Talk:Angular resolution

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Empirical?

...empirical diffraction limit is given by the Rayleigh criterion...

Thus it is worth bearing in mind that the Rayleigh criterion is an empirical estimate of resolution...

Although I'd agree that the diffraction limit is defined to provide a concrete comparison parameter, I wouldn't personally call it an empirical value. It has a quite precise derivation from mathematics rather than being based on data from observations or experiments, which is what the adjective normally invokes for me. Other thoughts? Objections to an edit? mh 02:17, 16 February 2006 (UTC)

Well.. Rayleigh is empirical none-the-less, from all my readings. It is Rayleighs idea of what the eye can resolve given an optical train before it. Sparrow should perhaps be given a mention as it is defined as: if intensity half way between Airy disks is equal or less than intensity of disk of lowest intensity, then two objects are resolved. Not so much to do with the eye as more digital sensors are used then perhaps more appropriate? Source of Sparrow def is D.J.Goldstein "understanding the light microscope" Acdemic press ISBN 0-12-288660-7. Goldstein calls Sparrow's "less arbitrary" than Rayleighs.--210.246.8.191 01:40, 6 October 2006 (UTC)
"It is Rayleigh's idea of what the eye can resolve" – yes, but results based on ideas should not be called empirical. Dawes made experiments with many observers and instruments to get his empirical result. Thus, the current wording "The formal Rayleigh criterion is close to the empirical resolution limit found earlier by the English astronomer W. R. Dawes" is correct.
A modern approach would include the ability of both the instrument and the observer and a confidence level: The critical distance is where the observer can tell with the given confidence whether the image contains one or two point sources. The critical distance would then depend on the signal-to-noise ratio and the amplitude ratio of the two point sources. – Rainald62 (talk) 10:39, 26 March 2011 (UTC)

Picture

This page needs a picture for what a resolution angle is. I made a picture that *doesn't* help at all. But this is an example of what a picture of resolution angle should look like:

I would greatly appriciate it if someone drew a simple picture explaining what the resolution angle is physically. Fresheneesz 22:45, 20 March 2006 (UTC)

I updated that picture ^ Fresheneesz 23:41, 20 March 2006 (UTC)

Human eye

The article says:

This factor is used to approximate the ability of the human eye to distinguish two separate point sources depending on the overlap of their Airy discs.

I don't see how this could be accurate: the human eye has a much worse angular resolution than this formula would allow. Look, let's do a rough estimate. The diameter of the pupil of the eye is at most 10 mm (half of that in typical enlightenment conditions). The shortest wavelength a human eye sees is about 300 nm. From this, we get the angular resolution 1.22 * 300 nm / 10 mm = 3.66*10^-5. But according to the eye article, a typicla human eye has an angular resolution of 7*10^-4, which is one magnitude larger. I belive there are two limiting factors: one is the imperfect shape or material of the lens and the cornea, the other is the packing of photosensitive cells in the retina.

On the other hand, I've read the statement that the size of the aparture is in fact a limiting factor (as well as the density of the cells on the retina) somewhere else too, so I'm in doubt. – b_jonas 20:57, 11 May 2006 (UTC)

There's a discussion of Rayleigh diffraction limit and the human eye (with refs) here (under the optometry stuff): http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/raylei.html#c2 164.54.53.165 17:22, 31 May 2006 (UTC)
Thanks. – b_jonas 13:55, 1 June 2006 (UTC)

Units?

What should be the units used for diameter, wavelength, and angle for the empirical formulas to work? They should be stated. From the lone example given, it appears that the diameter should be in meters? I did the calculations and the answer I got was 0.1196 so this has to be in meters in order to understand the answer given of 12cm.

In science discussions, never leave out the units.

As defined angular resolution should be in radians, that is, it is a dimensionless value. That means that it should not matter what the units are, as long as everything is expressed in the same units. Apparently you entered the wavelength in meters, and then you get D in meters.
However, the formula given under "Microscope case" produces a dimensioned value: a length. That cannot be an angular resolution. Something is wrong or inconsistent here.  --LambiamTalk 02:47, 7 October 2006 (UTC)

This article is called "Angular resolution". Other possible names, such as "Spatial resolution" and "Resolving power" redirect to this article. It strikes me that "Spatial resolution" is more general than simply another name for "Angular resolution". For example, it commonly refers to the number of pixels a computer display holds, or the detail that can be seen in a photograph. It is also in much more common use by people concerned with human vision (such as optometrists, ophthalmologists, and psychophysicists) than "Angular resolution". Moreover, the text of the article states that "Angular resolution describes the resolving power of a telescope". Apart from this being unrasonably restrictive--something I will correct--it means that "Resolving power" is defined (via redirect) as "Angular resolution", which is in turn defined as resolving power! This all needs to be clarified.

