B Fermat's principle and refraction -- Possible misunderstanding

[Philip Koeck]

You can measure it at any point on the axis and use the same point to compare phases of all the different cones. The variation will approach zero as the cones get closer to the axis. This approach to phase measurement applies to all wave measurements of all types of wave. It's the same principle as in surveying ; you don't have to take a reference height at sea level- you just choose some convenient reference location. It will still give you the right topography.

I'm trying to find out if I've missed something important.

Let's use the phase of the direct ray in the ideal image point as a reference, for example. At a specific time we choose the phase is zero, let's say.
Then a cone with a very small angle will have a phase very close to zero in a point on the axis very close to the ideal image point.
Do you agree so far?

What would the phase of a cone with large angle be in the point of intersection with the optical axis if this point is a distance λ/2 away from the ideal image point?

We're always assuming that all the rays come from a point-like object in phase, so they form a divergent spherical wave to the left of the lens.

 

[sophiecentaur]

Let's use the phase of the direct ray in the ideal image point as a reference, for example. At a specific time we choose the phase is zero, let's say.

The 'ideal image point' doesn't actually have any meaning for the direct ray because the direct ray passes through all points on the axis. It's just not necessary to chose any particular 'point'; you just choose an arbitrary one, say in the plane of one of the points on the axis where a chosen cone has its vertex.

There is also no need to choose a particular time because what counts is the phase differences. I can't think of a situation where the timing of a light wave can be actually measured to within one actual cycle of the wave. No one uses absolute time when describing or calculating the result of Young's Slits experiment - just relative phase.

What would the phase of a cone with large angle be in the point of intersection with the optical axis if this point is a distance λ/2 away from the ideal image point?

The phase will be + or - 180 degrees relative to the image point. If it were zero, it would coincide but it's 180 degrees early or late because of the different optical path length.

I think you may find it easier once you have decided to consider relative phases and nothing absolute. In practice, I think you would probably choose the reference plane as being where you get the 'best' image, with the least aberration of the image. (Median position of apexes of cones, perhaps.)

 

[Philip Koeck]

The phase will be + or - 180 degrees relative to the image point. If it were zero, it would coincide but it's 180 degrees early or late because of the different optical path length.

Okay. I think we completely agree. It looks like my assumption about Fermat's principle in the case of a lens with spherical aberration might be correct after all.

Thanks!

 

[Philip Koeck]

An observation and a question:
Rays are always orthogonal to wave fronts (I was told).
In the case of a lens with spherical aberrations there is an extended area where rays intersect and can't be orthogonal to the same wave front because they are oriented in different directions. (In the case of an ideal lens this only happens in a single point.)
See: https://www.cyberphysics.co.uk/graphics/diagrams/medical/aberration_spherical.png

I get the impression that the concept of rays sort of breaks down in this area around the intersections of the light cones, but are there still well defined wave fronts in this area?

 

[hutchphd]

Remember What you call "a wave" is not unique. The sum of an x-travelling plane wave and a y-travelling plane wave is a diagonally traveling plane wave (the assumption is the frequency for each). Is it two waves or one? The detailed description is up to you.
Locally any curved wavefront is "flat" and so the perpendicularity holds for an homogeneus medium ( I guess everything classical is homogeneous as you go down far enough in scale)
So I do not know how to answer the question. I think it is also true for different frequencies mixing so long as there is no dipersion (speed differences).

 

[Philip Koeck]

Remember What you call "a wave" is not unique. The sum of an x-travelling plane wave and a y-travelling plane wave is a diagonally traveling plane wave (the assumption is the frequency for each). Is it two waves or one? The detailed description is up to you.
Locally any curved wavefront is "flat" and so the perpendicularity holds for an homogeneus medium ( I guess everything classical is homogeneous as you go down far enough in scale)
So I do not know how to answer the question. I think it is also true for different frequencies mixing so long as there is no dipersion (speed differences).

