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Let us consider the situation where only we moved the camera.

The object and the light source remain in the same place. Thus the angle of the incident light is the same for every point.I have captured one image where 5 arrows marked as specular reflection (from one particular camera position) and rest of the portion could be treated as diffuse: enter image description here But when I change the camera position (but object position and light source position remain as it is) the specular portion becomes diffuse and vice versa in the image of that object. But we know that diffuse reflection is independent of camera ( only depends on normal and direction of incident light). My question is how diffuse reflection is failing to show same intensity from all direction?

I want to share my understandings why this is happening, don't know my postulation is right or not:

  1. When we see diffuse and specular reflection is happening on one single object then the definition of diffuse reflection isn't going to work but when only diffuse reflection is happening on one single object then definition of diffuse reflection has properly worked.

  2. When we see diffuse and specular reflection is happening on one single object then diffuse reflection is also dependent of camera like specular reflection.Then when the angle between viewing direction (camera) and reflection is increasing then specular portion gradually converging to diffuse.

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    \$\begingroup\$ Can you clarify how your question relates to game development? I suspect, if you try to implement this in your own game rendering code, you will find the answer becomes obvious, since you can toggle specular reflection on and off to see the difference it makes. \$\endgroup\$
    – DMGregory
    Dec 29, 2021 at 20:46
  • \$\begingroup\$ @DMGregory one thing tell specular intensity why not depends on angle between n and l? I mean suppose Case1: When angle between n and l are small. Then angle between r and n is also small. V is coincide with r. Case2: When angle between n and l are big. Then angle between r and n is also big. V is coincide with r. Case 1 has greater intensity than case 2? \$\endgroup\$
    – S. M.
    Dec 31, 2021 at 12:36
  • \$\begingroup\$ Once again, this is a question about the physics of light, not a question about developing games. Please post here when you have a question about developing games. \$\endgroup\$
    – DMGregory
    Dec 31, 2021 at 12:53

1 Answer 1

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First of all,

the angle of the incident light is the same for every point

The angle between the incident light and the normal is different. Even if the incident light comes form the same direction※, the normal is different for each point of the surface of the object.

※: In reality this would be an approximation admisible for a light source very far away. For computer graphics, we are talking about a directional light source instead of a spot or cone light source.


But when I change the camera position (but object position and light source position remain as it is) the specular portion becomes diffuse and vice versa in the image of that object.

I believe we have gone over the fact that the position of the specular highlights depends on the position of the camera, the object and the light source. Thus, moving the camera is sufficient to move the specular highlights. The last paragraphs of a prior answer are about that.


But we know that diffuse reflection is independent of camera ( only depends on normal and direction of incident light).

Right. What we are trying to model with diffuse reflection is light being scattered in every direction. This is why it is independent of the direction of the camera.

Every point with incident light has both specular reflections and diffuse reflections. There are no portions of the object that are getting specular reflection but no diffuse reflection, nor viceversa. There are no portions of the object that are inherently specular or diffuse. In fact, specular reflection and diffuse reflection happen together.

Instead, for a single point on the surface of the object with incident light:

  • Specular reflection dominates over the diffuse reflection.
  • Diffuse reflection dominates over the specular reflection.
  • Or somewhere in the middle. Because the decay of the specular highlight is not abrut.

The specular reflection is negligible everywhere except on the specular highlight. I mean, the specular reflection decays away from the specular highlight until it is virtually zero.

In consequence where you see an specular highlight, the specular reflection is much more intense than the diffuse reflection, so we do not perceive the diffuse reflection.


I want to point out that "specular highlight" and "specular reflection" are not the same thing. In the prior answer, on first paragraph, I took the care to define "specular highlight":

The specular highlight (…) is an specular reflection of the light source.

Emphasis added on "of the light source".

Other objects aside from the light source also appear in the specular reflection. But only the reflections of the light source are specular highlights.


While the way specular reflection decays attempts to model roughness. The diffuse reflection attempts to model scattering.

I quote form Wikipedia's article on Diffuse reflection:

Diffuse reflection from solids is generally not due to surface roughness. A flat surface is indeed required to give specular reflection, but it does not prevent diffuse reflection.

