Talk:Lambert's cosine law

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I corrected the statement that the radiance follows the cosine. I took out the telecommunications link and the section:

Sources

• Federal Standard 1037C

because it made no sense to me when I followed the link.

I am thinking this might cause a problem with some people's links, so if I did wrong, please fix it. I'm new to Wikipedia, so if I'm not following protocol, please let me know that too.

Paul Reiser 20:42, 30 Nov 2004 (UTC)

The protocol is be bold. Good job. For what it's worth, the link to Federal Standard 1037C is essentially a historical glitch. The original text of this page is from [1] (a sub-page of [2]. Frames suck). But it isn't especially relevant. Dbenbenn 02:54, 6 Jan 2005 (UTC)

Lambertian diffuse lighting model

The page Lambertian diffuse lighting model has a lot of the same information as this page. As far as I can tell, the Lambertian model simply posits the Lambertian cosine law. Should the pages be merged? Dbenbenn 02:06, 6 Jan 2005 (UTC)

I think they should, under "Lambert's cosine law". I will do that in a day or two, unless you do it first. Paul Reiser 04:06, 6 Jan 2005 (UTC)

I think the page Lambertian diffuse lighting model is quite a bit clearer than the Lambert's cosine law page. In the latter page, all that "flux", "total power", etc, makes things too complicated. "Apparent brightness" is so much easier to understand. Oleg Alexandrov 04:17, 6 Jan 2005 (UTC)

We could have the first paragraph be more intuitive, the second more technically precise with links to definitions. Maybe a diagram, too. --PAR.

I saw a nice explaination of Lambert's cosine law at this MIT class page. They do not talk at all about the observer's line of sight, as you do in the first paragraph of Lambert's cosine law. From that external link I think it follows that the power observed does not depend about the observer's line of sight, because light spreads in all directions equally. Does this contradict your first paragraph? I could be wrong, I don't know much about this. Oleg Alexandrov 19:25, 6 Jan 2005 (UTC)

Agreed. The lambert surface has nothing to do with the observer. It simply states that for a surface with constant radiance in all directions, it's power distribution over space has to go with cosine. The reasoning about cosθdΩ in figure 2 is simply wrong and covers the fact that cosine distribution of radiant intensity is a characteristic of the lambertian surface rather than observer angle. Constant radiance means equally obeserved brightness but that is just another story. This page is likely to be the source of all the confusions I've seen in computer graphics. --Xiaoyu87 (talk) 05:30, 4 July 2020 (UTC)

Thats ok, I seem to have problems explaining this clearly. If I can write a page that you approve of, maybe we will have a good page. The problem is that there are two quantities, both of which have units of photons/sec/cm^2/sr or energy/sec/cm^2/sr. The emission from the Lambertian surface is measured in these units, and that emission varies as the cosine of the angle from the normal. The observer measures radiance, which is also in these units and that is independent of the angle from the normal. Some people say "intensity goes as the cosine of the angle" which is correct for the emitted intensity, while others say "intensity is constant" which is true for the observed intensity. The thing about the observed intensity, is that as you vary the angle, and keep the area you are looking at constant, the solid angle the observer sees for that area decreases as the cosine of the angle, as does the number of photons/sec received, and so the ratio of the photons emitted from the area divided by the solid angle that the observer sees for that area has the two cosines cancel, and the observed photons/sec per solid angle is constant. I'll try to draw a diagram that illustrates this.Paul Reiser 22:21, 6 Jan 2005 (UTC)

OK, I understand things now. I think for people in computer graphics, who use Lambert's cosine law a lot (maybe more than people in other disciplines), what matters is the observed intentsity, that is, what is in the eye of the beholder. Maybe the article should be split into two clearly delimited parts, one being what the observer sees, and the other being what happens at the surface itself when the light strikes. And maybe what the obverver sees should be given priority. Or maybe not. I don't know. I am looking forward to your changes to the article. Oleg Alexandrov 22:46, 6 Jan 2005 (UTC)