Robert P. O'Shea 04:03, 25 October 2006 (UTC)

Additionally, I was looking for "Resolving Power" in a spectroscopic context, and was brought here to 'angular resolution'. There is no discussion about spectroscopic resolving power. AmberRobot 22:16, 8 March 2007 (UTC)

I agree with Robert, This should not be an automatic redirect for a search on "Spatial Resolution" as it is a major concept in Geography and more specifically Remote Sensing. There should be a link to the "Remote Sensing" entry. I spend an entire class period devoted to the concept in my Introduction to Geographic Information Systems Class and it is always an important factor in any GIS or Remote Sensing application or research. The current 'redirect' limits the concept to one specific field and it is a much more robust concept than this.

Thank you for your consideration and I apologize if this is not the correct forum to bring up such problems but I'm new to the Wikipedia World. Mike Sims

Optical?

This article seems to be optical-centric whereas "Angular resolution" covers more than "optical components", "optical device", or "optical wavelengths" (re: for example the radio spectrum.) It looks like a little cleanup could be done to correct that. 69.72.7.204 16:53, 16 January 2007 (UTC)

Define "f"

The first equation in section "Explanation" has a variable "f" which is never defined. From a similar equation in the "SLR cameras" section of Airy disc, I'm thinking that it is the distance from the lens to the film/image sensor (equivalently, i think, the focal length of the lens). Whether this or something else is what is intended, I think a definition for "f" should be added to the section. Thanks Bayle Shanks 00:36, 21 February 2007 (UTC)

Erroneus formula

Just spotted a small error on this page. Someone more Wikiable than me please correct it (my first ever Wiki contrib, I'm in a bit hurry now & sorry if I don't know all the etiquette...).

Anyway: The first formula is correct. The second should follow from the first by multiplying both sides by f (the system focal length). The left side doesn't add up. Delta l (the minimum spatial resolution) is actually TAN(theta) * f. (This is just basic trigonometry.) The Rayleigh criterion only gives us sine theta, not tan theta.

Therefore, to arrive at Delta l, we need to first inverse sine both sides of the first formula (to find theta in angular units), then TAN both sides and only then multiply by f.

Note that the current formula is approximately correct for small values of theta (i.e. high F-number / low numerical aperture / slow optics), as SIN(theta) is then roughly the same as TAN(theta). For a practical camera objective, F/3 or so, the error is about 1.36 %. Insignificant in most practical cases; therefore, the original formula remains a handy approximation. But let's have the correct formula there too, in case someone starts wondering.

192.100.124.219 18:45, 17 April 2007 (UTC) (Ville Nummela)

range resolution

The pulse compression article currently says

"Pulse compression is a signal processing technique mainly used in radar, sonar and echography to augment the range resolution as well as the signal to noise ratio."

The word "resolution" in that article was recently disambiguated from resolution to angular resolution.

The disambiguation page resolution claims that

"Angular resolution, the capability of an optical or other sensor to discern small objects"

which is certainly what pulse compression is all about.

Is "range resolution" really a specific kind of "angular resolution", as implied by the disambiguation page, and so we should talk about it here in the "angular resolution" article? Or is there a more appropriate article to point to? Or should we start a range resolution article? --75.19.73.101 (talk) 09:33, 15 December 2007 (UTC)

I see the same problem at Talk:Medical_ultrasonography#resolution. Drickey suggests the terms "spatial resolution" (which currently redirects to angular resolution) and "image resolution". Is "image resolution" what I want in the sonogram and in pulse compression radar articles? --68.0.124.33 (talk) 05:03, 4 January 2008 (UTC)

In my opinion, my answer to the "are angular resolution and range resolution the same thing?" question would be a definite no:

• a range resolution $\Delta r$ is measured in meters; it is related to distances. The notion of range resolution appears for example when measuring a distance using a radar, since the measure of the distance is not precise (the resolution is a measure of that precision).
• an angular resolution $\Delta \theta$ is measured in radians; it quantifies the precision of a piece of equipment to measure an angle, for instance the angle of arrival of some sound, radio or light wave.

But on the other hand, these two notions are certainly related. You can convert an angular resolution to a range resolution, for a given distance $r$ from the origin of the angle, by multiplying the angular resolution by $r$. Thus, when considering a rotating radar antenna, you can plot the received signal in a polar plot, the range coordinate $r$ being the range to the object having produced the reflected signal, and the angular coordinate $\theta$ being the angle of arrival of the reflected signal. The exact location of the object is unknown, but it lies in some "cell" of dimensions $\Delta r \times r\Delta \theta$ centered around the point of coordinates $(x = r\cos \theta; y=r\sin\theta)$. Thus $\Delta r$ and $r\Delta \theta$ are two spatial resolutions, both measured in meters; and if you consider the polar plot of the radar signal as an image, then you can call that a (specific case) of an image resolution. Phew! A drawing would help, but I'm too lazy to draw it. Hoping this helps anyway, Flambe (talk) 20:55, 31 January 2008 (UTC)

missing text

I noticed part of the description text below the picture box is missing, it says "Airy diffraction pattern generated by a plane wave falling on a circular aperture, such as" .. such as what? —Preceding unsigned comment added by 86.80.116.169 (talk) 15:33, 18 July 2009 (UTC)

Rayleigh Crit. Figure Needed

There should be a figure showing two overlapping Airy Discs spaced such that they just satisfy the Rayleigh criterion. This is a common figure and explains a lot in terms of how resolving power is defined. hovden (talk) 14:27, Jul 19 2010 (ET) —Preceding undated comment added 18:27, 19 July 2010 (UTC).

