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Assuming a light source, emitting white light,a red colored diffuse sphere lit up by the light, and a white plane below the sphere acting as the 'floor', we find the diffuse reflection of the sphere on the floor to be reddish in color, just like the sphere itself

Why does it have the exact same color as the sphere? The emitted light was white in color, so when the photons of the light source, collide and reflect from the sphere, should they not still be white, causing the diffuse reflection to instead be white? What causes the photons to take on a reddish hue?

In the case, of a black sphere as opposed to a red sphere, can we expect the same thing to happen? Will the diffuse reflection be grey-black in color as well? If, on the spot where the reflection is cast,a light grey soft shadow, lighter than the diffuse reflection were to exist, would the reflection of the sphere, 'overpower' that of the shadow, making the region darker as a result?

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When you shine white light at a colored object, the light that scatters off it is missing some of its wavelengths i.e., it is tinted: this is why it appears colored to our eyes.

So now you have colored light radiating away from that object, and when it strikes a nearby object and is scattered off, the scattered light is of course still missing the same wavelengths that got absorbed by the first object.

This tints the second object to a color similar to that of the first object.

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  • $\begingroup$ I see. So would light shining on a black matte sphere not cast any such diffuse reflection because all the light is absorbed off by the sphere? $\endgroup$
    – Hash
    Sep 1 '21 at 17:58
  • $\begingroup$ Yes, or nearly so- even the blackest matte paint still reflects some light. $\endgroup$ Sep 1 '21 at 18:01
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Surfaces do not reflect all wavelengths equally well. Objects that reflect red wavelengths better.. we simply call those “red objects”. Apples reflect the red wavelengths more and absorb the blue and yellow wavelengths more. Lemons reflect yellow wavelengths better and absorb more of the blue and red wavelengths. If objects reflected all wavelengths the same proportion, then there wouldn’t be colored objects.

This is different than how reflective it is overall. Something very reflective might reflect most of the light that hits it. Something dark and dull absorbs most light. If something is dark red and not shiny and not reflective, then we know two things about it: 1. It doesn’t reflect very much of the light that hits it, overall. Idk maybe it absorbs 80% of the light that hits it. 2. Whatever portion of light it does reflect, well that proportion is a bit higher for red wavelengths. Idk maybe it absorbs 60% of the red light that hits it and 95% of the blue and 95% of the yellow.

I’m not certain what you’re asking next. What we see will be the sum of the light that hits the surface. A grey shadow is only a shadow compared to some region next to it where there is light. You don’t really “add” a shadow to the situation. You add whatever light.

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  • $\begingroup$ Using what you have said, how do we explain something like a shiny black sphere? If its shiny, it should reflect most light, but if its black, it should absorb all the light that hits it. Would these two not contradict each other? Yet we find many shiny black objects IRL $\endgroup$
    – Hash
    Sep 1 '21 at 18:00
  • $\begingroup$ Yes that means two things: 1. they reflect a wide variety of wavelengths except weighted toward very low wavelengths. Thats why they can seem almost purplish, depending on exact wavelength profile. If they reflect a variety of wavelengths without being skewed towards the low wavelength they will be more white. 2. But you can have shiny of any color because of second effect $\endgroup$
    – Al Brown
    Sep 1 '21 at 18:59
  • $\begingroup$ 2. But also, shiny objects sometimes have a layered effect, where there is a mirror finish that can reflect bright light, almost white or like sun but not diffuse, only at exact angles, and underneath that there is a color (if pure black then no additional light from that layer). I.e. shiny black has no diffuse reflection only absorbtion for diffuse, and then as far as reflection has only incident reflection. Even though the nondiffuse light being reflected at you is white, you dont cognize and perceive it as white but as “shine” $\endgroup$
    – Al Brown
    Sep 1 '21 at 18:59
  • $\begingroup$ Shine vs color has to do with the nondiffuse aspect as I understand it. As you rotate it you get beams, but because they vary etc, you mind calls that shine not color. Usually technically those rays are white. That said you can have one color and a (nondiffuse) shine of a different color added. Like something dark blue with a reddish shine to it, less common but exists perhaps youve seen $\endgroup$
    – Al Brown
    Sep 1 '21 at 19:01
  • $\begingroup$ So essentially for a shiny black object, light can fall into either of two cases 1) Passes through mirror finish, gets absorbed by black diffuse sphere underneath appearing as black to us 2) Due to fresnel effect, viewing the sphere from grazing angle makes us see reflected light rays that couldn't quite pass through the mirror finish, but rather just reflected off of the finish, appearing shiny and white to us? $\endgroup$
    – Hash
    Sep 2 '21 at 6:25

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