Talk:Lambertian reflectance

message: To me, there's a mistake in the very beginning of the article:
"If a surface exhibits Lambertian reflectance, light falling on it is scattered such that the amount of light an observer sees, the surface luminance, is the same regardless of their angle of view."
If exhibiting a lambertian reflectance is the result of lambert's cosine law, then we can't really talk of "amount" of light. While I agree with the use of the term 'luminance', it implies a notion of brightness, and not really of quantity or amount. Lambert's cosine law says that the amount of light decreases with the cosine of the angle between the normal of the surface and the line of sight. But as the apparent surface the observer sees decrease with the same law, the brightness of the source remains constant.
However I may be wrong...
palleas - 04/03/2006
:Yes, you're right. I'll try to reword it. What the person who wrote that probably meant was that the amount of light an observer sees per unit area of the surface, as projected on the plane perpendicular to the observer's angle of view. That would be correct, but isn't exactly elegant. The description must not use the term "brightness", though. The use of "brightness" as a description of or measure of light is strongly discouraged, because it tends to lead to confusion between luminance, radiance, luminous and radiant intensity, irradiance, illuminance, etc. As discussed at brightness, the latter term should be used only to describe a person's subjective impression of light, not actual physically measurable "amounts" of light.--Srleffler 17:22, 3 April 2006 (UTC)
replies: text: I guess luminance (or radiance in energy units) is the term to use here, though it doesn't show clearly that there is the cosine factor in it. Maybe the article should have a first part that could be more 'instinctive', and say that a lambertian reflector is as bright whatever the line of view, and then a second part that explains it scientifically. I totally agree that the photometry terms are a hassle, and I'm french, so I have to deal with them in french and english, and as you can well guess, it's often not the same....--Palleas 9:10, 4 April 2006 (UTC)
thread_title: Untitled

message: The article states that the apparent brightness is the same regardless of the observer's angle of view, and that rotating the object doesn't change its apparent brightness.  Then it says the reflection is calculated using the normalized vector of the surface and the vector to the light source.  Now if angle has no effect, then why exactly are we taking it into account in our calculation, or what exactly is it that we're calculating?  I think the least that can be said is that the article needs to be re-worded.  I personally don't know how to interpret it; I don't know if this is the equation I need to use for emulating some sort of reasonable shading. Perhaps he meant that rotating the object does not change its apparent brightness if the light source rotates about the object with it?  Inhahe (talk) 06:09, 23 January 2009 (UTC)
replies: text: Your conclusion is correct. The shading of the surface remains the same regardless of the observer's angle to the surface. There shouldn't be any implication that the light or surface are changing with respect to each other. Xot (talk) 20:59, 21 June 2009 (UTC)
thread_title:  Doesn't make sense.. 

message: Shouldn't there be a factor of 1/\pi involved in the formula for I_D? Suppose that an equal amount of light was coming in from all directions, with I_L = 1. Then, integrating the formula for I_D given in the article over the hemisphere, you get that a total of 2\pi^2 light is going out, while only 2\pi came in. Note that the extra factor of 2\pi for the outgoing light comes from the fact that I_D light shines out in all directions equally, and that the surface area of a unit hemisphere is 2\pi. A similar argument gives the 2\pi for incoming light.
Gereeter (talk) 17:42, 21 January 2013 (UTC)
thread_title:  Constant Factor in Formula for Outgoing Light 

message: It would be nice to have more examples of approximately Lambertian surfaces. I've added snow and charcoal, with  templates.  Though non-Lambertian reflectance is a complicated, many-parameter property, depending on angles of incidence, reflection, not to mention polarization, it would be good to know if there is a rough measure in common use, a single number, to characterize the degree of deviation from Lambertian emission or reflectance.  Does anyone know of one? CharlesHBennett (talk) 07:43, 8 November 2017 (UTC)
thread_title:  More examples,  a simple measure of deviation from Lambert's law 

message: An example might be placed in the See Also links. Here is one suggestion: https://en.wikipedia.org/wiki/Deep_lambertian_networks which is used for image recognition as part of mapping. CvS 18:08, 3 July 2018 (UTC)
thread_title:  See Also Links 

message: urila  — Preceding unsigned comment added by Urila (talk • contribs) 02:27, 31 July 2018 (UTC)
replies: text: Wikipedia does not publish original research. --Srleffler (talk) 03:26, 30 October 2018 (UTC)
thread_title:  Backscattering yields a uniform full moon image. 

message: It would seem that a simple picture showing how incoming photons are reflected would be very useful, especially given the confusion seen in the notes above.  The easiest picture to show is the one in here that has been removed, I would suggest it be reinstated.  It shows a 2D right slice of the hemisphere above the Lambertian surface, with incoming photons from an arbitrary direction heading to a small area on the surface, and the relative distribution of photons reflected from it.  The lengths of the red arrows represent the proportion of photons (or Joules or Watts) reflected in each direction, according to Lambert's Cosine Law.  I find it useful, nay necessary, to visualize what is happening. Jack Hogan (talk) 07:51, 3 March 2023 (UTC)
thread_title: Pictures = 1000 Words

message: The description in the introduction is currently about photometry, but such surfaces star in radiometry, so it would be useful to start the article talking about flux density (E), which can be (Watts, lm or photons/s) per unit area - and can be called irradiance or illuminance - and take it from there.
The symbol for radiance/luminance is L.
Also, reflectance is not isotropic by definition, a microscopic Lambertian surface reflects just in the hemisphere above it. Jack Hogan (talk) 08:26, 3 March 2023 (UTC)
thread_title:  Lambertian Reflectance Is Used in Many Fields