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If we use yellow light on a blue paper, what will the paper look like?

Basically I have two different conflicting ideas in my head:

• If the light has only one specific wavelength then only this specific wavelength can be reflected at all. So if I use light of any color on the object, the object will always appear in that specific color.




• But on the other hand a colored object has that color because it absorbs light of any other wavelength and only reflects the part which gives it the respective color. This means that if I use any light on an object that has a different color, the object must appear black.

Probably neither of those are true and it's different? For example, I found an interesting image here which shows that red light lets a green paprika appear red. Can one explain what is happening exactly?

Your first statement is mostly true, an object can only reflect light of the wavelengths it is illuminated with. We can have fluorescence though, where light of a given wavelength or range of wavelengths is absorbed and light of a different wavelength is emitted (this is not actually reflection). In the absence of fluorescence, the object will reflect only (part of the) colors it is illuminated with.

You second statement is also mostly true, an object has the color it has because other wavelengths get absorbed. Your conclusion that an object must appear black when it is illuminated by light of a different color than that of the object, is not necessarily true, though it depends a bit on how you define the color of an object (if by color you mean monochromatic color, that would be true). Most of the light we see consists of a mixture of different wavelengths. Different mixtures may actually look the same, because of how the human visual system works. Colors that look different to us, may, and often do, contain common wavelengths.

Finally, if in your linked image you look at the green paprika under white light (which contains wavelengths over the full visible range), you see that it looks mostly green, but two shiny parts look white. In the green part we receive diffusely reflected light, while in the white parts we see a specular reflection, like a mirror. Here the incoming light makes an angle with the surface under which a lot of light gets reflected; in all other direction we lose all light that is not green. If we watch the paprika under a red light, we actually see that it looks mostly black, or at least pretty dark, except in those zones where the reflection is specular and is essentially the same color as the incoming light. This makes it look red.

As for the first question, if we think of colors spanning a broad range of wavelengths, yellow roughly corresponds to middle and low wavelengths, while blue corresponds to high wavelengths. That means that there is little overlap in wavelengths between the incoming yellow light and the light that the paper can reflect, so it will look very dark.

If by "yellow light", you mean a light with wavelength between 560 and 590 nm, and by "blue paper", you mean paper that reflects light only if its wavelength is between 450 and 490 nm, then yes, if the only illumination for a piece of blue paper is yellow light, it will appear black. But "yellow light" is generally light that has a mixture of wavelengths such that the average wavelength is between 560 and 590 nm, not light such that every photon has a wavelength in that range. That's why red and green light together looks yellow: when our eyes detect both red and green light, our brain perceives it as yellow.

Similarly, most "blue" paper does not reflect just blue light. Rather, it reflects light most strongly in the "blue" range of wavelengths, and/or reflects light whose wavelengths average out to be in that range.

So when mostly-yellow-but-also-has-a-little-bit-of-other-wavelengths light hits mostly-reflects-blue-but-also-weakly-reflects-other-colors paper, some light will be reflected. What color it looks like will depend on the exact composition of the light and the exact reflective properties of the paper. In the example you gave, the pepper reflects a significant amount of red light, but still looks darker than the red parts of the playing card. In green light, the pepper is bright green, while the red parts of the playing card are almost black. This implies that the pepper reflects green light more than it reflects red light, but still reflects some red light, while the red parts of the playing card reflects very little green light. This makes sense: the pepper is a natural object, while the playing card has an artificial dye specifically designed to be a particular color.

One of the things I thought about is how we perceive colours, especially when we have coloured objects. Whenever white light is shone on a red object, for example, the wavelengths that are not red are absorbed and red is transmitted to the eye. By shining yellow monochromatic light on blue paper, I assume that the colour seen will be black. This is because the wavelength that should be transmitted back from the blue paper should be around 400nm (wavelength of violet-blue) whereas yellow is closer to 700nm (wavelength of red-yellow). Since the wavelength of the yellow light is different in comparison to blue's it will be absorbed and black will be transmitted back to the eye due to the monochromatic nature of the light.

The two ideas stated are, in fact, correct on their own. However the conclusion that "This means that if I use any light on an object that has a different colour, the object must appear black." is very specific to monochromatic lights (a single wavelength emitting source) that do not have the same colour wavelength as the colour of the object. For more information, you may be interested in looking into colour subtraction as it explains more about the filtering process.