normal to a plane surface.
     
  9. Dec 25, 2014 #8

    phinds

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    Oh, it's going to be a mess all right. The focal point is totally irrelevant.
     
  10. Dec 25, 2014 #9
    I used the focal point to determine the normal at each of the points where my source ray reflected of the parabola wall.
    And the results are, as can be seen in my rough and ready image in yellow, if it hits the parabola walls twice it ends up parallel to the central axis. I almost didn't believe it.
    http://i.imgur.com/O8vREsQ.jpg

    So that would mean if I put a reflective parabola in a uniformly lit room, the wall directly in front of the parabola would be brighter?
     
  11. Dec 25, 2014 #10

    jbriggs444

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    You would have been right not to believe it. It is not true.

    Trace that ray backward. It came in parallel to the axis of the parabola. Accordingly it must be reflected through the focus. Keep tracing it. It came through the focus, so it must be reflected parallel to the axis of the parabola.

    A ray coming in from off-axis cannot reflect through the focus of the parabola. A ray leaving the parabola on-axis must have reflected through the focus. Accordingly, a ray coming in off axis can NEVER be reflected parallel to the axis, no matter how many times it is reflected.

    EDIT: Exception -- a ray that comes in and hits the focus without first having been reflected.
     
  12. Dec 25, 2014 #11
    If I trace the ray back and out through the focus then the angle of incidence and reflection aren't equal. Which must mean there is something very wrong with my parabola sketch, but I can't see it.
     
  13. Dec 25, 2014 #12

    DaveC426913

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  14. Dec 25, 2014 #13
    It's actually a concave spherical mirror, I didn't know how to sketch a parabola, so I thought it was a good enough approximation.

    Edit: i see that was a mistake
     
    Last edited: Dec 25, 2014
  15. Dec 25, 2014 #14

    phinds

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    OH, yeah! Definitely a mistake. You can almost certainly find a good parabola curve on the internet and print it out
     
  16. Dec 25, 2014 #15
    I can do that when I get home, but then is my ray drawing for the spherical mirror correct, where off axis light sources end up parallel to the central axis?
     
  17. Dec 25, 2014 #16

    phinds

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    No, I think you likely just happened to hit a good case.
     
  18. Dec 25, 2014 #17

    jbriggs444

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    For a spherical mirror it is clear that a light ray can trace out the outline of any regular n-gon. More generally, the number of reflections required to complete a circuit of a mirrored spherical shell need not be an integer. The case of exactly two reflections reflecting by 180 degrees would have the ray tracing out the outline of a square and would require a 45 degree hit on the mirror at each strike.
     
  19. Dec 26, 2014 #18

    Drakkith

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    Does this help?

    http://www.astro.virginia.edu/class/oconnell/astr511/im/paraboloid-3deg-offaxis-coma-in-out.gif

    http://www.astro.virginia.edu/class/oconnell/astr511/im/paraboloid-5deg-offaxis-coma.gif
     
  20. Dec 26, 2014 #19

    phinds

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    You've chosen near-parallel (to the axis) examples which somewhat mask how messy it can get (I think ... I haven't done the exercise)
     
  21. Dec 26, 2014 #20

    phinds

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    Oh, no ... I leave the hard work to you. I'll stick with the sideline kibitzing :)
     
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