Return to Answer

You must distinguish between reflection versus absorbtion-and-emission.

When incident light $I$ hits a surface, some portion is reflected $R$ and some is absorbed $A$:

$$I=R+A$$

• The reflected portion $R$ of the incoming light is spreading out depending on the surface roughness and texture. A rough surface reflects scattered light while a very plane and polished surface is shining, sending back the light as it came (a mirror for instance).

• The absorbed portion $A$ of the incoming light is partly kept inside the material by being converted into thermal energy $Q$ heating it up, and is partly re-emitted $E$: $$A=Q+E$$ The re-emitted light $E$ is what you are talking about. A material might absorb all wavelengths but only re-emits certain wavelengths depending on material and temperature (the emitted light is not only one specific wavelength but a fluent spectrum, where a part of the spectrum is more intense than others - that part at highest intensity drowns out other less intense parts and mainly determines the colour, but the resulting colour is a mix of them all).

The "outgoing" light $O$ that comes back from the surface, which is the light you see, is then a mix of what is reflected $R$ and what is emitted $E$:

$$O=R+E$$

Which-ever of these two effects is stronger will influence the appearance of the material surface the most. E.g. a completely black surface, that doesn't re-emit any light, will still look white (or shiny) if it reflects a lot by being finely polished.

When only a certain type of light (e.g. only light of red wavelengths) hits a surface, the reflection is only red (rather than being a mix of all, meaning white), and therefore this red reflection might have a larger visual impact. The object still at the same time absorbs some of the red light and re-emits some other colours of light, such as the green leaf does. But I am thinking that this green re-emission will now be mixed with a lot of red reflection and the resulting appearance will be redish.

You must distinguish between reflection versus absorbtion-and-emission.

When incident light $I$ hits a surface, some portion is reflected $R$ and some is absorbed $A$:

$$I=R+A$$

• The reflected portion $R$ of the incoming light is spreading out depending on the surface roughness and texture. A rough surface reflects scattered light while a very plane and polished surface is shining, sending back the light as it came (a mirror for instance).

• The absorbed portion $A$ of the incoming light is partly kept inside the material by being converted into thermal energy $Q$ heating it up, and is partly re-emitted $E$: $$A=Q+E$$ The re-emitted light $E$ is what you are talking about. A material might absorb all wavelengths but only re-emits certain wavelengths depending on material and temperature (the emitted light is not only one specific wavelength but a fluent spectrum, where a part of the spectrum is more intense than others - that part at highest intensity drowns out other less intense parts and mainly determines the colour, but the resulting colour is a mix of them all).

The "outgoing" light $O$ that comes back from the surface and is the light you see, is then a mix of what is reflected $R$ and what is emitted $E$:

$$O=R+E$$

Which-ever of these two effects is stronger will influence the appearance of the material surface the most. E.g. a completely black surface, that doesn't re-emit any light, will still look white (or shiny) if it reflects a lot by being finely polished.

When only a certain type of light (e.g. only light of red wavelengths) hits a surface, the reflection is only red (rather than being a mix of all, meaning white), and therefore this red reflection might have a larger visual impact. The object still at the same time absorbs some of the red light and re-emits some other colours of light, such as the green leaf does. But I am thinking that this green re-emission will now be mixed with a lot of red reflection and the resulting appearance will be redish.

You must distinguish between reflection versus absorbtion-and-emission.

When incident light $I$ hits a surface, some portion is reflected $R$ and some is absorbed $A$:

$$I=R+A$$

• The reflected portion $R$ of the incoming light is spreading out depending on the surface roughness and texture. A rough surface reflects scattered light while a very plane and polished surface is shining, sending back the light as it came (a mirror for instance).

• The absorbed portion $A$ of the incoming light is partly kept inside the material by being converted into thermal energy $Q$ heating it up, and is partly re-emitted $E$: $$A=Q+E$$ The re-emitted light $E$ is what you are talking about. A material might absorb all wavelengths but only re-emits certain wavelengths depending on material and temperature (the emitted light is not only one specific wavelength but a fluent spectrum, where a part of the spectrum is more intense than others - that part at highest intensity drowns out other less intense parts and mainly determines the colour, but the resulting colour is a mix of them all).

The "outgoing" light $O$ that comes back from the surface, which is the light you see, is then a mix of what is reflected $R$ and what is emitted $E$:

$$O=R+E$$

Which-ever of this two effects is stronger will influence the appearance of the material surface the most. E.g. a completely black surface, that doesn't re-emit any light, will still look white (or shiny) if it reflects a lot by being finely polished.

When only a certain type of light (e.g. only light of red wavelengths) hits a surface, the reflection is only red (rather than being a mix of all, meaning white), and therefore this red reflection might have a larger visual impact. The object still at the same time absorbs some of the red light and re-emits some other colours of light, such as the green leaf does. But I am thinking that this green re-emission will now be mixed with a lot of red reflection and the resulting appearance will be redish.

