Talk:Numerical aperture

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[edit] Merge proposal

The two definitions should be merged into a single page, replacing this disambiguation page. The definitions are all fundamentally the same--the NA formula for a fiber in terms of the indices of refraction theoretically should be the same as the NA as defined by the acceptance angle of the fiber (which would be the "correct" optical definition). Unfortunately, it seems that the federal government has confused this in their standards, making the approximate theoretical formula the definition of NA for optical fibers. This ambiguity could all be addressed in a single page.--Srleffler 06:48, 16 November 2005 (UTC)

[edit] symbolic syntax

Is it so common to write out f/# as a variable name? I took a class and we used f_\#, which seems more like usual mathematical conventions. Potatoswatter 22:58, 13 March 2007 (UTC)

f/# is by far the most common notation, and is officially the standard notation in photography (e.g. ASA standard PH2.12-1961 American Standard General-Purpose Photographic Exposure Meters). Note, though, that this notation is not a variable name in the usual sense, which helps to explain its odd form. One writes that the f-number of a camera lens is f/2.5, not that some variable equals 2.5. Additionally, though it looks like a fraction it is best not thought of that way since when one writes f/2.5 one means that the f-number is 2.5, not that one should consider the focal length divided by 2.5 (which would be the diameter of the entrance pupil).
If one is doing a lot of mathematics with f-numbers in a science or engineering class, it is convenient to assign a more conventional variable. N is commonly used, but f_\# seems like a good notation as well.
The notation is discussed at f-number#Notation, and there is a brief description of the history of this odd notation at f-number#Typographical standardization.--Srleffler 03:51, 14 March 2007 (UTC)
The notation # is horrible. I would change it to ~D~. dima 07:35, 15 June 2007 (UTC)

english?

this article is a very hard read.. I feel like Im reading a math book or an issue of scientific american... most of the people on here are not math prodigy's.. anyway to rewrite this in laymans terms..lol `Tracer9999 02:39, 23 August 2007 (UTC)

[edit] Light gathering capacity

"Lenses with larger numerical apertures also collect more light and will generally provide a brighter image." - and, moreover, a better signal-to-noise ratio. I'd like to know a bit more about this - how exactly does light-gathering vary with NA? This is of increasing interest in microscopy, where we've recently seen improvements in NA at the top end from 1.40, the state of the art 5-10 years ago, to 1.45, now common although far from standard, and recently 1.49, at sell-a-graduate-student-into-slavery prices. The improvement in resolution is small, about 6%, but people reckon the better lenses make a significant difference, so i assume it's to do with light-gathering capacity. Or it's a big con.

The 1.49 lenses are marketed specifically for Total internal reflection fluorescence microscopes; i don't know if that's because the TIRF physics is especially sensitive to NA, or because TIRF microscopes are just such bastards generally that you need a great lens to get useful data.

-- Tom Anderson —Preceding unsigned comment added by 128.40.81.185 (talk) 16:04, 26 October 2007 (UTC)

[edit] Working f-number

Anyone who reads this article AND the f-number article will get confused by a couple of things:

  1. This article uses the optical convention, m = − o / i, while the f-number article uses the photograph convention, m = o / i ignoring the inversion of the image. So the two articles give definitions of working f-number with opposite sign conventions.
  2. Neither article explains adequately what happens in macro photography. In that case, the image plane is significantly behind the focal point. Thus the light-gathering capacity is not adequately measured by the f-number, which assumes they coincide. In this case, the angle subtended by the entrance pupil, seen from the image plane, is less than the angle subtended from the focal point. They differ by a factor 1 + m (in the photographic sign convention).

I can clarify point 1 fairly easily, but I'm not sure what the best way to explain macro photography is. As I understand it, working f-number is needed in macro photography because the "light gathering" happens further behind the focal point... but maybe it ALSO matters that the light from the object is not collimated? Can someone help with this? ǝɹʎℲxoɯ (contrib) 00:08, 25 January 2008 (UTC)

