Can there be more than one path for which Fermat's least time principle is obeyed?


Yes, there are situations in which more than one optical path can obey to Fermat's Least Time Principle. For example, you can imagine a reflecting ellipsoid in which you have a single, homogeneous medium. By geometrical construction, every path starting from a focus, going straight to mirror, being reflected and going back to the other focus has the same geometrical length and, since the medium is always the same, also the optical path is the same. Therefore, every optical path of this type will obey to Fermat's Least Time Principle.

  • $\begingroup$ This doesn't make sense to me. None of those paths would be the shortest path between the two points, and indeed, light traveling between those two points would not be taking any of those paths. Rather, light traveling in a different direction that happens to get blocked by an obstacle and reflected would take one of those paths, and only one of them. In every component of the path where there is no obstacle, again, light would be taking a unique path. So I don't see how this is an answer at all. $\endgroup$ – user541686 Dec 11 '16 at 11:53
  • $\begingroup$ This isn't answering the question. While it's true that every ray emanating from the first focus will obey Fermat's Least Time Principle, each ray will only have one path available to it that fulfills FLTP. $\endgroup$ – Dancrumb Dec 11 '16 at 14:24
  • $\begingroup$ @Dancrumb I thought that the question was about possible paths. I haven't written that a single ray will pass through more than one path. $\endgroup$ – JackI Dec 11 '16 at 15:20
  • $\begingroup$ @Mehrdad I implicitly assume not to analyze the path that goes directly from one focus to the other. I haven't said that a ray of light will pass through more than one of these paths, but just that there are more paths available. $\endgroup$ – JackI Dec 11 '16 at 15:50
  • $\begingroup$ In the link in the first comment to the question, however, there are better answers than mine. $\endgroup$ – JackI Dec 11 '16 at 15:52

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