You must distinguish between reflection versus absorbtion-and-emission.

When incident light $I$ hits a surface, some portion is reflected $R$ and some is absorbed $A$:

$$I=R+A$$

• The reflected portion $R$ of the incoming light is spreading out depending on the surface roughness and texture. A rough surface reflects scattered light while a very plane and polished surface is shining, sending back the light as it came (a mirror for instance).

• The absorbed portion $A$ of the incoming light is partly kept inside the material by being converted into thermal energy $Q$ heating it up, and is partly re-emitted $E$: $$A=Q+E$$ The re-emitted light $E$ is what you are talking about. A material might absorb all wavelengths but only re-emits certain wavelengths depending on material and temperature (the emitted light is not only one specific wavelength but a fluent spectrum, where a part of the spectrum is more intense than others - that part at highest intensity drowns out other less intense parts and mainly determines the colour, but the resulting colour is a mix of them all).

The "outgoing" light $O$ that comes back from the surface, which is the light you see, is then a mix of what is reflected $R$ and what is emitted $E$:

$$O=R+E$$

Which-ever of these two effects is stronger will influence the appearance of the material surface the most. E.g. a completely black surface, that doesn't re-emit any light, will still look white (or shiny) if it reflects a lot by being finely polished.

When only a certain type of light (e.g. only light of red wavelengths) hits a surface, the reflection is only red (rather than being a mix of all, meaning white), and therefore this red reflection might have a larger visual impact. The object still at the same time absorbs some of the red light and re-emits some other colours of light, such as the green leaf does. But I am thinking that this green re-emission will now be mixed with a lot of red reflection and the resulting appearance will be redish.

You must distinguish between reflection versus absorbtion-and-emission.

When incident light $I$ hits a surface, some portion is reflected $R$ and some is absorbed $A$:

$$I=R+A$$

• The reflected portion $R$ of the incoming light is spreading out depending on the surface roughness and texture. A rough surface reflects scattered light while a very plane and polished surface is shining, sending back the light as it came (a mirror for instance).

• The absorbed portion $A$ of the incoming light is partly kept inside the material by being converted into thermal energy $Q$ heating it up, and is partly re-emitted $E$: $$A=Q+E$$ The re-emitted light $E$ is what you are talking about. A material might absorb all wavelengths but only re-emits certain wavelengths depending on material and temperature (the emitted light is a fluent spectrum, where a certain part of the spectrum is more intense - that part at highest intensity drowns out other less intense parts and determines the colour).

The "outgoing" light $O$, which you see, is then a mix of what is reflected and what is emitted:

$$O=R+E$$

Which-ever of this two effects is stronger will influence the appearance of the material surface. E.g. a completely black surface, that doesn't re-emit any light, will still look white (or shiny) if it reflects a lot by being finely polished.

When only a certain type of light (e.g. only light of red wavelengths) hits a surface, the reflection is only red (rather than being a mix of all, meaning white), and therefore this red reflection might have a larger visual impact. The object still at the same time absorbs some of the red light and re-emits some other colours of light, such as the green leaf does. But I am thinking that this green re-emission will now be mixed with a lot of red reflection and the resulting appearance will be redish.

You must distinguish between reflection versus absorbtion-and-emission.

When incident light $I$ hits a surface, some portion is reflected $R$ and some is absorbed $A$:

$$I=R+A$$

• The reflected portion $R$ of the incoming light is spreading out depending on the surface roughness and texture. A rough surface reflects scattered light while a very plane and polished surface is shining, sending back the light as it came (a mirror for instance).

• The absorbed portion $A$ of the incoming light is partly kept inside the material by being converted into thermal energy $Q$ heating it up, and is partly re-emitted $E$: $$A=Q+E$$ The re-emitted light $E$ is what you are talking about. A material might absorb all wavelengths but only re-emits certain wavelengths depending on material and temperature (the emitted light is not only one specific wavelength but a fluent spectrum, where a part of the spectrum is more intense than others - that part at highest intensity drowns out other less intense parts and mainly determines the colour, but the resulting colour is a mix of them all).

The "outgoing" light $O$ that comes back from the surface, which is the light you see, is then a mix of what is reflected $R$ and what is emitted $E$:

$$O=R+E$$

Which-ever of this two effects is stronger will influence the appearance of the material surface the most. E.g. a completely black surface, that doesn't re-emit any light, will still look white (or shiny) if it reflects a lot by being finely polished.

When only a certain type of light (e.g. only light of red wavelengths) hits a surface, the reflection is only red (rather than being a mix of all, meaning white), and therefore this red reflection might have a larger visual impact. The object still at the same time absorbs some of the red light and re-emits some other colours of light, such as the green leaf does. But I am thinking that this green re-emission will now be mixed with a lot of red reflection and the resulting appearance will be redish.