2
$\begingroup$

Statement: "In optics, you can take the example of a concave mirror: the optical path chosen by the light to join two fixed points A and B is a maximum."

The statement gives the impression that the light ray travels the path with maximum optical length (maximum time) between any two fixed points when reflected off the surface of a concave mirror. (Usually, a light ray takes the path of least optical length, such as the case when reflected off a flat mirror or a refraction between mediums with positive refractive index). However, if you use a ray simulator you can configure an infinite number of point pairs so that the optical path is not maximised, as I have found out. So the statement can't be correct. right?

A generative AI answer for the query above (namely Bing AI):

The statement is false. The optical path chosen by the light to join two fixed points A and B is not always a maximum when reflected off a concave mirror. It depends on the position and shape of the mirror, as well as the position of the points A and B. There may be other ray paths that have longer or shorter optical paths than the one chosen by the light. The correct statement is that the optical path chosen by the light to join two fixed points A and B is stationary with respect to variations of the path when reflected off a concave mirror. This means that a small change in the path does not change the optical path length significantly. This is a consequence of Fermat’s principle of stationary time, which states that light travels between two points in such a way that the time taken is stationary with respect to variations of the path.

Original link to the statement: https://physics.stackexchange.com/a/144362/366787

Someone (earlier I also erroneously thought so) can say "The statement is made regarding symmetrical points off the axis and for such points, the ray should contact the pole of the mirror in order to traverse both points and such that, it is the maximum distance that the ray can traverse assuming only one reflection off the mirror." But such an assumption is wrong because we can always introduce kinks to the path of the ray like zigzag or bending to increase path length hence it is still an inflexion point. Legendary Richard P. Feynman had this say on a similar scenario in one of his lectures at Caltech :

"Actually, we must make the statement of the principle of least time a little more accurately. It was not stated correctly above. It is incorrectly called the principle of least time and we have gone along with the incorrect description for convenience, but we must now see what the correct statement is. Suppose we had a (flat) mirror. What makes the light think it has to go to the mirror? The path of least time is clearly AB. So some people might say, “Sometimes it is a maximum time.” It is not a maximum time, because certainly a curved path would take a still longer time! The correct statement is the following: a ray going in a certain particular path has the property that if we make a small change (say a one percent shift) in the ray in any manner whatever, say in the location at which it comes to the mirror, or the shape of the curve, or anything, there will be no first-order change in the time; there will be only a second-order change in the time. In other words, the principle is that light takes a path such that there are many other paths nearby which take almost exactly the same time."

Further explanation: The question is not about whether or not such rays have the ability to form usable images for practical applications, but about the stationery action of a light ray reflecting off a concave mirror. It is about whether the stationary action of such a ray is minimum, maximum or in-between (saddle point). For a plane mirror, for example, any ray travelling from point A to B through point C (Point C must be on the surface of the mirror) takes always the geometrically shortest path no matter where you geometrically configure those two points. (The point C automatically selected by the ray will be on the shortest path or rather only the ray that reaches B will be the one that reflected off point C on the shortest geometrical path). So what about a concave mirror? My guess is the stationary action of a ray being minimised, maximised or saddle point depends on the configuration of the mirror and the three points. So the statement "In optics, you can take the example of a concave mirror: the optical path chosen by the light to join two fixed points A and B is a maximum." is poorly conceived.

Improve this question
$\endgroup$
9
  • 1
    $\begingroup$ Note that using AI for physics questions is terrible idea. $\endgroup$
    – joseph h
    May 11 at 4:45
  • $\begingroup$ I actually agree with it in this case, because it at least gives a starting point to the question @Duke William $\endgroup$
    – Hopeful Whitepiller
    May 11 at 4:47
  • 2
    $\begingroup$ @joseph h I agree that it is a terrible idea, but not so terrible regarding that they give instant answers, whereas in a forum you have to wait for an unknown time for some benevolent member kindly take interest in answering. No disrespect intended. $\endgroup$
    – Duke William
    May 11 at 4:54
  • $\begingroup$ You'll have to excuse us all for not immediately answering. The moment I saw this post I wanted to go straight home, since answering from my office won't do because of the possibility of work-related stuff. That's the least all "benevolent members" can do. Now, sarcastic comments aside on your and my part, there are on average five to ten members that read each post at this time of day/night, so having an expectation that you'll get a quick response and that members don't take an interest, comes off as disrespectful and slightly entitled. No disrespect either. Thanks. $\endgroup$
    – joseph h
    May 11 at 5:29
  • 1
    $\begingroup$ Does this answer your question? When is the principle of stationary action not the principle of least action? $\endgroup$
    – Miyase
    May 11 at 6:21