How about this: "The observed brightness at a point depends on the angle at which the light strikes that surface, but not on the angle of view. Specifically, the brightness is proportional to the cosine of the angle between the light source and the surface normal." Dbenbenn 23:10, 6 Jan 2005 (UTC)

This would be nice! One could also maybe add after your text: "Therefore, a point appears brightest if the light strickes "head on" at that point, and dimmer if the light strikes under an angle...." Just some thoughts. Oleg Alexandrov 23:18, 6 Jan 2005 (UTC)

Done. I ended up removing the following text. Perhaps some of it should go back in?

... is the statement that the total power observed from a "Lambertian" area element is directly proportional to the cosine of the angle θ made by the observer's line of sight and the line normal to the area.

This means that the area element will be just as bright no matter what angle it is viewed from. The total amount of power that an observer sees will be proportional to the brightness multiplied by the solid angle subtended by the area element. The area element will have a maximum solid angle subtended when it is viewed "head on" (i.e. θ=0) and will become smaller as the angle is increased until it is zero when θ=90 degrees. That means that the total power observed will be maximum when the area element is viewed head-on, and will drop to zero when viewed edge-on. The brightness (or power per unit solid angle) will be constant, however. The sun, for example, is almost a Lambertian radiator, and as a result the brightness of the sun is almost the same everywhere on an image of the solar disk.

When an area element is radiating as a result of being illuminated by an external source, the flux (energy/time/area) landing on that area element will be proportional to the cosine of the angle between the illuminating source and the normal. For a Lambertian reflector, the light reflected from this source will be the same in all directions, so the radiance seen by any observer will then be proportional that incident flux which will be proportional to the cosine of the incident (not the observing) angle.

Dbenbenn 00:25, 7 Jan 2005 (UTC)

I made a few word smithing changes to the derivation of the observer-angle independence of the radiance of a Lambertian surface. I think the existing material was solid but it was missing a few little hints that would have made it easier for me to digest on the first reading. I'm a Wikipedia neophyte, so I apologize if my edits or this comment are misplaced. Hope it helps! Ryan-feeley (talk) 08:22, 19 January 2019 (UTC)

Lambertian radiators?

The text I removed above talks about "Lambertian radiators", such as the sun. And some of the links to this page expect there to stuff about Lambertian radiators here. As far as I can tell, a Lambertian radiator is simply one that emits light uniformly in all directions. I don't see where the cosine comes into it. Dbenbenn 02:23, 7 Jan 2005 (UTC)

On the rewrite

Now I understand! I have just one problem with the article. The people in computer graphics I think use this law much more than anybody else. For them, the Lambert's law is:

"The brightness of a point on a surface is proportional to the cosine of the angle betwen the incident ray at that point and the surface normal at that point".

The way you wrote the article, this is an afterthought, put in the very last paragraph, and stated there as a conclussion of the very long theory you developed in this article. So, for phisisists your article will be interesting, for computer graphics people, it will be kind of not helpful. I don't know how to reconcile these. Oleg Alexandrov 05:33, 7 Jan 2005 (UTC)

The reflection law is a special case of the emission law, so thats why I did it that way, plus I have a physics bias. Putting the reflection part in the first section doesn't sound like a bad idea to me. Paul Reiser 16:25, 7 Jan 2005 (UTC)

Image:LambertCosineLaw.png

Hey Paul,

Would you consider splitting Image:LambertCosineLaw.png into two images (the top and bottom). Then if you remove the "Figure 1" and "Figure 2" captions, we can just put them in text, with code like

[[Image:LambertCosineLaw.png|thumb|right|411px|Figure 1: Emitted intensity]]

Dbenbenn 14:28, 7 Jan 2005 (UTC)

Ok, I will split it up and stick it in somewhere using the above format. I didn't like the captions anyway, and this gives more freedom to change. Paul Reiser 16:25, 7 Jan 2005 (UTC)

I split them, but do you know how to center the captions?Paul Reiser 17:11, 7 Jan 2005 (UTC)

Why do you want the caption centered to start with? I think left-aligned looks just fine. There is some info about captions at Wikipedia:Picture tutorial, but not what you want (at least I could not see centered captions).

