Parabola incoming ray not parallel to axis

[sophiecentaur]

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.

 

[jbriggs444]

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.

 

[sophiecentaur]

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.

 

[jbriggs444]

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.

 

[sophiecentaur]

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.

 

[Drakkith]

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.

It took me a few re-reads to understand this, but you're basically saying that a parabola has no "opposite parabola wall", right?

 

[jbriggs444]

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.

The envelope is in tatters. Its supposed result is wrong. Thank you for revealing my mistake.

I had managed to convince myself that since an infinitesimally thin ray that passes through one focus of an ellipsoid will converge on a back and forth path running the long way through the ellipsoid (true) that any nearby and nearly parallel ray will have a path that similarly converges (plausible sounding) and that this means that a parallel ray will have a path that passes ever closer and closer to the foci as it converges to that center line path (provably false).

 

[sophiecentaur]

Drakkith: Both what you wrote initially and my thought processes are ancient history (to my aging brain at least) so I can't remember. (Sad old sod) My thought was, of course totally correct, incisive and brilliant haha.
That diagram on Post 30 says it all, really - except for the case of rays arriving very near the principle axis. These will reflect (only once) off the surface and will miss the wall of any finite sized parabola so they'll be lost out in another direction. I must say, I never considered this before and I am glad you guys pushed my mind in this direction; cheers.

jbriggs: My intuition told me it couldn't have been right - but you can't always trust intuition. I just remember that a ray through one focus will pass through the other one. So, up to a point, that will apply to an object or real image in the vicinity of one focus: it will appear at the other focus. I guess that's the theory behind the ' pink holographic pig' illusion.

I will have to reflect on this, some more.