# Geometric Optics Approximation - validity

• I
How is the "geometric optics approximation" exactly defined? Given all the source of visible radiation's parameters, all the apparatus, instruments, screen, etc, specifications, how can I know if, e. g. there will be diffraction, interference or other wave properties or if I'll be able to describe the radiation's behaviour only with the geometric optics law? Usually, in the textbooks, I find this definition: "...when the wavelength is small" but compared to what? Or I find "... when the wavelength is small compared to the (smallest of the) object's dimensions", which is not even correct.

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• tech99

## Answers and Replies

tech99
Gold Member
I think there are many effects caused by the wave nature of light, so it depends on the problem, but I think useful pointers might be as follows:-
1) Are you expecting a beam to remain parallel to a great distance. It will start to diverge beyond a distance of lambda/2xDiameter^2.
2) Are you wanting to focus a lens at a great distance. The furthest you can form a focus is as in 1).
3) Do you want to reflect light from an object whose size is less then half a wavelength? Reflection drops off very rapidly below this size.
4) Do you want to send a beam through a hole with a dimension smaller than half a wavelength? Transmission is very small for this case.
5) Are you interested in the light behind or grazing an obstruction? Diffraction.
6) Is the problem clearly one of synchronous multiple sources? Interference.
7) Are you using polarisers? Waves are involved.

I think there are many effects caused by the wave nature of light, so it depends on the problem, but I think useful pointers might be as follows:
....
Thanks for the answers, I'll discuss all of the situations you examined, in case. For the moment let's consider this one:
4) Do you want to send a beam through a hole with a dimension smaller than half a wavelength? Transmission is very small for this case.
Let's say we have a plane wave, 500 nm wavelength, hitting a 1 mm hole in the cover of a spyglass' objective. Will I have to use the wave nature of light or geometric optics is good enough?

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tech99
Gold Member
Thanks for the answers, I'll discuss all of the situations you examined, in case. For the moment let's consider this one: Let's say we have a plane wave, 500 nm wavelength, hitting a 1 mm hole in the cover of a spyglass' objective. Will I have to use the wave nature of light or geometric optics is good enough?--
lightarrow
The hole is 2000 wavelengths in diameter, so light will pass through easily.
If we take D^2 / 2xlambda this gives a result of 1m (if I am correct). So if the hole is more than 1m from the each end of the spyglass, I think you can rely on geometric optics.

Andy Resnick
Science Advisor
Education Advisor
How is the "geometric optics approximation" exactly defined?

The exact definition of 'geometrical optics' corresponds to the limiting case λ→0; from this approximation one can derive the 'eikonal equation' from Maxwell's equations, each ray corresponds to a differential cone of radiance that is transported along a certain curve.

As long as the electromagnetic field behaves *locally* as a plane wave, the geometrical model is generally adequate. This approximation breaks down over regions where the intensity changes rapidly (boundaries of shadows or at the focus, for example). So, for your example in post #3, it's unclear if you can use geometrical optics or not- are you interested in, for example, the far-field diffraction pattern?

• DrDu
The hole is 2000 wavelengths in diameter, so light will pass through easily.
If we take D^2 / 2xlambda this gives a result of 1m (if I am correct). So if the hole is more than 1m from the each end of the spyglass, I think you can rely on geometric optics.
I wrote "spyglass" but I'm not sure is the exact term in English. We could also use a binoculars with one objective completely covered and the other with a 1mm hole in the cover: if you look at a bright star (excluded the Sun for obvious reasons ) through it, you will see the standard Airy disks of diffraction. It's not a plane wave and not even monochromatic as in my previous example, but the essence is the same.
That criterion for the Geometric Optics Approximation (let's call it GOA from now) cannot be totally correct.
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The exact definition of 'geometrical optics' corresponds to the limiting case λ→0; from this approximation one can derive the 'eikonal equation' from Maxwell's equations, each ray corresponds to a differential cone of radiance that is transported along a certain curve.

As long as the electromagnetic field behaves *locally* as a plane wave, the geometrical model is generally adequate. This approximation breaks down over regions where the intensity changes rapidly (boundaries of shadows or at the focus, for example). So, for your example in post #3, it's unclear if you can use geometrical optics or not- are you interested in, for example, the far-field diffraction pattern?
Probably my previous post answeres part of your question. My question to your, instead is: how do I know if the wavelenght is little enough? The "limiting case λ→0" doesn't seem to give practical criteria for GOA: is visible radiation's wavelenght little enough? Or it have to be little respect to what?
Then you wrote another interesting thing: "This approximation breaks down over regions where the intensity changes rapidly". You mean changes in space, I assume. How much can the intensity vary in space?
In general, my question is: how can I say if GOA is applicable, given a specific experimental setting?
Certainly this will also depend on the level of precision I intend to reach, but let's say that I simply want to distinguish from clearly visible wave effects and tipical geometric optics effects in an high school laboratory.