Oh, if you really really want it, I think you can hard-code centered caption in html instead of using Wiki markup. Just look at the html source code of the page, and figure out where to insert a <center></center> thing. Oleg Alexandrov 17:54, 7 Jan 2005 (UTC)

Thanks, that worked fine. Its just that in scientific publications, single-line captions are centered, multi-line are not. I expanded the captions, so they stayed left-justified.

Terminology

I have adjusted the terminology used on this page to agree better with correct optics usage. Note that "brightness" is an ambiguous term, and should not be used in any scientific context, except when talking non-quantitatively about human perception of light. "Intensity" was also a problem here, since it is also ambiguous, and the sense in which it was used on the page conformed to none of the common standards. Intensity was used here to mean luminance and/or radiance. Intensity much more commonly means any one of luminous intensity, radiant intensity, irradiance, or illuminance. See also intensity (which is the same as irradiance). I am in the process of updating the definitions of some of the linked terms, so if the definitions are not clear right now they may be soon. --Srleffler 07:04, 15 November 2005 (UTC)

I strongly object to the use of the luminous terms - the article should not give the impression that this is a phenomenon which involves the response of the human eye for its validity or understanding. It should be done in radiance units, and every time the article offers an intuitive explanation in terms of a human observer, I think, with a little thought, it could be written so as to be technically correct, but not confusing to someone who doesn't understand the difference. I will do this soon, unless somebody agrees with me and does it sooner. PAR 17:57, 15 November 2005 (UTC)

Sounds fine to me. I used photometric units because I thought they lent themselves better to intuitive explanations, and also because, based on the date, I assumed Lambert originally formulated his law in photometric units. I did keep the radiometric units for the long example at the end. It is important to me that whatever units are used, the units and the explanation must be technically correct. Incorrect use of these units already leads to a lot of confusion.--Srleffler 04:39, 16 November 2005 (UTC)

It seems that the term "radiance" is used improperly here. If you consider the radiance definition, you will see that the radiance of a Lambertian reflector is equal in all directions (because it already takes into account the cosine between the surface normal and the observer line of view). This contradicts with the sentence "In mathematical terms, the radiance along the normal is I photons/(s·cm2·sr) and the number of photons per second emitted into the vertical wedge is I dΩ dA. The number of photons per second emitted into the wedge at angle θ is I cos(θ) dΩ dA." Midnighter84 (talk) 23:12, 17 March 2011 (UTC)

Moon example

An anonymous editor altered the moon example today, completely reversing the sense of it. I have commented it out in the article for now, pending confirmation of one version or the other by someone with knowledge of this subject. The old text was:

For example, if the moon were a Lambertian reflector, one would expect to see its reflected brightness appreciably diminish towards the outer edge, or limb. The fact that it does not diminish illustrates that the moon is not a Lambertian reflector, and in fact tends to reflect more light into the oblique angles than a Lambertian reflector would.

and the new text is:

For example, if the moon were a Lambertian reflector, one would expect not to see its reflected brightness appreciably diminish towards the outer edge, or limb. The fact that it does not diminish illustrates that the moon is nearly a Lambertian reflector.

--Srleffler 05:48, 6 February 2006 (UTC)

Yes the anon edit was wrong. The increased angle of incidence near the limb will cause a decreased amount of light/area falling on the surface at the limb for any kind of reflector. For a lambertian reflector, that decreased amount of light will have the same brightness when viewed from any angle, but its still decreased from the larger light/area falling on the surface normal to the sun (i.e. away from the limb). The fact that the decrease in brightness is less than expected means the moon is not a perfect lambertian reflector. PAR 20:10, 6 February 2006 (UTC)

Despite the discussion above, I think the text on the page is still wrong. The light falling on the moon's surface per square metre must fall continuously to zero as you approach the terminator (the border between the illuminated and the dark parts of the moon's surface). Therefore the apparent brightness (however you define it) must also fall to zero continuously. The fact that the terminator appears to be a discontinuity must be because of the non-linearity of human perception.

