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Is the unit for L = luminance cd / m2 ?
yes, indeed. But other common units are footlambert with abbrev. fl , 1fl = 3.426 cd/m2 or the nit, 1nit = 1 cd/m2
- Jurgen Jacobs -
No. Not all meanings of each term are synonyms for the other. --Srleffler 07:42, 15 November 2005 (UTC)
- Yes, same thing is true: they are distinct quantities. One based on radiant flux, the other on luminous flux. One an energy-based measure, the other based on ability to simulate the human eye to perceive brightness. Dicklyon 05:33, 19 April 2007 (UTC)
I'm wondering if the TV signals materal should be split off into its own page, perhaps Luminance_(video). It seems to be a completely different concept from the physical quantity.--Srleffler 07:42, 15 November 2005 (UTC)
I split the article. The TV contents are now at Luminance (video), and links have been adjusted to point there where appropriate.
- I'm OK with the split, but the concept is really very close conceptually. The video luminance or "luma" signal, or the luminance (L or Y) in a colorspace such as Luv or Lab or XYZ, represents the luminance of the scene, pretty nearly, with some exposure normalization and nonlinear encoding. It's still the same concept.
Luminosity is not a synonym for Luminance
In the original version of the page, it was stated that The word luminance, a synonym for luminosity, means "the quality of being luminous; emitting or reflecting light and also Luminance is also a technical term in photometry. OK, those are both sort of true. But now that the article of about the latter term, the former statement is no longer operative. I suggest we remove the false suggestion that photometric luminance is sometimes called luminosity. Dicklyon 06:52, 19 May 2006 (UTC)
After someone added a weighted RGB expression for luminance, I made a section to explain exactly what it meant. Then, srleffler removed it, saying it's the wrong sense, because of the separate article Luminance (video). I half agree, but it also seems we're missing the opportunity to tie these two senses together, which is what I had tried to do in that section by specifying the exact meaning of the proportional relationship between photometric luminance and a colorspace luminance. This might not have been the best way or place to go about, but it's a possible worthwhile bit of info. Suggestions? Dicklyon 04:59, 16 November 2006 (UTC)
- NB. At least according to the articles, that formula applied to luma (video) rather than luminance (video). Those articles warn about the risk of confusing the two. The fact that the section that was added here appeared to confuse the two was a factor in my decision to delete.
- And you were correct, as I later verified in trying to find the right numbers. Dicklyon 16:03, 16 November 2006 (UTC)
- Is the video usage at all related to the photometric usage, or is it just a coincidence of terminology? (Yes they are both measures of amount of light, but there are a lot of measures of amount of light. They are not all the same, and should not be confused even when they happen to have the same name.--Srleffler 07:54, 16 November 2006 (UTC)
- Yes, the video and photographic meanings are the same. It's basically a colorimetry term related to the photometry term, which is applied in all forms of color reproduction. It's the "Y" term in XYZ space, or proportional to it in systems where the gain is a free variable.
- I think the thing to do is get rid of luminance (video), and to talk about that concept in both the luminance article and the luma article, so that the relations can be drawn where relevant. Dicklyon 16:03, 16 November 2006 (UTC)
- I'm not sure what you mean by "the same". The photometric term denotes a quantity that has dimensions of luminous flux per unit area per unit solid angle. The dimensions of the video quantity are unclear to me. Is it a measure of the actual luminance per pixel? If so, what is the conversion to cd/m²? Or, is it actually a misnamed measure of luminous flux, luminous intensity, or illuminance? (I trust that it is at least proportional to all of these, for all choices of R, G, and B component.)
- I definitely favour merging Luma (video) and Luminance (video), although it's not clear which title is best. They were one article until not long ago. Someone split them. I agree with discussing it in this article, but how it is presented depends critically on the answer to the question above.