3 Answers 3

Reset to default
1
$\begingroup$

If I understand properly your question, there is a very simple example: the elliptical mirror. It is obvious that the foci $F_1$ and $F_2$ are conjugate since the optical path is constant: neither minimum nor maximum. This is a well-known property of the ellipse: the distance $F_1MF_2 = 2a$ is constant if $M$ is on the ellipsoid.

If $M$ is inside the ellipsoid, $F_1MF_2 < 2a$ and If $M$ is outside the ellipsoid, $F_1MF_2 > 2a$

Now, consider a curved mirror tangent to the ellipsoid (any shape, but tangent) at a point $H$: the ray $F_1HF_2$ verifies Descartes' law for this new mirror: it is a path actually followed by light.

If this tangent mirror is entirely inside the ellipsoid, for a point $M$ close to $H$ on the mirror, the distance $F_1MF_2$ is always smaller than $F_1HF_2$: we are therefore dealing with a maximum.

If this mirror is entirely outside the ellipsoid, for a point $M$ close to $H$ on the mirror, the distance $F_1MF_2$ is always greater than $F_1HF_2$: we are therefore dealing with a minimum.

If this mirror is on one side outside the ellipsoid and on the other inside, we are dealing with a saddle point.

EDIT : I specified that the tangent mirror in $H$ is not necessarily a plane mirror.

Hope it can help and sorry for my poor english.

Improve this answer
$\endgroup$
11
  • $\begingroup$ In other words, you are saying for an elliptical mirror maximum and saddle points for optical length is possible right? This means that It is a valid example where stationery action becomes maximum or saddle point action rather than the usual least action we experience in classical mechanics, although it may be due to convention. That's why Fermats original wording, the principle of least time was later weakened for wide use as "time that is "stationary" with respect to slight variations of the path" $\endgroup$
    – Duke William
    May 11 at 12:25
  • 1
    $\begingroup$ Yes, that is what I think. When we use the principle of least action (to obtain Lagrange's equations for example), we never calculate the terms of order 2: we simply say that the variation is zero at first order. $\endgroup$
    – Vincent Fraticelli
    May 11 at 12:45
  • $\begingroup$ I do not understand this "If this tangent mirror is entirely inside the ellipsoid". How can a straight line (plane) be entirely inside a convex domain such as an ellipsoid? $\endgroup$
    – hyportnex
    May 11 at 14:31
  • 1
    $\begingroup$ @DukeWilliam A visualization of the concave-mirror-fermat-time case is on my website. The initial view is an ellipse shaped reflector, with point of emission and point of reception at $F_1$ and $F_2$. I refer to that as 'the critical case'. There is a slider to modify the shape of the ellipse, keeping the point of emission and point of reception fixed. Less concave than the critical case: Fermat time is a minimum; more concave than the critical case: Fermat time is a maximum. Scroll to: picture 6 Reflection and Fermat time $\endgroup$
    – Cleonis
    May 11 at 17:40
  • 1
    $\begingroup$ @DukeWilliam One more point to address: why allow meandering? Here is, I think, the context of that. Given a light emitter A and a detector B: we can conceive of a hypothetical variation space such that the light can meander in any way as it travels from A to B. Within that hypothetical variation space the path along a straight line is uniquely the path of shortest time. In order to apply Fermat's stationary time to propagation from A to B in free space one allows meandering paths. If you wouldn't allow the hypothetical meandering then you wouldn't have a variation range to choose from. $\endgroup$
    – Cleonis
    May 12 at 4:18
0
$\begingroup$

It is possibly helpful to stress that [within the set of (virtual) paths between 2 points A and B] there is never$^1$ a path of maximal time/optical length in the Fermat's principle. It is always possible to construct an even longer (virtual) path for the variational principle.