When the moon is full, the terminator coincides with the limb. Therefore the apparent brightness must fall smoothly to zero as you approach the limb.

I am not sure that the moon is a good example at all. --194.81.223.66 13:22, 19 December 2006 (UTC)

The 'New Text' in this context is correct: the moon is a good example of a lambertian scatterer. The brightness of a lambertian scatterer does not appear to diminish with increase in observer angle and therefore appears equally bright across the whole disk (As does the Sun in one of the above examples). The point here is that the ignoring the observer angle is misleading. Let us take two observers, one on the earth and another on the moon. The earthbound observer shines a torch on the moon and illuminates an area that subtends a fixed solid angle. It does not matter where this light is shone on the moon, it always subtends the same solid angle. To the observer on the Moon, depending at what latitude/longitude (i.e. where the normal from the surface is pointing with relation to the torch) the area subtended by this solid angle will change, therefore the irradiance changes. For the observer with the torch, there will be no change in apparent brightness as a function of torch beam position on the moon disk. Here's the point that is missing from previous arguments: depending on the angle between the normal of the surface and the observer, the observer will see an area proportional to the inverse of the cosine of that angle for a given solid angle viewcone. For example, the Earth-bound observer shines the torch at edge of the disc of the moon, the area in the solid angle of the Earthbound source tends to infinity, so irradiance tends to zero, but the observed area from the same point tends to infinity, so the overall response is unchanged; Similarly, for whichever point on the moon the moon-bound observer is standing, for a given irradiance, it does not matter at what direction that observer looks at the ground, it will always appear the same brightness as again, a given unit solidangle of viewcone subtends a larger area as angle of observation increases.Jpddowning (talk) 14:08, 27 April 2015 (UTC)

shorter is better

I prefer shorter articles. So I prefer not merging.

Examples

I came here looking for information -- and didn't find it. As far as I can tell, laser light is non-lambertian on account of the photons being in phase. I'm interested to know if there are other sources (perhaps LEDs? electric arcs?). While it's an interesting definition, it should be possible to know what it distinguishes between. If all the known light in the universe is Lambertian, then it's not a useful distinction ... sittingduck 21:06, 1 Mar 2006 (UTC)

Yes, laser light is not lambertian. Nor is a flashlight. Lots of other sources of light are not lambertian, and most surfaces are not pure lambertian reflectors. Pure lambertian light is not all that common, but is widely used as an approximate model for diffuse reflection and emission of light. Combining some lambertian reflection with some specular reflection creates a good model for many surfaces. This approach is used in computer graphics, etc.--Srleffler 23:21, 1 March 2006 (UTC)

First WIKI edit ever

I dunno what I am doing with this WIKI thing... I guess this is the right way to add a comment to the bottom here.

Anyway, I am a computer graphics guy and would like to comment on this discussion. I see what apears to me to be a gaping hole in the definitions offered here. First Diffusion is a reflective property of a surface normal. Scattered reflectivity of a kind defined by the shader being used. Radience seems different. Radience would be more along the lines of luminosity - although it may be a superset that includes various reflective properties in some sciences. Luminous or radient properties are defined by rays which originate from the calculated normal for a smoothed surface. Diffusion is calculated from a light source. For example Lambertian Diffuse is basically the just the cosine of the angle between the surface and the light.

The gaping hole I mentioned is for a solid deffinition of Diffuse or Diffusion as it applies to Computer Graphics. The algo being used to "scatter" the reflected light is not important to this general definition. But I think should be included in any of the definitions of diffuse shading models such as Phong, Lambert, Minnaert, OrenNayer, Generic Occlusion Shaders, Generic Translucency shaders (where the diffusion is of refracted light and not reflected light), And likewise Sub Surface Scattering which is also a kind of "Diffuse" property.