I agreed initially, but now I'm not so sure.

Let's take away the time dependency and discuss only the spatial part of the wave (the "wave shape"), like a snap shot of an actual moving wave.
I'll write "wave" anyway, but I mean "wave shape".

In two or more dimensions 2 plane waves in different directions (no matter what the wave lengths are) add up to a pattern that is periodic in a discrete set of direction. You get a 2-dimensional crystal not another plane wave, I would say.
In 3D the same is true for 3 waves etc.

Do you agree?

PS: I'm thinking of simulating wave propagation through a lens with spherical surfaces using multi slice. Might be interesting to see the wave fronts.

 

[valenumr]

Consider a light starting at A in media 1 and going in and out a media 2 (say shaped as a disk) with relative index of refraction n to arrive at point B (in media 1).
Fermat's principle says that the path taken by the ray between points A and B is the path that can be traversed in the least time.
That could be misunderstood by thinking that the "least time" path from A to B and through the disk is the path taken by the light ray, and the one following Snell's law at the surfaces separating the two media.
That's not true. Actually there are paths (orange broken line) that do not follow Snell's law that produce a smaller time than the one produced by the path following the Snell's law (blue broken line).
View attachment 285787
Actually Fermat's principle says that the time of the path taken by light (following the Snell's law) is a local minimum with respect to nearby paths.
That doesn't prevent the fact that there could be a global minimum for the time following a different path.
I'm posting this note because I was myself victim of the misunderstanding.
Any comments? Any suggestion?

(in the simulation n=1.8, A(-1.6, 0), B(1.3, 0.3), r=1)

If I understand correctly, the orange path is faster because the index of refraction is high, and the traversal through the m2 medium is shorter. Interesting, I have to think about it more, but maybe you're just describing rainbows.

 

[hutchphd]

Do you agree?

Care must be taken here.
The concept of "rays" of light is the classical approximation. But like any quantum object, the light actually samples all paths and predominant paths are those where the phase is stationary We can then improve the approximation by looking at the phase along the path and "apply lipstick to the pig". But there is not really a single path.
My statement about about synthesis of rays is really just a restatement of Fourier's theorem where the individual plane wave components correspond roughly to "rays". Any particular situation involves the net sum (integral) of all the various components. I fear you pushing the approximation too far, and so I do not want to give a simple definitive response.

 

[Philip Koeck]

Care must be taken here.
The concept of "rays" of light is the classical approximation. But like any quantum object, the light actually samples all paths and predominant paths are those where the phase is stationary We can then improve the approximation by looking at the phase along the path and "apply lipstick to the pig". But there is not really a single path.
My statement about about synthesis of rays is really just a restatement of Fourier's theorem where the individual plane wave components correspond roughly to "rays". Any particular situation involves the net sum (integral) of all the various components. I fear you pushing the approximation too far, and so I do not want to give a simple definitive response.

For me the main question is what happens in the "waist" of the light beam coming from a lens with spherical aberration.
Ray tracing would indicate that there are intersecting wave fronts there, whereas if I think of simple wave propagation I find it difficult to see how there can be more than one set of wave fronts. I can imagine something like one converging aspherical wave that "folds over" and turns into a diverging wave after the image plane.
Maybe both pictures are wrong in the end.

 

[sophiecentaur]

It looks like my assumption about Fermat's principle in the case of a lens with spherical aberration might be correct after all.

I wonder why you are being so specific here? Afair, Fermat's principle is universal. If it were not, there would be some logic that says it wouldn't be valid to use it anywhere that you didn't already know that it was valid.

 

[hutchphd]

I can imagine something like one converging aspherical wave that "folds over" and turns into a diverging wave after the image plane.

But we know that any finite lens, even without spherical aberration, will be diffraction limited when forming a point image. This comes from the unavoidable wave nature of stuff. In your head you can draw rays and it is a very useful exercise in many circumstances, but reality is far more complicated but hence richer: the archtypal two-edged sword.