I returned the formula at f-number to the conventional form. I had the correct sign there originally but another editor changed it. I think he forgot that m is typically negative. I made some other adjustments there. Of particular note: someone had asserted there that the NA was being defined. In fact the reverse is true: the working f-number is defined as 1/(2NA).
I'm not sure I understand point 2. Is the light-gathering capability of the lens in macro photography correctly described by the working f-number? I would have thought so. Do you assert otherwise? I thought the working f-number took into account the displacement of the image plane from the focal plane. This displacement is not particular to cameras, but is generally true of finite-conjugate optical systems (where working f/# is used). --Srleffler (talk) 01:00, 25 January 2008 (UTC)
I also don't understand point 2. The 1+m (or 1-m) factor should take care of it. As to the sign convention, I agree we should keep our text in agreement with out cited sources. I don't recall if it was I who changed it, but I would generally prefer to use the positive m convention in the f-number article, because that's what I see in photography sources, and conversely the negative m in the the numerical aperture article, because that's what's used in optics. Dicklyon (talk) 01:17, 25 January 2008 (UTC)
For the sign convention, it's fine with me now that it's consistent! I should change angle of view too, since I used the opposite convention there.
For point 2, I agree with you guys that the formula is correct. I'm just confused about the explanation given in the article. As I see it, the light-gathering ability is reduced due to displacement of the image plane, which I think you agree with me on. But this article gives a different explanation:
“ The f-number describes the light-gathering ability of the lens in the case where the marginal ray before (or after) the lens is collimated... In optical design, the finite distance between the object and the lens must often be considered. In these cases, the working f-number is used instead. ”
The article seems to claim that it's the decreased distance to the object plane which explains the working f-number, whereas we think that it's the increased distance to the image plane. Which is right? ǝɹʎℲxoɯ (contrib) 01:21, 25 January 2008 (UTC)
I'd say the latter makes more sense, even though they're probably equivalent. And I wouldn't call it the "light-gathering ability"; it's that the gathered light gets spread out more when the image plane is further from the exit pupil. In fact, the NA should be a ratio involving exit pupil and size, which are not very closely related to focal length; at the focal plane, the ratio comes out the same; at other image planes, it's less obvious to me what happens. Anyone have a generally correct formula for this? Is it really just dependent on m? Perhaps so. Dicklyon (talk) 04:07, 25 January 2008 (UTC)
Right, it's not the light-gathering ability per se, good point. How would the two notions of working f-number be quivalent? When you say "the NA should be a ratio involving exit pupil and size", what size are you referring to? As I see it, the NA should be the ratio of exit pupil to image distance... but I may be missing something here. The current text bothers me the more I stare at it. I don't really see what collimation has to do with it. ǝɹʎℲxoɯ (contrib) 04:27, 25 January 2008 (UTC)
Sorry, fingers outraced brain. I meant to type "ratio between exit pupil size and distance". I think the collimation really just means the image plane has to be moved; too indirect for my taste, too. Dicklyon (talk) 04:34, 25 January 2008 (UTC)
I think I probably wrote that (I haven't checked), and that what I had in mind was not that the change in object position causes the difference in the working f-number, but rather that the object not being at infinity is the sign that you need to consider working f-number rather than the regular f-number. A photographer doesn't know where the image plane is relative to the rear focal plane, but she does know where the object is. I agree that it is the change in image plane position that explains why we need the working f-number.--Srleffler (talk) 05:54, 25 January 2008 (UTC)

I'm not too sure about the version of the formula with the absolute value of m. That works only if working f-number is not used in cases where a lens has positive magnification (in the usual sign convention). Is the formula inapplicable there? I don't recall. At the least, this form of the equation appears to be original research, unsupported by any citation.--Srleffler (talk) 05:09, 25 January 2008 (UTC)

Hmm... I don't think numerical aperture would really make sense in the case where you have a virtual image. I don't really see it as OR, but just an attempt to use consistent terminology and conventions across the optics and photography articles. There may not be any other encyclopedias or other works that are so broad in scope, so it's a challenge to use consistent notation across overlapping but non-identical fields of study. ǝɹʎℲxoɯ (contrib) 05:17, 25 January 2008 (UTC)
Yeah, I figured that was too clever to get by you. There are a few sources that might back this up. How would you handle a mirror imaging system if not this way? Dicklyon (talk) 05:20, 25 January 2008 (UTC)
What is the problem with mirrors, exactly? Single mirrors also have negative magnification whenever they form a real image.--Srleffler (talk) 06:06, 25 January 2008 (UTC)
Oh, sorry, I was confusing the mirror situation with something else. Never mind, I retract my supposed need for an absolute value there. Still, it wouldn't hurt; might be more clear; or we should at least mention the alternate convention if we don't do that. Dicklyon (talk) 06:10, 25 January 2008 (UTC)