My email address for further disscussion or whatever is Tesselator@gmail.com and my name is James Dean Prentice III. You can also reach me during the day at the Kyoto Institute Of Science and Technology in the Computer Science Department by just asking for Jim sensei. :)

Welcome! Yes, this is the right way to add a comment. Only one change: add four tildes ("~~~~") at the end of your comment. The software will replace this with your ip address or username and a timestamp, to make it easier to see which comments are from whom.

I think you may be confused because you're looking for information on computer graphics, and you have ended up at an article on physics. This article deals with the physics of diffuse reflection from an idealized "Lambertian" surface. The companion article on Lambertian reflectance talks a bit about computer graphics applications. These articles are still "under construction", and may end up getting merged into a single article eventually. There are also articles on the Phong reflection model and Phong shading. Some of the material you are looking for may be summarized at 3D computer graphics.--Srleffler 21:24, 4 May 2006 (UTC)

photons vs energy

What is the difference between photons vs energy in this context, for thinking about the distribution of light output vs angle?

Please add a linear plot of angle vs photons/energy/light emitted.

What is the half-power angle for Lambertian sources? This would be good material to add to the article, because this is the way many current light sources are rated.

It seems like the article should reference/link to LED SSL devices, since Lambertian is often mentioned for some high-power LED emitters. -96.237.10.106 (talk) 12:40, 3 June 2009 (UTC)

In this case it seems like the half-power angle would be 60 degrees, since cosine of that angle is one half. Personally, I donät know why they are making a difference between energy and photons, since a photon basically is energy quantized, i.e. one photon of a certain frequency v contains an energy E that is given by the Planck relation (or the Planck–Einstein equation) E = hv. Maybe they just prefer to keep a unitless number and instead try to obtain the probability distribution for the reflection angle of the photon. --Kri (talk) 01:11, 7 February 2011 (UTC)

Peak intensity?

I'm trying to work out what the peak flux density is. That is if, a Lambertian radiator is radiating 1 W, what is the radiant intensity in the normal direction? I think it is 2W/sr. I get that from this:

((pi*0.0000001^2)*(1/3) / (4/3*pi*.5^3)/0.0000001^2 sr = 2 sr

That is, all the power fills a sphere of radius 0.5. If we have a small element at 1 unit away in the normal direction, the left quantity is the volume of the cone to it, the right quantity is the subtended angle. Does that sound right? —Ben FrantzDale (talk) 21:54, 15 November 2010 (UTC)

If the peak luminous intensity is $I_{max}$, the total radiated luminous flux from the surface will be

$F_{tot}=\int \limits _{0}^{\pi /2}\,\int \limits _{0}^{2\pi }\cos(\theta )I_{max}\,\sin(\theta )\,\operatorname {d} \phi \,\operatorname {d} \theta =2\pi \cdot I_{max}\int \limits _{0}^{\pi /2}\cos(\theta )\sin(\theta )\,\operatorname {d} \theta =2\pi \cdot I_{max}\int \limits _{0}^{\pi /2}{\frac {\sin(2\theta )}{2}}\,\operatorname {d} \theta =\pi \cdot I_{max}$

(where sin(θ) is the determinant of the Jacobian matrix for the unit sphere), so the peak intensity will be 1/π of the total radiated luminous flux per square radian according to my calculations. --Kri (talk) 01:52, 7 February 2011 (UTC)

Great. Thanks! —Ben FrantzDale (talk) 14:50, 7 February 2011 (UTC)

Hm, maybe this is something to add somewhere in the article. --Kri (talk) 23:41, 9 February 2011 (UTC)

I agree; it should include units: that must be a factor of $pi$ steradians...

