If an object absorbs a light, it emits back the same wavelength (if electrons get excited by a particular frequency, surely they must emit back that frequency or several lower frequencies). We say the object has the colour which it does not absorb (or which it reflects). Isn't this contradictory?
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2 Answers
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There appears to be a confusion here regarding the idea of absorption. In one sense, every photon which interacts with an object or medium is “absorbed”. Some of that absorbed energy is re-emitted, generating the reflected (or transmitted) light, usually of the same color. But some of the energy is not re-emitted; it is lost, converted into another form such as heat. This lost energy is often called the “absorbed” light, in the second sense that you referred to. So indeed, the color of an object is due to the light energy that was not lost, or absorbed, in the object.
To summarize, in the “energy is lost” sense, you are right, the light is not absorbed in the kind of reflection you would normally think of. But in the sense that every photon which reflects from an object is absorbed and subsequently re-emitted, then the light is “absorbed” during the interaction. This is the source of the confusion: there are two concepts of “absorption”. One refers to absorption as a necessary step of the reflection interaction, and the other refers to absorption as a loss of energy into the object.
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1$\begingroup$ What role does actual reflection play in this?( Reflection which allows me to see another object in that object.) That doesn't require absorbtion, right? $\endgroup$– triple ACommented Feb 20, 2018 at 14:18
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1$\begingroup$ @tripleA I edited my answer to try to clarify. Hopefully that helps! $\endgroup$– GilbertCommented Feb 20, 2018 at 14:28
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In short, the fact that the absorbed photons can be re-emitted at lower frequency removes the "contradiction". Thermal channels exists other than radiative ones.
Moreover, these photons are randomly emitted in all directions.
As such, the light reflected to the eye/detector will have a specific spectrum that differs from that of the illuminating source.
Eye and perception are by far complicated as, among other things, we tend to operate correctios for tint and white
--Think of a white wall illiminated by a candle. Only under very unnatural conditions we could say that the wall is yellow because that's the colour it reflect to us, most likely we would compensate and still perceive it as white --
but the above suffice to explain why not all wavelengths are reflected/reemitted back the same, unless the body is perfectly white (indeed not absorbing).
In short, while in isolated atoms is true that absorption and emission energetically coincide, this is not true when different relaxation channels are available, such as in molecules and denser states. Note that even in case of isolated atoms, their "complementary colour", i.e. the atomic absorption lines, can be determined, right because emission occurs in all directions.
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$\begingroup$ So when I see a black body, every frequency of light gets transformed into heat dissipations? $\endgroup$– triple ACommented Feb 20, 2018 at 14:24
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$\begingroup$ @Triple A. A black body emits by virtue of itself. It is not illuminated. In this case the colour of the black body is still determined by the light spectrum impinging to your eyes. If you shine light at it, all lights gets absorbed by definition and indeed redistributed as heat. Eventually thus change the T (colour) of the black body. $\endgroup$ Commented Feb 20, 2018 at 15:17
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$\begingroup$ Ohh. And for glass like bodies, they just refuse to absorb energy? What properties make them do so?(Surely they must also have states to which they can absorb energy) $\endgroup$– triple ACommented Feb 20, 2018 at 15:26
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$\begingroup$ This is why I try to keep it simple.... First of all not all electromagnetic radiation are associated with colour. Glass absorbs but it does in the UV. IR is also absorbed but is directly associated to thermal effect. Indeed for transparent bodies their colour is generally the most transmitted one. You can have thin film of materials that if the layering fits can show different colour when seen in reflection or in transmission. Obviously mirrors and opaque bodies are extreme case of this. A green piece of a common bottle absorb in the red, part of which is lost as heat part of it can be .. $\endgroup$ Commented Feb 20, 2018 at 15:35
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$\begingroup$ Thanks Alchimista. Yes it helped me a lot.And I did like your answer but since I am new they don't count my upvotes publically. $\endgroup$– triple ACommented Feb 20, 2018 at 17:10