--

$^1$ If space is 1-dimensional $d=1$, one can consider back-and-forth paths. If one rules out back-and-forth paths, then assume that $d>1$, so that paths are not spacefilling.

Improve this answer
$\endgroup$
4
  • $\begingroup$ If A and B located symmetricaly in other words if A an B are mirror points of the main axis, then shouldn't the optical path (where point c is located at pole of the mirror), be geometrical maximum for a circular arc mirror? $\endgroup$
    – Duke William
    May 11 at 12:07
  • $\begingroup$ The main point is that the variational principle is over the set of virtual paths; not the set of actual paths. $\endgroup$
    – Qmechanic ♦
    May 11 at 12:14
  • $\begingroup$ I think my original post is explicitly about actual paths as it also involves optical paths and action. Virtual path is usually helpful when the image forming is considered. In my question, point C is on the surface of the mirror whereas for virtual paths point C is formed behind the mirror. $\endgroup$
    – Duke William
    May 11 at 12:36
  • $\begingroup$ Well, the referenced post is about the variational principle and hence about virtual paths. $\endgroup$
    – Qmechanic ♦
    May 11 at 12:41
0
$\begingroup$

The answer for a concave, rotational symmetric mirror:

The geometric path is the shortest path between A and B and a mirror point C on the surface, where the length is measured by multiples of the wavelength. Point C can be determined by the reflection principle: The path triangle has to be perpendicular to the tangent plane in C and both rays against the tangent plane in C have to be equal. Variation of C in a small circle around the point of reflection gives a quadratic variation of the distance of C.

This proof follows the same principles as for plane mirrors. The shortest path is the shortest time of course by the constancy of the speed of light c in vacuum for all wavelengths.

I have to admit, that the reflection pinciple for the shortest straight path between two points and a point of any concave manifold is not quite transparent to me by elementary geometry. Needs an afternoon of minimum detection in differential geometry.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Thanks, but the question is about whether actual rays reflected off a concave mirror follow the maximum time path hence the path with maximum geometrical length always. Regarding plane mirrors light rays "reflected off the surface" travel the least distant path (true when you consider only the rays reflected off, but not rays directly reach). But what about concave mirrors? Is it maximum action, least action or saddle point action or it differs according to the choice of mirror and points? I think we consider stationary action minimum when ray takes the path with the least time. $\endgroup$
    – Duke William
    May 11 at 7:08
  • $\begingroup$ Its the invariant length by wavenumber principle in general, that results in the minimum principle, if there exists one. Saddle principle would be relevant, if the mirror is convex-concave with curvature zwero in one point or everwhere, but such things dont make images, except for a laser ray of diameter zero. E.g an elliptic concave mirror maps tiny circular spot sources to ellipses, On a screen perpendicular to the ray, a s should add. As everybody knows, only rotation-paraboloid mirrors are devices, that are able to map points to points and are in use widely for telephoto lenses. $\endgroup$
    – user365522
    May 11 at 8:46
  • $\begingroup$ @ Roland F Thanks, can you give a definitive answer to the question? If a light ray connecting any two fixed points and the surface of a concave mirror follow the shortest geometrical path or maximal geometrical path or path in between? I am guessing that it depends on the choice of points and the mirror. (I understand that such rays might not be able to form coherent images.) $\endgroup$
    – Duke William
    May 11 at 9:19
  • $\begingroup$ The extreme of a concave mirror is an ellpsoid. Any two points connected over a point on the surface form a possible path of light with the conditions said.Take your points on the two focal points of any ellipse inside, then any path in this plane is a light path. The same is true for the rotation-paraboloid. Any point has an image over all straight paths. On the hyperboloid, the image is on the back somwehre inside the other sheet of the cone section, if 'reflexion' is defined accordingly. $\endgroup$
    – user365522
    May 11 at 10:41

Not the answer you're looking for? Browse other questions tagged or ask your own question.