- If it's not clear what I'm getting at, think about a CRT display and an LCD display (for sake of argument), each with a region of the same physical size displaying the same luma. Will the photometric luminance be the same between the two? The LCD spreads the emitted light into a narrower angular cone. --Srleffler 23:04, 16 November 2006 (UTC)
- What I meant by "the same" is that the luminance in video is, or is intended to be, a weighted integration of the emitted or reflected spectral energy of the display, using the same spectral weighting as the photometric luminance. There is an absolute "Y" value that has the same units as luminance, but usually that's ignored and all values are taken relative to whatever arbitrary luminance is defined as white. In this sense, for any given display system, the video luminance is proportional to the actual display luminance, with an unknown constant of proportionality. So it's the same concept, but "uncalibrated" with respect to absolute units; or "normalized" to a reference white level, if you prefer. Dicklyon 23:59, 16 November 2006 (UTC)
- In the encoding of video signals, luma is used and this is different than luminance/photometry. The term "luminance" is often misused in video to refer to luma, while people get it confused with luminance/photometry. In video however, luminance/photometry is used in defining standards such as monitor brightness. i.e. SMPTE RP 166 I believe defines 35footlamberts (I forget how many nits this is) for evaluation of television pictures. DCI (digital cinema) specifies 14fL, theatrical film projection 16fL (open gate; when you have film in the projector, the film blocks some light so white becomes ~14fL).
- Sometimes the color science (i.e. CIE) definition of luminance comes into play with video (if you consider digital cinema to be interrelated to video). DCI uses a color space with CIE's XYZ tristimulus components. (from what I understand) These components represent relative luminances, not absolute luminances. i.e. if your scene is shot in daylight, it really makes sense to code the image in relative luminances and not absolute luminances (because the projector isn't as bright as the sun).
- (Ok re-reading my message.) Basically I think that there should be three articles. 1- Luminance (photometry). 2- Luminance (colorimetry/color science). These numbers specify relative luminances, not absolute luminances. These numbers are most similar in concept to luminance and not illuminance, luminous flux, and luminous intensity. 3- Luma (video). These numbers are derived from #2 (luminance in color science), but they use a different 'order of operations' for pragmatic performance/cost reasons. Glennchan 21:50, 19 November 2006 (UTC)
- This is exactly what we've done in recent days, right? Dicklyon 00:15, 20 November 2006 (UTC)
- Yeah I think so. The luminance (colorimetry) article should be worked upon, so it is a clear explanation of luminance in the context of color science. It feels too video oriented, and doesn't have to be. Well I'm confused by it, that's basically what I'm getting at. Glennchan 01:19, 20 November 2006 (UTC)
I think this could do with some examples. As I understand it, luminance describes how bright things look, but I'd like some verification before I make edits. Do the following all sound right?: If my monitor displays all white, then its luminance is x. If I put a 1-stop (50%) ND filter over it, I see its luminance as x/2. If the monitor displays all black except for one pixel, the luminance of that pixel is still x. If I have a point source, its luminance is infinite; if I put a diffuser around it with finite size, the luminance would become finite, say y. If the area of the diffuser were doubled, the luminance would fall by half. —Ben FrantzDale (talk) 11:38, 11 June 2008 (UTC)
- I believe that's correct, subject to some question about the nature of the diffuser; instead of saying a diffuser, you can take an "average luminance" over the area, even when there's a delta function in it. Dicklyon (talk) 15:50, 11 June 2008 (UTC)
- I'm not an expert on this but I believe luminance is a good indicator of how things look under certain circumstances. Visual perception is pretty complex. See also brightness. I think luminance is a good indicator of how bright a relatively distant, isolated object will appear. Replies to your statements:
- If my monitor displays all white, then its luminance is x.
- I wouldn't use a monitor as an example, due to confusion of terminology. The video industry uses the term "luminance" in several ways, not all of which exactly correspond to the optical concept.
- If I put a 1-stop (50%) ND filter over it, I see its luminance as x/2.
- If the monitor displays all black except for one pixel, the luminance of that pixel is still x.
- I think so, at least for an ideal monitor.
- If I have a point source, its luminance is infinite;
- There are no point sources. Luminance is never infinite. This is not a trivial objection; it is fundamental.
- if I put a diffuser around it with finite size, the luminance would become finite, say y.
- If you put a diffuser around a small source, you decrease the luminance, because the surface area of the new combined source is larger than that of the small source within it.
- If the area of the diffuser were doubled, the luminance would fall by half.
- Only if the initial source is also diffuse, with the same emission profile as the diffuser. For example, both could be Lambertian.