In general, I find the biggest source of confusion with radiometry and photometry is that it is so easy to inadvertently assume Lambertian surfaces, which suddenly makes lux and nits measures of the same thing, for example. If you don't realize you are making the Lambertian assumption, both seem to be measures of perceptual brightness of a surface. Given I am much more comfortable with the notion of power than steradians, I kept being drawn to lux over nits when I started working with this stuff.

If I understand correctly, then, a computer monitor that is 300 nits and perfectly Lambertian would have an emittance of

300 nits π sr = 942.5 lx

in the normal direction. —Ben FrantzDale (talk) 12:29, 10 February 2011 (UTC)

From what I understand, nit is the unit for intensity per square meter (of radiating material), which is only measured in one specific direction (for example the normal direction), while lx is the unit for emittance, which is the total emitted flux per square meter that you would catch if you would collect all the light that is emitted in all directions. Hence it's not possible to talk about 942.5 lx in any specific direction. And nit and lux would not measure the same thing even if you would have a Lambertian surface.

For example, if a computer monitor that has an intensity per square meter of 300 nit would be Lambertian, it would have a total emittance of 942.5 lx, as your calculations showed. And even though it would be perceived equally bright from all directions (which a Lambertian surface is), the intensity per square meter would change and be lower when viewed from an angle instead of viewed directly from the front, according to the Lambert's cosine law. Consequently, one have to divide the intensity per square meter by cosine of the angle it is seen from, to get the perceived brightness. This is true for all surfaces, not only for Lambertian. --Kri (talk) 14:10, 10 February 2011 (UTC)

I misspoke. I meant 300 nits in the normal direction for a Lambertian emitter implies 942.5 lx. For comparison, an old LCD monitor, which emits primarily in the normal direction and is dim off axis, might achieve 300 nits, but would produce <942.5 lx total. —Ben FrantzDale (talk) 14:38, 10 February 2011 (UTC)

It should be noted that it is not wrong to say that $F_{tot}=\pi \,I_{max},$ and omit the steradians, since the radian (and hence also the steradian) is unitless and actually = 1 (see Radian#Dimensional analysis). It may be a point to write it out in the last step as it is done right now in the article, but it is not a result of calculation but only for clarification of what we are actually measuring. It's a good thing to add the derivation to the article though. --Kri (talk) 00:45, 12 February 2011 (UTC)

Yes, radians and steradians are dimensionless (unless you start getting into geometric algebra, in which I think there are algebras that give angles a different "grade" than scalars, but that's way out of scope ;-) ). Especially since the "units" of those two things differ only by steradians, it seems useful. —Ben FrantzDale (talk) 14:26, 14 February 2011 (UTC)

Really nice example, to explain this. I added it to the article. I did change the numbers to 100 nits, just to make the math even more transparent - it's still in the range of a typical computer screen. I was tempted to continute a bit further, to get it even closer to something most readers can relate to, but i'm not sure if it'll get too far of topic:

Assuming 30% transmitivity of the LCD, the backlight would need almost 100 lm, which modern LEDs could supply using around 1 W power. Note however, if screens were perfect Lambert emitters, they would have a viewing angle of 180°. Especially notebooks will save power, by having a reduced viewing angle, ie. being non-lambertian.

Do you think it's to farwinded? Tøpholm (talk) 00:16, 12 March 2011 (UTC)

I think it can be a good idea to mention what you wrote about LED screens and power saving for notebooks. --Kri (talk) 03:14, 15 March 2011 (UTC)

The talk about sin(θ) being the determinant of the Jacobian matrix for the unit sphere, seems completely out of place... The sin(θ) part is there, because the area of the integration surface element is smaller at lower polar angles θ. In other words, the differential solid angle is $\mathrm {d} \Omega ={\frac {\mathrm {d} S_{r}}{r^{2}}}=\sin \theta \,\mathrm {d} \theta \,\mathrm {d} \varphi .$ See solid angle or spherical coordinate system for reference. Tøpholm (talk) 00:54, 12 March 2011 (UTC)

