If I have, for example, a white light, consisting of many wavelengths of the whole visible spectrum (from 400nm to 700nm) focusing on a surface, and the surface reflects only the frequency near to 600nm, so I will see the object as yellow (600nm).

If I have, for example, a pure yellow light, with a single wavelength of 600nm and it focuses on this same surface, I will still see the surface as yellow.

Until this part, that is ok.

But now, if I have a white light, but this time it is not consisted by the whole visible spectrum, but only by red (700nm), green (550nm) and blue (400nm) wavelengths. Our brain still understand this as being white light. If this light is reflected on the same yellow surface, it will reflect only the frequencies near to 600nm, but this time there is no frequency near to 600nm. So there is no reflected light, and I will see the surface as black? The first white light will show me a yellow surface, and the second white light will show me a black surface?


Thank you both for the clarification. So that is what I imagined. The main reason for this question is that some time ago I was working in a 3D game for fun, and I made a simple shader from tutorials I saw. The way I calculated light reflection color was simply making a element-wise multiplication between two vectors, they represented the light color and the surface reflection color. So, for example, if I had a yellow light (in RGB, with a range of 0 to 1, it would be a vector yv=(1,1,0) ), and a surface with a green color gv=(0,1,0), the reflected light color would be green too, as (1,1,0)*(0,1,0)=(0,1,0), this supposing that the surface normal vector and the light ray vector are parallel to each other.

But in the real world, that yellow light represented by yv, would be a light with two wavelengths (red and green). But if it was a pure yellow light, with a narrow band (like a sodium lamp), the reflected light on the green surface would be almost null, and the surface would appear as black instead of green.

  • $\begingroup$ The crucial concept that's missing here is the color rendering index of lighting sources. The narrow and combination in your second example is white, but it has a poor CRI, and objects will tend to look unnatural under that illumination. $\endgroup$ Commented Dec 11, 2019 at 1:08

2 Answers 2


I feel this is a little too short for an answer, but "yes." You are correct.

Now this goes against intuition, so it helps to remember two details:

  • "White light" is something that doesn't actually exist in physics. Our concept of white light is a heuristic based on the astonishing things our brain does with color. In your scenarios, the brain is being fooled into thinking both lights are the same because it simply lacks the capability to realize they aren't.
  • In most cases, objects have very wide ranges of reflection. They're rarely a nice clean notch at 600nm, so you usually get some reasonable behaviors. But if you did have something that reflected at 600nm and only right around that frequency, it would indeed be black under the second light.
  • Having seen effects like this (involving sodium lamps and airhead wrappers), I can say that when you do have an unexpected color effect due to these strange lighting conditions, your brain gets really confused. It's a very uneasy feeling looking at something that you know has a bright range of colors and perceiving it as grey-scale because the lighting is tricking your visual centers.

For an excellent example of this watch the video from Smarter Every Day Laser vs. Balloons. They're having fun trying to figure out which balloons are going to pop faster or slower based on color. But at around 2:50 they get confused. They have white balloons in their hand, but because of the laser safety goggles they are wearing (which have very exacting transmittance), they didn't realize they were white.


The other answer in detail describes white light and some part of reflection, so I will only answer your question's not so intuitive part.

You are not defining it, but I assume you are talking about a surface, that only emits yellow wavelength photons. And you are not saying it, but I will assume you are talking about an object that needs incoming light to be visible (so it is not like a star that emits light even without receiving any).

Now the very important thing to understand is, that this object can appear yellow because of two main reasons:

  1. Reflection

-specular reflection

it reflects (elastic scattering) only yellow light, whatever wavelength light is shone on it, if the light that is shone on it includes yellow wavelength photons, those will be reflected, the rest (non-yellow wavelength or non-visible) will be absorbed and not re-emitted in visible range (or just re-emitted as yellow wavelength). This is a special object that only reflects yellow light, like gold.

Why are most metals gray/silver?

-diffuse reflection

And, when a colored object has both diffuse and specular reflection, usually only the diffuse component is colored. A cherry reflects diffusely red light, absorbs all other colors and has a specular reflection which is essentially white (if the incident light is white light).


This is usually how a normal wall appears yellow.

  1. Absorption and re-emission

it emits yellow wavelength photons, whatever wavelength photons shine on it, that means, that this object's surface absorbs different wavelength photons, but only emits yellow wavelength. Yes, it is possible, for the material on the surface to absorb different wavelength photons, excite the electrons, and these electrons when they relax, they do it in cascades, or multiple photon emission, including a visible yellow wavelength photon. This is usually in part too how a normal wall appears yellow.

And if you take that last approach, you could have a series of transitions between A and D through state B and another series of transitions between A and D through C and in addition to doing one series or the other series there could be a superposition of both.

Path taken by electron returning to ground state

  • $\begingroup$ why the downvote? $\endgroup$ Commented Dec 11, 2019 at 16:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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