- Luminance is directional, and sources need not emit the same luminance in all directions (and generally do not). A diffuser will change the angular distribution of radiation as well as the emitting area. Examples will need qualifications to deal with the directional aspect. A Lambertian surface has the property that it emits (or reflects) with equal luminance in all directions.--Srleffler (talk) 16:47, 11 June 2008 (UTC)
- I agree with all that. I think the basic idea of trying to confirm his intuitive understanding was about right. It would be more precise and objective to speak of the light received by a pixel in a sensor, covering some area, rather than putting an eye and a diffuser. A finite pixel area is a good way to average a luminance that includes a point (delta function). Dicklyon (talk) 20:23, 11 June 2008 (UTC)
Thanks for all of the responses. I have one other example I'm not sure about:
Suppose I have an LCD monitor rated at 300 nits. For a given area of the monitor, it radiates a certain number of photons per second (which defines its luminous emittance). If it were an ideal Lambertian surface, it would look "equally bright" from any angle. (Luminous intensity goes by cos θ, but in a camera imaging this surface from different angles, the image area also goes by cos θ, so each photoreceptor would count the same number of photons per second regardless of θ.)
Now suppose the monitor isn't Lambertian, but rather has half the luminous emittance—half the number of photons per second per area—by way of just not sending photons off at grazing angles, so it looks the same when sitting in front of it, but goes to black when you walk by. Would this monitor have the same luminance as the first or half the luminance. Or, that a nonsensical question but it has the same luminance in the viewing direction but zero luminance in the grazing directions?
- It would have the same luminance in the viewing direction and zero luminance in the grazing directions. Luminance is a function of θ (and also of position on the source)--Srleffler (talk) 02:21, 3 December 2010 (UTC)
- Thanks. That clarifies things a lot. That helps get away from the naive assumption that everything is Lambertian. Obviously, if you slip into that assumption, then luminance and luminous emittance seem like interchangeable descriptions of the perceived brightness of a surface. I think that's a big source of confusion, at least it was for me. —Ben FrantzDale (talk) 18:58, 3 December 2010 (UTC)
- Not really. Besides the fact that it's in French, it's not clear to me what it illustrates. At the least, we would need some text explaining it.--Srleffler (talk) 23:56, 29 July 2008 (UTC)
Why use 'd', you should use the actual 'backwards six-thing' that is the correct symbol for differential (like in the 'Radiant Flux' article. People might think it is not infinitessimal... 22.214.171.124 (talk) 19:28, 28 February 2018 (UTC)
Comment removed from article.
Please check, "The light at the image plane, however, fills a larger solid angle so the luminance ... " Isn't this meant to be, "The light at the image plane, however, fills a SMALLER solid angle so the luminance ... " The demagnified image is smaller (and thus brighter), but the luminance remains the same because when measured over the same original un-demagnified area it is the same amount of light (lossless lens). If agree, then please fix. —Preceding unsigned comment added by 126.96.36.199 (talk • contribs) 12:16, 22 August 2009
- No, it is correct as it is. The image is smaller, and has higher illuminance, but the light arriving at the image plane has a larger angular spread than that leaving the (larger) object.--Srleffler (talk) 19:21, 22 August 2009 (UTC)
- You used the term "brighter", apparently as a synonym for "illuminance". Where the article uses that term, it is as a synonym for "luminance". Generally, luminance is the closest physical quantity to our perception of "brightness" of an object. The optical surfaces in your eye form an image on your retina, but the light received by each receptor is limited not just by the size of the receptor, but also by the angular acceptance of the eye itself. If light is spread over a larger solid angle, less of it gets into your eye to be focused onto the retina. For an ideal optical system, the focusing of light into a smaller area at the image is exactly compensated by the fact that the light is spread over a wider range of angles, so less of it is received by the imaging system.--Srleffler (talk) 19:31, 22 August 2009 (UTC)
I still don't understand how luminance can be invariant through a loss-less optical system. Luminance is a density-measure: it has the area in the denominator. If you spread the same amount of lumens over a larger area, it no longer has the same luminance; double the denominator, and the ratio falls by a half. So I don't see how the luminance can be preserved in a divergent (demagnified) path. If two luminance-figures are equal, but one has a larger area in the denominator, then it also has to have a larger candela-figure in the numerator, right?
- Luminance has area in the denominator, but it also has solid angle. When the area increases, the solid angle subtended decreases; this is related to étendue and to the Lagrange invariant, both of which are invariant in ideal optical systems.
- Putting it more practically, if your optical system images the source with a magnification of two, so that the image is four times the area of the object, the rays in image space will be shallower than the corresponding rays in object space, by a factor of two. The solid angle subtended in image space is then a quarter that in object space, and the total luminance is the same.--Srleffler (talk) 00:53, 19 March 2011 (UTC)
I don't see the full text that was discussed, but the error seems to be that the luminance is not the total amount of light but the light per phase space. David R. Ingham (talk) 15:25, 24 May 2014 (UTC)
"if you form a demagnified image with a lens, the luminous power is concentrated into a smaller area, meaning that the illuminance is higher at the image. The light at the image plane, however, fills a larger solid angle so the luminance comes out to be the same assuming there is no loss at the lens."