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Drakkith
Staff Emeritus
Science Advisor
Probably my previous post answeres part of your question. My question to your, instead is: how do I know if the wavelenght is little enough? The "limiting case λ→0" doesn't seem to give practical criteria for GOA: is visible radiation's wavelenght little enough? Or it have to be little respect to what?

Is your object of a comparable size to the wavelength of visible light? If so, you need to use something better than the geometric approximation. The closer you get to the size of the wavelength, the larger the error is in your approximation. I'm not sure if there is a hard cutoff between the two.

Then you wrote another interesting thing: "This approximation breaks down over regions where the intensity changes rapidly". You mean changes in space, I assume. How much can the intensity vary in space?

Focus light down to a point using a lens. The geometric approximation has a severe jump in intensity between the area where the light is focused and the area just outside this region. In the real world you will instead get a gradual dropoff in the intensity, with the entire pattern being one of an airy disk instead of a single spot.

In general, my question is: how can I say if GOA is applicable, given a specific experimental setting?
Certainly this will also depend on the level of precision I intend to reach, but let's say that I simply want to distinguish from clearly visible wave effects and tipical geometric optics effects in an high school laboratory.

There's no single thing you can look at that can tell you this. You can see wave effects even at large scales given the right setup (such as a Newton's rings experiment, or just looking at the rainbow pattern on a bubble). You simply have to learn enough to be able to know when a specific setup can use the geometric approximation and when it can't. But the explanation that the approximation breaks down when the size of your object/aperture is within about an order of magnitude of the wavelength of the light is a good rule of thumb.

• Charles Link
Is your object of a comparable size to the wavelength of visible light? If so, you need to use something better than the geometric approximation. The closer you get to the size of the wavelength, the larger the error is in your approximation. I'm not sure if there is a hard cutoff between the two.
And if the object is not of comparable size to the wavelength of visible light, what can you say? In my example the object is a star, or the 1 mm hole in the objective's cover. 1 mm is 2000 times the wavelength of light. Can I use GOA or not? The experimental answer is not, for what I wrote previously.
You have given a criterion to establish when GOA is not valid but you haven't given a criterion to establish when it is.
Regards.

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Lord Jestocost
Gold Member
A criterion when ray optics can be applied reasonably is given by the so-called Fresnel distance.

A criterion when ray optics can be applied reasonably is given by the so-called Fresnel distance.
Thanks.
Fresnel distance is:
F = a2/2λL
a = aperture' size
L = distance from aperture and screen
λ = wavelength.
In my example:
a = 10-3 m;
λ = 5*10-7 m;
L = (e. g.) = 10-1 m, objective's focal lenght (the first real image of the star is created in the objective's focal plane).
Then F = 10. It's not >> 1 so we are not allowed to use GOA, is it?

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Charles Link
Homework Helper
Gold Member
When you focus the light from a star with a focusing optic, if you have perfect focusing, (e.g. paraboloidal type mirror etc.) the diffraction limited spot size ## \Delta x \approx \frac{1.22 f \lambda}{D} ##. Geometric optics tells you it focuses to a point. I'm not sure how to apply the above Fresnel criterion for this case, but if you are able to measure the finite ## \Delta x ## with some high resolution receiver, such as a focal plane array of very small photodiodes , your measurements will show that it is not focused to a point. ## \\ ## Alternatively, if you use spherical type optics, the focused spot size will be even larger because of lens aberrations. In this case, for spherical optics with fairly large diameters, diffraction theory really isn't necessary to compute the larger focused spot size. Highly refined geometric optics calculations that computes the angle that the emerging rays take off of the spherical optics will show a considerably larger focused spot size due to the spherical optics. The spot size due to the (spherical) lens aberrations will be far in excess of the diffraction-limited spot size. (For a small diameter spherical focusing optic, the geometric ray trace will incorrectly give the result that the focusing results in a nearly perfectly focused point. For that case, you need diffraction theory to compute the spot size). ## \\ ## For the paraboloidal mirror, regardless of diameter, the geometric optics tells you (incorrectly so) that the rays focus to a perfect point if they are incident on-axis. For the paraboloidal mirror, you need diffraction theory to get the correct answer for the focused spot size. ## \\ ## For additional detail, see https://www.physicsforums.com/insights/diffraction-limited-spot-size-perfect-focusing/

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Lord Jestocost
Gold Member
Then F = 10. It's not >> 1 so we are not allowed to use GOA, is it?