 

[Philip Koeck]

I wonder why you are being so specific here? Afair, Fermat's principle is universal. If it were not, there would be some logic that says it wouldn't be valid to use it anywhere that you didn't already know that it was valid.

I agree completely that Fermat's principle always applies.
My problem is if I can make a specific conclusion from it.

My question is all about a derivation of the lens aberration function.
It's for the transmission electron microscope, but the principle is the same for all wave optical systems.
For electrons with about 100 keV the scattering angles are tiny, so that leads to some approximations which might not be valid for light though.

Here's the derivation:
https://www.researchgate.net/publication/358235022_The_Lens_Aberration_Function

In the first part, where I treat only defocus, but assume an ideal lens, I use the ideal image point as a reference point for the phases. If the object point is a source of a spherical wave then there is a convergent spherical wave after the lens and all rays are in phase in the image point.
Fermat's principle says that a ray is defined as a path where the optical path length is a local extremum and this is consistent with the above.
This extremum must be a saddle point in this case.

In the second part I consider spherical aberration, but choose zero defocus. I other words I want to find the phase difference between a scattered beam at angle θ and the direct beam in the ideal image plane. (Just to clarify: These 2 rays don't hit the image plane in the same point.)
Here I use the point of intersection of the scattered and the direct beam as reference point.
So this is point that's closer to the lens than the ideal image point if spherical aberration is positive.
It's clear that Fermat's principle applies to both these beams.
What's not so clear is whether one can conclude that the two beams are in phase in this point.
My derivation depends on this assumption.

One could think of the specific case where the surface of the lens is smooth, for example spherical.
This leads to a continuum of intersection points between the direct ray and the scattered rays with different angles. This might help as pointed out by hutchphd.

PS: I'm thinking of calculating or simulating a specific case, such as focusing of a plane wave by a lens with one plane and one spherical surface, but that's obviously not a general answer.

 

[Philip Koeck]

I agree completely that Fermat's principle always applies.
My problem is if I can make a specific conclusion from it.

My question is all about a derivation of the lens aberration function.
It's for the transmission electron microscope, but the principle is the same for all wave optical systems.
For electrons with about 100 keV the scattering angles are tiny, so that leads to some approximations which might not be valid for light though.

Here's the derivation:
https://www.researchgate.net/publication/358235022_The_Lens_Aberration_Function

In the first part, where I treat only defocus, but assume an ideal lens, I use the ideal image point as a reference point for the phases. If the object point is a source of a spherical wave then there is a convergent spherical wave after the lens and all rays are in phase in the image point.
Fermat's principle says that a ray is defined as a path where the optical path length is a local extremum and this is consistent with the above.
This extremum must be a saddle point in this case.

In the second part I consider spherical aberration, but choose zero defocus. I other words I want to find the phase difference between a scattered beam at angle θ and the direct beam in the ideal image plane. (Just to clarify: These 2 rays don't hit the image plane in the same point.)
Here I use the point of intersection of the scattered and the direct beam as reference point.
So this is point that's closer to the lens than the ideal image point if spherical aberration is positive.
It's clear that Fermat's principle applies to both these beams.
What's not so clear is whether one can conclude that the two beams are in phase in this point.
My derivation depends on this assumption.

One could think of the specific case where the surface of the lens is smooth, for example spherical.
This leads to a continuum of intersection points between the direct ray and the scattered rays with different angles. This might help as pointed out by hutchphd.

PS: I'm thinking of calculating or simulating a specific case, such as focusing of a plane wave by a lens with one plane and one spherical surface, but that's obviously not a general answer.

For the specific example of two parallel rays, one of which is on the optical axis, focused by a lens with one flat face and one spherical face I've shown numerically that they travel the same optical path length from where they enter the lens through the flat face to the point where they intersect on the optical axis.
A general demonstration still eludes me.