Should that be "The light at the image plane, however, fills a smaller solid angle"? Oops - just saw previous section, so please disregard or delete this one. Though perhaps this is evidence that the comment in the article requires slightly more explanation, e.g. using the term "angular spread" to distinguish it from the area of the image. I confess I'm still a bit confused because if the demagnified image looks smaller at my eye than the source then I would have thought that the angular area subtended by the image at my eye is smaller than the angular area of the source. The explanation seems to involve the derivative ("spread") rather than area itself.
- It's not the solid angle subtended by the image at your eye that matters. You're right that it's about the angular spread of the rays. If you have optics that produce a demagnified image of the source, the cone of rays from the last optical element to that demagnified image will inevitably have a larger cone angle than that of the cone of rays from the source that are collected by the first element in the optical system. It turns out that the product of the cone angle and the height of the image is an invariant—it can't be changed by any ideal optical system. This is related to the Lagrange invariant. Image area and solid angle of the rays are proportional to the squares of the image height and cone angle. Their product is also an invariant, called the étendue.--Srleffler (talk) 22:16, 7 September 2013 (UTC)
Yes, as worded above it is correct. But that should be "it can't be changed by any ideal ideal optical system" in the last paragraph above. Real optical systems have loss and tangle the phase space. David R. Ingham (talk) 15:34, 24 May 2014 (UTC)
About the claim "If you form a demagnified image with a lens, the luminous power is concentrated into a smaller area, meaning that the illuminance is higher at the image"; I think that the illuminance is lower, not higher. The fact that the light points (actually not infinitesimal) get closer doesn't mean that the power increases, in fact the points get smaller too, giving less power and so less illuminance. With an extreme demagnifying lens you will have technically zero illuminance, as perceived by a lux meter. Also, when you light a fire concentrating (magnifying) the sunlight on a leaf then you are getting more power in the same little area (leaf), this is more illuminance (strictly more irradiance, infrared and so). Grausvictor (talk) 19:00, 19 July 2015 (UTC)
- That is nonsense. The same amount of power is concentrated into a smaller area. By definition that is higher illuminance/irradiance. Your statement about "light points" is just wrong. There are no "light points" that behave as you describe. An extreme demagnifying lens produces very high illuminance, not zero. This is exactly the case of using a lens to start a fire: extreme demagnification creates very high illuminance.--Srleffler (talk) 01:40, 20 July 2015 (UTC)
- At the end I understood what it's happening. I'm not using the word "demagnified" as probably the optics experts do. For me and for some people, "demagnified image" suggests the contrary of using a normal biconvex magnifying glass, it suggests using a biconcave demagnifying lens, and we probably are wrong. That's the origin of my confussion. Perhaps the expression "demagnified image" could be explained a little in the article, or stress that it's a normal lens, just for the layman. Grausvictor (talk) 09:20, 27 July 2015 (UTC)
I find that the units for Luminance are not useful. I prefer the radiometry units of W/cm^2/str. This is easily tractable through magnification calculations. (i.e. doubling the area halves the solid angle). I assume that this is only different from the photometry units by multiplying the power (w) by the eye response function. I would recommend adding something about this to the definition description to make it more usable. — Preceding unsigned comment added by 188.8.131.52 (talk) 00:32, 25 September 2013 (UTC)
- The radiometric quantity has its own article: Radiance. The luminance can be obtained by taking the product of the spectral radiance times the eye's response function (the luminosity function), and integrating over the visible spectrum.
- The choice of whether to use radiance or luminance is not really a matter of personal preference. Which is appropriate depends on the application. If you want to know how bright something will look to a human observer, luminance is much more useful than radiance.--Srleffler (talk) 01:34, 25 September 2013 (UTC)
luminous flux|luminous power
"The luminance indicates how much luminous power will be detected by an eye looking at the surface from a particular angle of view." This is ambiguous, at best. It can easily be read as total power reaching the eye from all directions. How about Illuminance on the retina or luminous flux per solid angle? — Preceding unsigned comment added by David R. Ingham (talk • contribs)
The statement "Luminance is invariant in geometric optics." is not general enough.