The Fresnel distance is defined to estimate the distance (propagation distance of the light wave after passing the aperture) up to which the concepts of ray optics are valid:
DF = a2/λ
where a is the aperture diameter and λ the wavelength. (DF = 2 m in your case)
Divergence becomes important if the propagation distance L > DF. As in your case L < DF, it is justified to apply ray optics. Have a look at:
Ray Optics I - Galileo

tech99
Gold Member
I think the boundary between the Fraunhofer and Fresnel regions in front of any aperture (lens, mirror or hole) is called the Rayleigh Distance and, although ill defined, is about D^2/(2xlambda). Lord Rayleigh's paper is available on-line at the following URL:-
http://idea.uwosh.edu/nick/rayleigh.pdf

The Fresnel distance is defined to estimate the distance (propagation distance of the light wave after passing the aperture) up to which the concepts of ray optics are valid:
DF = a2/λ
where a is the aperture diameter and λ the wavelength. (DF = 2 m in your case)
Divergence becomes important if the propagation distance L > DF. As in your case L < DF, it is justified to apply ray optics. Have a look at:
Ray Optics I - Galileo
Thanks for the link to that document; however the problem is: how can this criterion be useful in my case, since I can clearly see the Airy disks looking at the star with a binoculars?
Regards.

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Charles Link
Homework Helper
Gold Member
Perhaps @Drakkith has a good answer in the 3rd paragraph of post 8 above.

Lord Jestocost
Gold Member
...how can this criterion be useful in my case, since I can clearly see the Airy disks looking at the star with a binoculars?

You are right! The Fresnel distance is no criterion in your case. When using a converging lense behind the aperture, you produce - so to speak - Fraunhofer diffraction conditions. The image of a distant point source of light formed on a screen placed in the focal plane of a converging lense will thus be no point as the geometrical optic treatment would suggest, but an Airy disc pattern.

Charles Link
Homework Helper
Gold Member
You are right! The Fresnel distance is no criterion in your case. When using a converging lense behind the aperture, you produce - so to speak - Fraunhofer diffraction conditions. The image of a distant point source of light formed on a screen placed in the focal plane of a converging lense will thus be no point as the geometrical optic treatment would suggest, but an Airy disc pattern.
It might be helpful for the OP @lightarrow to realize that the lens of his eye is the converging lens and the retina of his eye is the screen on which the Airy disc forms. Alternatively, the light (parallel rays) emerging from the binoculars could be focused with an additional lens on a screen and the Airy disc viewed in that manner instead of by looking into the binoculars.## \\ ## Otherwise, (but the light intensity would be very low=too low), a screen could be placed far away from the binoculars, and the emerging light viewed on the screen in the far-field with no lens. Again, it would be an Airy disc pattern from a single star. ## \\ ## Additional note: The Airy disc first gets formed from the objective lens of the binoculars or telescope inside the instrument, and the eyepiece lens then puts the Airy disc image at infinity, because the Airy disc image in the focal plane of the objective lens is also the focal plane of the eyepiece lens. It then requires one additional lens, such as the lens of your eye, along with a screen, e.g. the retina of your eye, to view the pattern. ## \\ ## And alternatively, a single objective lens with a screen in the focal plane could be used to observe the Airy disc, without any eyepiece lens. To see the Airy disc image on the screen more easily in this case, it might also help to use a magnifying glass, (or an eyepiece lens), to view the image on this screen.

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There's no single thing you can look at that can tell you this. You can see wave effects even at large scales given the right setup (such as a Newton's rings experiment, or just looking at the rainbow pattern on a bubble). You simply have to learn enough to be able to know when a specific setup can use the geometric approximation and when it can't. But the explanation that the approximation breaks down when the size of your object/aperture is within about an order of magnitude of the wavelength of the light is a good rule of thumb.
I think this is rather a wise interpretation. I have just a tiny doubt on your last sentence, but I still have to elaborate something I've found which I'll discuss in the next days (if you will still be interested :-) ).
Anyway, I don't agree with the sentence which can be found in some textbooks of Optics that GOA happens when "wavelength is little in comparison to the size of the apparatus". E. g. :

Hecht - "Optics" - second edition - 1987- chapter 3, introduction:
https://i.postimg.cc/MGg1KHmG/IMG_20180922_103520.jpg

Born and Wolf - "Principle of Optics" - fourth edition, 1970, chapter III, paragraph 3.1:
https://i.postimg.cc/mgtxjDNf/IMG-20181022-141149.jpg

Regards.

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lightarrow

I found something in italian language, a little part of which I have roughly translated, at hand, into english language. There certainly are a lot of mistakes, of which I apologize.
The source is:

http://www.sagredo.eu/
"Corsi universitari"
"Lezioni di astronomia"
file: p2c09rf-1.pdf
paragraph: "L'approssimazione dell'ottica geometrica".

This is my translation:
pag1
https://i.postimg.cc/FRJcsXTz/IMG-20181021-135429.jpg

pag2
https://i.postimg.cc/bw1Kn7WD/IMG-20181021-133049.jpg

pag3
https://i.postimg.cc/hPdJBD0f/IMG-20181023-141632.jpg

pag4
https://i.postimg.cc/L4Z64KPd/IMG-20181021-163740.jpg

It is interesting the equation (O9.9), pag 3.

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