I mentioned the determinant of the Jacobian matrix since that is what enters in the integrand when one switches from one coordinate system to another, essentially from cartesian coordinates to spherical coordinates as in this case. Only that here r=1, so r²sin(θ) (which is the determinant of that specific Jacobian) becomes just sin(θ). But maybe it is wrong to use it in this case since the surface we integrate over (that of the unit sphere) is two-dimensional while it is contained in a three-dimensional room, and that the method only happens to work in this case. So yes it can be removed, or maybe replaced by something that explains where it comes from. --Kri (talk) 03:02, 15 March 2011 (UTC)

error in first paragraph

"It has the same radiance because, although the emitted power from a given area element is reduced by the cosine of the emission angle, the size of the observed area is decreased by a corresponding amount."

... shouldn't it be "increased by a corresponding amount"? —Preceding unsigned comment added by 216.228.112.21 (talk) 23:19, 6 April 2011 (UTC)

The sentence is ambiguous. I think it's trying to say "the observed angular size of the area is decreased by a corresponding amount". I think you are saying "the observed area subtended by a given angle is increased by a corresponding amount". Both are right, of course. —Ben FrantzDale (talk) 11:10, 7 April 2011 (UTC)

For a Lambertian surface, "the apparent brightness of the surface to an observer is the same regardless of the observer's angle of view". This means that the energy per solid angle is constant and does not depend on the angle of the viewer. The original statement was correct, the current statement is wrong.

This means, for example, that to the human eye it has the same apparent brightness (or luminance). It has the same radiance because, although the emitted power from a given area element is reduced by the cosine of the emission angle, the area subtended by a given solid angle from the viewer is increased by a corresponding amount. Therefore, its radiance (power per unit solid angle per unit projected source area) is the same.

The use of the word increased (which I highlighted above) is wrong. I suggest changing this to the following

This means, for example, that to the human eye it has the same apparent brightness (or luminance). It has the same radiance because, although the emitted power from a given area element is reduced by the cosine of the emission angle, the apparent size (solid angle) of the observed area, as seen by a viewer, is decreased by a corresponding amount. Therefore, its radiance (power per unit solid angle per unit projected source area) is the same.

Q Science (talk) 23:32, 12 August 2011 (UTC)

First Paragraph, again

It has the same radiance because, although the emitted power from a given area element is reduced by the cosine of the emission angle, the actual area of the surface visible to the viewer is increased by a corresponding amount (1/cosine).

How actual area of visible surface is increased, when observer is looking at surface at more oblique level? This is strange.

disambiguate angle and solid angle ?

In figure 1 and the associated paragraph d-Omega is variously referred to as both an angle and a solid angle. The result is it's not clear in Figure 1 if the segments are of equal angle or equal solid angle. In a 3-dimensional situation it doesn't make much sense that they would be both equal angles and equal solid angles, because the solid angle of a band around the hemisphere, of equal angle, increases as you approach the horizontal. I'd be very grateful if someone could clear up the confusion.

Araphid (talk) 10:34, 19 March 2014 (UTC)

Absolutely. Geometrically, what d-Omega, d-Omega_0 D-A and d-A_0 refer to is not clear at all. Without further explanation it appears that the number of photons an observer sees is inversely proportional to the observing objects surface area. This is confusing.

Perhaps a figure clearly showing what dA_0 applies to and a rephrasing of the variables in geometrically general terms would go a long way to make this more article usable for calculations. — Preceding unsigned comment added by 132.195.104.241 (talk) 15:09, 4 December 2017 (UTC)

Backscattering yields a uniform full moon image.

There is no true photo of light scattered from a surface that obeys Lambert's cosine law.

The mean backscattering from the full moon is directed back to the sun because the scattering dipoles on the moon oscillate in a plane perpendicular to the coming sunlight.

Any calculation that does not take the direction of the polarizing dipoles into account will not be correct.

Backscattering yields a uniform moon image, as well as the earth's image and the images of all the planets and their moons.

