1
$\begingroup$

I'm coming from learning art, and I'm trying to figure out the way light perception works.

If I got it right, in order to see the surface, it has to reflect rays of light in viewer's eyes, but the thing is: what if the surface is facing the light source, but it's tilted away from viewer's eyes?

For example we have two surfaces ("A" and "B"), "A" is facing the source light at 90 degrees and another ("B") is facing viewer's eyes at 90 degrees angle, while still facing the light source, although its value would be perceived darker by a viewer. How come we see the "A" surface as a lighter surface rather than "B"?

Light perception

Given the picture, even if the surface "A" gets hit by more rays than "B" does, it looks like surface "B" is more capable of reflecting more rays in viewer's eye than "A" does. So why one still perceives "A" as a lighter surface?

$\endgroup$
1
$\begingroup$

It depends if the surfaces are producing specular or diffuse reflections. Specular reflections are produced by polished metals, and this is the case that gives us the rule about angle of incidence equaling angle of reflection. Diffuse reflectors scatter the light in all directions, like most of the surfaces we see around us every day.

But certainly the perceived brightness of a surface will depend on the angle we view it from. If you're currently indoors in a room with painted walls, depending how the room is lighted you might notice that the part of a wall facing directly toward you looks brighter than the parts you're viewing at an angle, for example.

$\endgroup$
0
$\begingroup$

In the surface is a mirror then you will see the mirror image of the light source but only for suitable angles. If the surface is a diffusor then you will see an intensity that ideally varies like Lambert's cosine law. https://en.m.wikipedia.org/wiki/Lambert%27s_cosine_law In general the behaviour is a combination of the two.

$\endgroup$
0
$\begingroup$

Here is a more QM explanation:

When a photon interacts with an atom, three things can happen:

  1. elastic scattering, the photon keeps its energy and changes angle

  2. inelasstic scattering, the photon gives part of its energy to the atom and changes angle

  3. absorption, the photon gives all its energy to the atom and electron

Now there are visible and non-visible wavelength photons, but in your case, only the visible ones are more important, so I am going to talk about those (except for absorption and re-emission in which case the absorbed non-visible light might be re-emitted partly as visible wavelength light).

In your case, the visible light we see from an object is mostly from its surface, there are two cases for visible light coming from the surface of the object:

  1. specular reflection (mirror or shiny surface), this is elastic scattering

  2. diffuse reflection, (not mirror, faded), this is absorption, and re-emission

Now it is very important to differentiate between specular reflection and diffuse reflection, because:

  1. in the case of specular reflection, a mirror or a shiny metal, you will always see the mirror facing towards you lighter, then the mirror facing a little bit away from you

  2. in the case of diffuse reflection, like a wall, you might either see the wall facing towards you lighter or darker then the wall facing a little bit away from you, this is the case you are referring to

Now you are saying that it is possible that a wall that is facing towards you is appearing darker then a wall that is facing a little bit away from you.

Yes it is possible, because walls are typically creating diffuse reflection. In the case of diffuse reflection, most of the visible wavelength photons are absorbed by the surface atoms in the lattice, and then re-emitted. Now in this case, the photons do not keep their original angle of incidence, and do not keep their relative angle nor their relative phases nor their energies. That is why in this case you do not get a mirror image, but a faded reflection.

Now the reason why the brightness of the wall facing towards you might be darker then the wall facing a little bit away from you is because even though more light should be reflected towards you from the wall facing towards you, the photons on that wall will have diffuse reflection, in which case the angle of reflection is random. It has nothing to do with the angle of incidence.

Now there is one more thing in the case of diffuse reflection. With absorption and re-emission, it happens, that even non-visible wavelength photons get absorbed by the surface and then they are re-emitted in multiple steps, in different wavelength photons, some of which will be visible light. That is why that a wall facing away from you (but towards the light) might absorb more non-visible photons by the surface, and then those photons get re-emitted in multiple steps and some of them as visible light creating an effect as if the wall was brighter.

In this case, like a wall, or a material that has a lattice structure, it is basically depending on the structure of the lattice atoms, how they reflect the photons, in random ways, but these random ways might even create certain phenomenons, where the wall facing you will appear darker.

You know, there are those materials, where from a certain angle, you can see a picture, but not from a certain other angle. Now i just saw a car painting, that was made so, that from different angles, the car's certain parts appeared in different colors, and if you moved, the car appeared to change colors. These are just examples of the same thing.

You can create certain materials, with certain surfaces, where the surface, viewed from a certain angle, will reflect light so that it will appear in a certain color, and then from another angle in a certain other color.

Now in your case these colors could be a darker color for the wall facing you, and a brighter color facing a little bit away from you.

But normally, in case of a wall, that you are talking about, yes it is possible even without these special surface materials, to have an effect, where the diffuse reflection will make the wall facing you appear darker, and the wall facing a little bit away from you brighter, just because the angle of reflection is random and has nothing to do with the angle of incidence.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.