Subject and image resolution

I feel like something is missing on this page. For telescopes and microscopes, the usual focus of attention is the objective since both of those are specialized devices. On the other hand, for e.g., a cell-phone camera focused at infinity, the diffraction limit (as I understand it) is preventing you from making the pixels and optics ever smaller. That is, for a telescope, the given formulat doesn't tell us anything about the image sensor. That can't imply that if I keep the entrance pupil diameter fixed, increase the power, and shrink and move the sensor closer to the lens, that I can keep the same angular resolution? (Unless that in so doing the image-space NA will increase?)

What about the case in which you want exceptional angular or spatial object resolution on a small sensor? Shouldn't there be an equation involving the image- and object-space NA? —Ben FrantzDale (talk) 20:06, 23 September 2010 (UTC)

Aperture versus pupil diameter?

The formulas all say "aperture diameter"; is it really that or is it entrance- or exit-pupil diameter? On the one hand, I find it usually best to think of a lens as a black box with an entrance pupil and an exit pupil and a power. On the other hand, if you contrived a system with pupils much larger than the aperture, there would have to be issues... in the limit, I am imagining an aperture on the order of λ in diameter but magnified so that the pupils are much bigger. Geometric optics would tell you that this would work, but zooming in on the light hitting the aperture, it would loose all of its directionality as it went through by diffraction. This implies that the actual aperture diameter matters; on the other hand, NA is determined by entrance-pupil diameter, so that implies that this article means "entrance pupil" when it says "aperture". Which is it? —Ben FrantzDale (talk) 20:06, 23 September 2010 (UTC)

I think it is the entrance pupil for the subject side and the exit pupil for the image side. Suppose you have a lens with two positive meniscus lenses around the aperture so the entrance and exit pupils are bigger than the aperture. While it may be that the angular resolution you can squeeze through the aperture is the fundamental limit, the positive lenses magnify the aperture so given two chief rays with an angle θ between them, the first element makes them converge, so θ gets bigger. I suspect (although I haven't done the math) that this all works out so that the entrance-pupil diameter prescribes the subject angular resolution, then after the first element the corresponding angle is magnified just as the aperture is smaller than the entrance pupil.
In the case that we are interested in comparing different cameras, if we hold the entrance-pupil diameter but shrink the format size, pixel size, and focal length together, I think we wind up with constant angular diffraction limit (constant image resolution in terms of pixels), but by decreasing the f/# and so increasing the image NA, the optics have to "do more" to avoid aberrations and stay diffraction-limited.
In the case of an image-space telecentric lens, the resolution equations have a singularity in that the entrance pupil is at infinity with finite NA, so the angular resolution is infinite (although the spatial resolution must be finite).

Following up on my initial question, I am wondering about what happens when you are working at finite conjugates—that is, when you are focused at a subject that isn't at infinity. There, the working f/# is different from the image-space f/#, etc. In the limit of a close subject and a big sensor, you should have a diffraction-limited microscope; in the limit of a distant subject and a small sensor, you have the f/#-based resolution equation, but what about when both the sensor and subject are small? I like to think of a camera lens as a black box with an entrance pupil on one side and an exit pupil on the other, but that suggests the resolution on one side is independent of that on the other side (unless they are coupled by pupil diameter and Lagrange invariant/etendue). My initial thought is that as long as the lens can columnate light from the entrance pupil, then the entrance and exit pupils are decoupled enough that your final image resolution depends only on image-space NA, but that seems like it must break down in the case of a microscope. Where am I going wrong? —Ben FrantzDale (talk) 15:05, 13 December 2011 (UTC)

Shortest visible wavelength

In the specific case of the microscope, it is noted that the shortest visible wavelength is 450 nm. In my experience, the human eye can see violet light (for instance of 411 nm) with ease. This is probably why (invisible) UV is called ultra-violet and not ultra-blue. Unfortunately the numerical example is hard-coded into a picture and not easy to change. What would be a good way to correct this? Changing the text to read: One of the shortest visible wavelenghts? It seems unnatural to me to put it like that. -Victor Claessen 8 November 2010

Spatial Intensity Distribution

Would it be possible to create a sub-section in this article and define the concept "spatial intensity distribution". I'm not a scientist myself, but the phrase keeps popping up in astronomical literature and I don't understand what it means exactly. It occurs in this abstract on astronomical interferometry which is being used as a reference in the Betelgeuse article. If one googles the string "spatial intensity distribution", 81,700 results are produced, yet nowhere have I found a definition that a lay person would ever understand.

I don't know if such a concept warrants a sub-section in itself. The reason for my request is that I'd like to blue link such a sub-section from the Betelgeuse article. Thanks for any help you can provide.--Sadalsuud (talk) 11:18, 14 June 2012 (UTC)