Back-scattering from any surface will not depend on the surface inclination angle to the coming light, and a curved surface will look uniform. It is, therefore, not surprising that true images, that obey Lambert's cosine law, are difficult to find.

In the calculation of Lambert's cosine law, there is a hidden assumption that the scattering dipoles oscillate in all random directions in space. This is not the case with the moon, and probably with other examples. Urila (talk) 13:07, 29 October 2018 (UTC)

urila — Preceding unsigned comment added by Urila (talk • contribs) 00:37, 31 July 2018 (UTC)

    The Scattered light is considered in the literature as a diffusive light, light that passed a number of scattering events before it left the scattering material. Diffusely scattered light must obey Lambert's Cosine scattering law. In the case of unidirectional light scattered backward from a surface of a sphere, the meaning is maximum scattering intensity in the middle of the sphere surface, and a decline to zero toward the periphery by the cosine law. 
    The full moon looks uniform. People continue to assume that the light is diffusely scattered from it.
    More than that. The nearly uniform sphere image is common to all the planets and their moons, including the earth as observed from the moon. Out of thousands upon thousands of true photos, there is no single true photo that obeys Lambert's Cosine law. The only photos that do obey the law are rendered photos, photos that are at least partly simulated.
    Contrary to all that, if the scattering is assumed to be mainly a single event, then all the scattering dipoles are directly stimulated by the light radiation on the illuminated scattering material. Then scattering by them must be coherent, and then the full moon and all the other illuminated bodies, with similar illumination geometry, must be uniform, at least approximately. The full moon tells us that single event scattering is dominant. Maybe with small corrections of multiple scattering.   
    Why is the single event dominant? It seems that the effect is geometrical and statistical. If we consider one event scattering, two event scattering, multiple event scattering, then the event probability will decline with the increasing number of scatterings. The single event has a probability of at least 50% and it is the strongest event.

Urila (talk) 09:40, 16 May 2020 (UTC)

References

The opposition surge and Lambert's Cosine Law

The opposition surge, the significant increase of 180 degrees backscattered light, is a well-established and well-documented phenomenon. See "Opposition surge" in Wikipedia, and the references therein. On the other hand, there is no single true photo that obeys Lambert's Cosine law, "Lambert's cosine law" in Wikipedia. The first figure in the article, "Lambert's cosine law" in Wikipedia, is an erroneous artist view.

Urila (talk) 09:12, 9 October 2020 (UTC)

R

In optics, Lambert's cosine law says that the radiant intensity or luminous intensity observed from an ideal diffusely reflecting surface or ideal diffuse radiator is directly proportional to the cosine of the angle θ between the direction of the incident light and the surface normal; I = I0cos(θ).[1][2] 2409:4042:2C1E:5C14:C0C4:2FFF:7ADA:503E (talk) 08:12, 29 May 2022 (UTC)

The Symbol for radiance/luminance is L

In the interesting "Details of equal brightness effect" section Radiance (a certain number of photons/s per meter squared per steradian) is shown with the symbol I and Io. This is confusing because with SI conventions I is used for intensity (e.g. photons/s per steradian). The correct symbol for radiance/luminance is L. Jack Hogan (talk) 08:47, 3 March 2023 (UTC)

Lambertian Scatterers section Inaccurate

"When an area element is radiating as a result of being illuminated by an external source, the irradiance (energy or photons /time/area) landing on that area element will be proportional to the cosine of the angle between the illuminating source and the normal"

This is only true for a directional light source, like a light bulb. Is it true on the peak of a mountain range with an overcast sky, for instance? Of course not. The point is that while the input to a Lambertian reflector can be from any random direction, its output is omnidirectional in the hemisphere above it in the proportion suggested by the cosine law. What matters for the input is the number of photons/s that fall onto the surface of interest, regardless of where they came from. And for that irradiance/illuminance the output is predetermined by the cosine law. Jack Hogan (talk) 09:15, 3 March 2023 (UTC)