The conservation of luminance is a case of Liouville's theorem (Hamiltonian), because Maxwell's equations, and most other fundamental physical theories, can be stated in Hamiltonian form. So this sentence should read something like "Luminance is invariant in lossless coherent optics." In a sense, it is invariant under any process that does not change the entropy of something else. However the tricky case is diffuse reflection, which effectively decreases luminance by tangling the phase space. The other complicating point is that the definition given here may not be identical to a general measure of phase space. Perhaps someone will find a reference with a more general but equally accurate statement. David R. Ingham (talk) 15:58, 24 May 2014 (UTC)
- Generally these concepts are taught in terms of ray optics, not waves and phase space. Do you have sources for the alternative approach? It sounds interesting. Dicklyon (talk) 18:46, 24 May 2014 (UTC)
I edited the health effects section. More information on this would be good. I took out the sentence about heating being proportional to the product of luminance and time, because it isn't really true and doesn't tell the whole story. The weighting of wavelengths used in calculating luminance is not correct for determining heating of the retina. Red light will produce more heating than one would expect due to luminance. Near infrared light can quite effectively heat (and damage) the retina, but provides no luminance at all.
Stating that local heating is proportional to the product of luminance and time is also a bit misleading. Damage is clearly not proportional to this product in general. Exposure to a low level of light for eight hours is clearly not the same as exposure to 288,000 times as much luminance for a tenth of a second. We can talk about heating, but if we do we need to go into more detail, so as not to leave the impression that the product of luminance or radiance and time is all that matters.--Srleffler (talk) 03:33, 25 May 2014 (UTC)
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Confusing or Incorrect text?
Current text: "In this case, the solid angle of interest is the solid angle subtended by the eye's pupil."
The unit of luminant flux is the candela = cd = lumen/steradian
The unit of luminance is candela per square meter = cd/m^2 = lumen/steradian/m^2
On the one hand, the text states:
"The luminance indicates how much luminous power will be detected by an eye looking at the surface from a particular angle of view. Luminance is thus an indicator of how bright the surface will appear. "
Yet the definition of luminance is candela per unit angle per square meter, and there is nothing in this definition which presupposes the angle of interest is that "subtended by the eye's pupil", which angle it should be noted will change depending on the distance of the object.
Luminance refers to light subtended by an angle and reflected from a source over an area (cd/m^2), but there is no decent explanation for how this is a reasonable measure of an objects' brightness.
To give a specific example: the luminance of an "incadenscent light bulb" is given as 130 kcd/m^2, which is 130 kilo-lumens per steradian per square meter. How does one understand this geometrically when looking directly at a lightbulb? Where does the square meter come into play?— Preceding unsigned comment added by 2601:643:8104:a620:bdf1:61b3:1e91:c6eb (talk • contribs) 13 August 2017 (UTC)
- Luminance can be kind of hard to get your head around. One thing that helps is to understand that in an ideal optical system with no absorption, luminance is conserved. This is related to the conservation of etendue, and the article on that has some explanation you may find helpful. Because luminance is ideally conserved, you can calculate it at any point in the system, by considering the area through which the light passes and the angular spread (solid angle) of the light. The luminance of the object will, assuming perfect imaging, be the luminance of the image it forms in your eye. You can consider the area of the object and the solid angle subtended by your pupil, or the area of your pupil and the solid angle subtended by the object. You'll get the same result either way. Yes, the solid angle changes according to the distance to the object. Luminance tells you how bright an object appears. An object further away appears dimmer, because a smaller angle is subtended therefore less light passes through your pupil and contributes to forming the image.--Srleffler (talk) 04:44, 14 August 2017 (UTC)
- Regarding the light bulb: I don't see where you get that number, but supposing you had a lightbulb with such a luminance to answer your question I would first need to know whether it was a clear or a frosted bulb (it matters!). Assuming a frosted bulb, the square meter would be the area of any given part of the surface of the bulb. If you consider a small element of area on the surface (think of it as a pixel in an image of the bulb), you could use the luminance, the area of the surface element, and the solid angle subtended by the pupil of your eye to determine how bright that part of the bulb's surface would look to you. If the surface of the bulb has uniform luminance and if the frosted bulb is a perfect (Lambertian) diffuse emitter, then the cos θ in the formula cancels and all parts of the bulb's surface would look equally bright. For a clear bulb, you would be looking at the filament, and the relevant area would be on the surface of the filament. A clear 100 W bulb will have much higher luminance than a frosted one.--Srleffler (talk) 05:01, 14 August 2017 (UTC)