A light ray travelling from one medium to another scattering inside the object

A light ray travelling from one medium to another scattering inside the object. This picture is taken from the article THE PBR GUIDE BY ALLEGORITHMIC - PART 1.

So, some of the light gets refracted and scattered back out the surface. And the distribution of the direction and intensity of that light is close to uniform. So we model it as being the same from every direction.


If we consider an object with a mirror finish (i.e an object with very little micro-imperfections, i.e an object with very low roughness, i.e. a very smooth object), the object will act as a mirror. And in a mirror we can see sharp specular reflections of the environment (including objects other than the light source).

I continue quoting from Wikipedia:

A piece of highly polished white marble remains white; no amount of polishing will turn it into a mirror. Polishing produces some specular reflection, but the remaining light continues to be diffusely reflected.

However, even mirrors have some diffuse reflection. It just happens to be very, very, hard to notice. But we can make it more evident by making repeated reflections, by reflecting a mirror in a mirror.

Mirror in mirror reflection

Picture taken from the video What Color is a Mirror by VSause.

We observe that the repeated reflections are less bright, and have a tint. Usually a green tint, due to the chemistry of the glass.

By the way, the the video includes a picture that suggest diffuse reflection happens due to roughness. So, I'll reiterate that we attempt to model scattering as diffuse reflection.


  1. When we see diffuse and specular reflection is happening on one single object then the definition of diffuse reflection isn't going to work but when only diffuse reflection is happening on one single object then definition of diffuse reflection has properly worked.

The specular reflection, diffuse reflection (and ambient and any other components according to the lighting model) contribute to the final shading of every pixel.

But away from the specular highlight, the contribution of the specular reflection is negligible. Leaving only the contribution of the diffuse reflection (and ambient reflections or any other component according to the lighting model) noticiable.

On the other hand, on the specular highlight, the specular reflection is very intense. As a consequence we do not notice the contributions of the diffuse reflection (and ambient reflection and so on).

That is, the definition of diffuse reflection holds. But it does not account for how its contributions is combined with the others. For any lighting model, the contributions of specular reflection, diffuse reflection, (and any other component) must be combined to make up the final shading of the pixel. This is usually a weighted sum.


  1. When we see diffuse and specular reflection is happening on one single object then diffuse reflection is also dependent of camera like specular reflection.Then when the angle between viewing direction (camera) and reflection is increasing then specular portion gradually converging to diffuse.

The final result of combining the reflection is dependent on the camera. Because the specular reflection is dependent on the camera. But the diffuse reflection, considered on its own, isn't dependent on the camera.

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  • \$\begingroup\$ when move the camera then specular portion becomes diffuse and diffuse portion becomes specular, so how can we say diffuse is independent of camera? \$\endgroup\$
    – S. M.
    Dec 30, 2021 at 10:39
  • \$\begingroup\$ "But it does not account for how its contributions is combined with the others "---- what do you mean? \$\endgroup\$
    – S. M.
    Dec 30, 2021 at 12:24
  • \$\begingroup\$ in your answer everything looks good. But one thing doesn't clarified, when diffuse reflection is happening, it's intensity same from all viewing direction, but when camera is moving how diffuse portion (previously said) intensity changing?? What is the logic behind this. \$\endgroup\$
    – S. M.
    Dec 30, 2021 at 13:28
  • \$\begingroup\$ (continue) so how the diffuse reflection definition isn't violating? \$\endgroup\$
    – S. M.
    Dec 30, 2021 at 13:39
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    \$\begingroup\$ @Ponting For a given point, the intensity of the diffuse reflection is the same regardless of the viewing direction. But - for the same point - the intensity of the specular reflection changes as the viewing direction changes. So, from some viewing directions the specular reflection - for that same point - is greater than the diffuse reflection; and from some other viewing direction the specular reflection - for that same point - is lesser than the diffuse reflection. The light we see from that same point is a combination of those two (and any other components). \$\endgroup\$
    – Theraot
    Dec 30, 2021 at 13:45

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