Parabola incoming ray not parallel to axis

  • #1
zimbabwe
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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.

What about light coming in not parallel to the central axis. What path do these rays take?
 

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  • #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
zimbabwe
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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
zimbabwe
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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
zimbabwe
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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
zimbabwe
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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
zimbabwe
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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.
 
  • #13
zimbabwe
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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
zimbabwe
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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
zimbabwe
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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
 
  • #26
zimbabwe
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Is there a time limit on editing previous posts? Or at least to mark that I've realized I've written something that doesn't make sense

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.

This should be changed to, "all light in front of the parabola that is parallel to the central axis will go through or has originated from the focal point of the parabola. "
 
  • #27
zimbabwe
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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.

Assuming the parabolic walls are long enough won't a parabola give you the same result, but only for rays parallel to the central axis. All rays parallel to the central axis will reflect through the focal point, continue through the focal point and hit the opposite parabola wall, and they will end up parallel to the central axis going back to their source.
 
  • #28
sophiecentaur
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Assuming the parabolic walls are long enough won't a parabola give you the same result, but only for rays parallel to the central axis. All rays parallel to the central axis will reflect through the focal point, continue through the focal point and hit the opposite parabola wall, and they will end up parallel to the central axis going back to their source.
The geometry of the reflections is the same, however far out you take the paraboloid.
A paraboloid reflector will give you the same sort of effect as a spherical reflector as in a telescope (just with different aberrations). It will produce an image of the distant scene (perhaps the sky), which is in focus in the focal plane of the paraboloid. Each source (star) will turn up at a different point in the image plane. It will never return a beam of light, back the way it came.
 
  • #29
phinds
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Is there a time limit on editing previous posts? Or at least to mark that I've realized I've written something that doesn't make sense
Yes there is only a brief time in which you can edit your posts. Something like 4 hours, I think
 
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  • #30
zimbabwe
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It will never return a beam of light, back the way it came.

I don't understand this statement. In my parabola it shines the light rays back from whence they came.

P1040591.jpg
 
  • #31
sophiecentaur
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Assuming the parabolic walls are long enough won't a parabola give you the same result, but only for rays parallel to the central axis. All rays parallel to the central axis will reflect through the focal point, continue through the focal point and hit the opposite parabola wall, and they will end up parallel to the central axis going back to their source.
I see where you're coming from here but a paraboloid shape will never allow you to be in that position. The sides are constantly diverging (just less and less). The shape that will let this happen is an Ellipsoid and rays will go through one focus - hit the walls and then arrive at the other focus (at the other end). But it is a closed shape and you can't send rays into it. I have never worked it out but I imaging that rays command through a window in the side may pass through both foci (severals times) before some of the energy emerges back out of the window.
I don't understand this statement. In my parabola it shines the light rays back from whence they came.
Your diagram is ok but that's only when the incident light is along the axis. (a singular special case and similar to why you tend to get redeye when people look at the camera flash - but not when they look away). For any other angle, the image is not on the principal axis and so the emerging rays will not be in the same direction as the incoming. If it were like you say, then all paraboloids would reflect light back for all angles and that would be the best way to make high viz fabric and radar reflectors. In fact, it's always done with corner cube -based reflectors. That suggests, to me, that I must be right, as, also drawing even a rough sketch seems to confirm.
I suggest that you repeat the above construction with the light off axis and see where it takes you. Just one off-axis ray is enough to prove my point.
 
  • #32
jbriggs444
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For a point source at finite distance emitting light parallel to the axis of the paraboloid, it is clear that the twice-reflected returning ray will miss. It will be offset on the other side of the paraboloid.

For a source at infinity, such a ray will hit -- in the sense that the twice-reflected ray will go back parallel to the incident ray and that in an extended geometry that allows sources at infinity we may regard the source as being characterized purely by its direction and not by its offset from an axis pointing in that direction. In this sense, a paraboloid is analogous to an ellipse with one focus at infinity. All light rays coming from that focus will be reflected back to that focus.

I have worked it out for the ellipsoid with light rays coming in through a window (a long time ago after seeing a reference). They get trapped and never reflect back out. The path converges onto a stright line going the long way from one end of the ellipsoid to the other.
 
  • #33
sophiecentaur
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For a point source at finite distance emitting light parallel to the axis of the paraboloid, it is clear that the twice-reflected returning ray will miss. It will be offset on the other side of the paraboloid.

For a source at infinity, such a ray will hit -- in the sense that the twice-reflected ray will go back parallel to the incident ray and that in an extended geometry that allows sources at infinity we may regard the source as being characterized purely by its direction and not by its offset from an axis pointing in that direction. In this sense, a paraboloid is analogous to an ellipse with one focus at infinity. All light rays coming from that focus will be reflected back to that focus.

I have worked it out for the ellipsoid with light rays coming in through a window (a long time ago after seeing a reference). They get trapped and never reflect back out. The path converges onto a stright line going the long way from one end of the ellipsoid to the other.
But what about light arriving off the direction of the principal axis? That has been my point. In the case of two plane mirrors at right angles, (consider the 2D example cos it's easier), angles of incidence and reflection will produce a 180 diversion in the path of a ray at any angle but, when reflected in a parabolic surface, the normals at the points of reflection are not the same so the outgoing ray cannot be parallel with the incoming.
Your point about rays becoming offset is only relevant if the 'beam' is not wide enough to cover the whole mirror when it arrives. The beam will be inverted, as will an image (when it's a real one), seen in the mirror.

Do you really mean that? You are implying that it would be a 'black hole' (not a relativistic one). In practice, of course, a small window and a poorly reflecting interior is a very good 'black hole device' and it can be used as a good standard 'black level' for TV and other imaging work (they normally just use a rectangular box with matt black paint). How can there not be a way out if the reflections are 'perfect'? It must surely depend upon the actual angle of incidence and the position (and size!!) of the window. I suspect that your calculation method was a ray tracing algorithm and that the errors for multiple reflections may have added together to give the result you got. I doubt that you would get that result from an analytical method.
 
  • #34
jbriggs444
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But what about light arriving off the direction of the principal axis? That has been my point. In the case of two plane mirrors at right angles, (consider the 2D example cos it's easier), angles of incidence and reflection will produce a 180 diversion in the path of a ray at any angle but, when reflected in a parabolic surface, the normals at the points of reflection are not the same so the outgoing ray cannot be parallel with the incoming.
Agreed, totally.

Do you really mean that?
A perfectly reflecting ellipsoid (of which, as you point out, there are none) would indeed be a "black hole". Yes, that is what I am saying.

Trace a light ray. For simplicity, choose one that hits a focus at one end. It reflects and goes through the focus near the other end each time the path gets closer and closer to a straight line going the long way through the ellipsoid. Trace it back the other way. Each time it reflects, it will be going farther and farther away from the straight line. Eventually it will be going at a wide angle from the straight line and will start converging back to the straight line the long way through the ellipsoid.

Barring a happy coincidence shortly after having entered the ellipsoid, a ray cannot come back out, having first come in.

No need for a computer simulation. This is easy back of the envelope work.
 
  • #35
sophiecentaur
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Perhaps you could post this back of an envelope for me. The calculation would have to include the formal ellipse shape. I can't imagine a simple way through this.
 

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