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I am currently learning about Optics using an MIT OpenCourseWare Course. I have in particular a question about the application of Fermat's principle in Lecture 3, i.e. deriving the shapes of parabolic reflectors and ellipsoids of refractive material for a given focal point. The lecture notes could be found here, and the derivations in question are on the first slides.

For example, consider the derivation of the paraboloidal reflector (3rd slide), he does this derivation at around minute 6 in the video.

In short, he says that all rays getting reflected and hitting the focus point F should be the same length by Fermats principle that the optical path length (or the time taken by the photons/wave fronts) should be minimal.

But I doubt that reasoning and have certain objections about his arguments.

1) First you cannot measure distance for points in infinity (as $\infty + c = \infty$), but if we replace infinity with some unspecified far away point, then we can put some plane behind F and measure distance from this plane onwards (as the rays before hitting that plane have all the same distance). So this is easily fixed.

2) In my opinion Fermats principle is just applied for the ray that goes directly through F and is reflected in the exact opposite direction (i.e. the ray on the optical axis), and that it must have this path is easily seen with Fermats principle. But the reasoning that all the other paths must have the same length (or must take the same time) is not Fermats priniple as I see it, as I see it an additional argument must be in order here, I can image that we might require that the photons that start at the same time (or the wavefronts thereof) should all meet in F, hence should all take the same time (or path length).

Am I right? Is the author a little bit sloppy here and are additional arguments in order? I have the feeling that Fermats principle is not enough, or said more provocative that it is not the crucial part of the argument (mainly that all paths should have the same length), but it just gives the magnitude of the common path (i.e. $2f$ in the lectures).

PS: A student seems to ask a similar question at around minute 11, but the answer does not seem to address it properly in my opinion (and btw. also thread infinity as a point from which we can measure distance).

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I think this is explained clearly in the Feynman Lectures Vol I Section 26-4 (online at here). Feynman first discusses how a lens can focus a point source $P$ to a given focus $F$. He reasons that light will only take the shortest path and that usually this means only one ray can go from $P$ to $F$. However, if we arrange it so that there are many paths which all take the same amount of time, then the light will be able to take all of them, and you will get focussing behaviour.

Now to discuss a parabolic mirror. The geometrical definition of a parabola is the locus of points equidistant between a focus $P'$ and a line $LL'$:

Diagram of parabolic mirror taken from Feynman lectures

(Image taken from Feynman lectures because I'm too lazy to make my own)

For a far away source, we can treat the rays as being parallel and having taken equal times to travel to a plane $KK'$. Clearly the minimum possible path length it hit the mirror and reach $P$ is the central ray in this figure, but the point is that every other ray takes the same amount of time. Thus, these are all minimum time paths from a distant point source and so they all reflect perfectly to $P'$.

As a remark, the absolutely correct thing to do would not be a parabolic dish, but a segment of an ellipse. An ellipse has is the locus of points whose distances from two foci is constant, so one if you had an ellipse with the point source at one focus and your detector at the other all light would be captured. But as the ellipse gets stretched out the approximation that the incoming light is parallel becomes better and better. This is reflected in that the ellipse segment begins to look more and more like a parabola.

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Fermat's principle does not say that the optical path length is always minimal. It only says that all rays take a local minimum of the length, i.e. that you can't in first order decrease the path length by making small deviations from the path. For instance, a ray that hits the mirror at an acute angle and comes back at that same angle would not be locally minimal, because you could make the path shorter by having it hit at a slightly steeper angle. What actually happens instead is that the ray exits at the supplementary angle. In this case, changing the point the ray hits can't to first order change anything about the path-length, because any reduced length in the ingoing path would be offset at least as much by increased length in the outgoing path / vice versa.

Mirror rays

You might also argue that even a ray bouncing vertically off a mirror is not locally minimal: couldn't you just have it turn around slightly earlier, before it actually hits the mirror? Well, that's actually topologically forbidden because there must be a fixed phase between the ingoing and outgoing ray. You'd need to shorten the path length by an entire wavelength, but again Fermat's principle only considers the locally shortest path, and within the fixed phase it is the shortest path (point in the $U(1)$ group).


Generally just a local extremum, in fact. Only, maximal optical-path lengths don't tend to occur natually, because you can always increase the path length by introducing small local detours.

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  • $\begingroup$ I do not understand "a ray that hits the mirros at an acute angle and comes back at that same angle would not be locally minimal" - but is not that exactly what is happening, i.e. the incident angle equals the outgoing angle??? And (considering an isotropic medium) the path between two points and a reflection point on the mirror is globally the smallest path, so what you mean with local minimum... $\endgroup$ – StefanH Apr 5 '17 at 16:56
  • $\begingroup$ Not actually, the outgoing angle is really the supplementary angle to the ingoing. Or, equivalently, the angle of the outgoing measured from the opposite side is the same as the ingoing, which is what's meant by the handily short “outgoing = incident”. — I added an illustration to clarify what I mean. $\endgroup$ – leftaroundabout Apr 5 '17 at 17:07
  • $\begingroup$ But you cannot move the point, the contraint that the path has to go through the reflection point is implicit in Fermats principle (otherwise a straight line between both points would be even shorter) so you cannot move the reflection point freely as you did???? $\endgroup$ – StefanH Apr 5 '17 at 17:47
  • $\begingroup$ “Otherwise a straight line would be shorter” – correct, that's what I adressed in my second paragraph; the reason this isn't a problem is that omitting the mirror entirely would not be a local change (you would need to “jump” out of the mirror plane by an entire wavelength), whereas just moving the point within the mirror plane is possible in arbitrarily small steps. $\endgroup$ – leftaroundabout Apr 5 '17 at 19:13
  • $\begingroup$ Okay, but what has this to do with my question? And in what sense gives what you write the form of a parabolic reflector? $\endgroup$ – StefanH Apr 6 '17 at 8:44

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