Parabola incoming ray not parallel to axis

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Main Question or Discussion Point

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?
 

Answers and Replies

  • #2
phinds
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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.
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
What about light coming in not parallel to the central axis. What path do these rays take?
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.
 
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  • #3
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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?
 
  • #4
phinds
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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?
Why don't you draw one and see?
 
  • #5
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Would I do it by having the angle of incidence equal to the angle of reflection at each point of the parabola wall?
 
  • #6
phinds
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Would I do it by having the angle of incidence equal to the angle of reflection at each point of the parabola wall?
Uh ... isn't that how reflection WORKS ?
 
  • #7
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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.
 
  • #8
phinds
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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.
Oh, it's going to be a mess all right. The focal point is totally irrelevant.
 
  • #9
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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?
 
  • #10
jbriggs444
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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.
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.
 
  • #11
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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.
 
  • #12
DaveC426913
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  • #13
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I am not familiar with how you constructed this shape. Are you sure it's a parabola?
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
 
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  • #14
phinds
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Edit: i see that was a mistake
OH, yeah! Definitely a mistake. You can almost certainly find a good parabola curve on the internet and print it out
 
  • #15
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find a good parabola curve on the internet and print it out
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?
 
  • #16
phinds
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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?
No, I think you likely just happened to hit a good case.
 
  • #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.
 
  • #18
Drakkith
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What about light coming in not parallel to the central axis. What path do these rays take?
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
 
  • #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)
 
  • #20
phinds
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Feel free to provide double-digit angle examples then...
Oh, no ... I leave the hard work to you. I'll stick with the sideline kibitzing :)
 
  • #21
Drakkith
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Oh, no ... I leave the hard work to you. I'll stick with the sideline kibitzing :)
How did you even quote that? I deleted it like 20 seconds after posting it, lol.
 
  • #22
phinds
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How did you even quote that? I deleted it like 20 seconds after posting it, lol.
Hey, I am FAST !!!

Actually, it's trivial. You posted it so I got an email with the contents and I just quoted your other post and exchanged content. Sneaky, huh? I've often thought of doing that to embarrass someone but generally I allow my better self to intervene. In your case I was willing to make an exception. :-p
 
  • #23
sophiecentaur
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The nearest thing to what you want, in practice is probably the Corner Cube Reflector. See this wiki link.
I can't think of a curve that will give you the right result.
 
  • #24
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Does this help?

http://www.astro.virginia.edu/class/oconnell/astr511/im/paraboloid-3deg-offaxis-coma-in-out.gif
What program did you use to make this? I made my own attempt on paper.
I think the answer I was looking for is, all light parallel to the axis goes through the focus, even if the light source is off axis. Yes it's obvious but it didn't make sense till I drew it.

parabola.jpg
 
  • #25
sophiecentaur
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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.
This is basically not right. All the light coming parallel with the axis will end up passing through the Focus but, after it passes through the focus all but a minute fraction will go towards the source, the light will diverge after the 'image' is formed. The same goes for off axis arrivals, but the focus will be offset from the principle focus and then disperses again. Perhaps the OP meant something different, though.
The only time that all light is reflected back to its source is when it strikes a plane mirror along the normal.

Many correct points have been made in this thread but the OP is not correct
 

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