I am trying to read up and understand a lot about how normal looking objects like chairs are visible to us as compared to fluorescent substances. This question says that -

"It is often said that substances, objects have color because they selectively absorb all color of sunlight except one. The wavelength that is not absorbed reaches our eyes and we perceive it as "color". This "color" phenomenon is often described as light of a certain wavelength being reflected. I think it is more appropriate to call it scattering"

So, I guess we see a blue chair as blue since it scatters/reflects light of blue wavelength. And this paper says that the main difference between fluorescence and scattering is that - "The wavelength of fluorescence emission is generally independent of the wavelength of the exciting light. In contrast, the wavelength of light scattering increases with increasing wavelength of the exciting light".

So my question is, if we increase the wavelength of light incident on a chair, will it start looking a different color/invisible, as this statement says?, whereas this would not happen for a fluorescent substance?

And apart from this (if this is correct!), is there any major difference between normal color of a object, and fluorescence? The original curiosity in my mind was that why do we use fluorescent proteins like GFP in Biology? I presume that fluorescent substances are brighter and hence easier to see, but I may be completely wrong on this.

  • $\begingroup$ Green Fluorescent Protein stands out when illuminated with a shorter wavelength that is invisible (UV) or that is filtered away. $\endgroup$
    – user137289
    Apr 2, 2018 at 0:29
  • $\begingroup$ @Pieter, yes! but how do fluorescent objects look different from normal? If I had an orange non-fluorescent protein, would it make sense to use that in the place of GFP? $\endgroup$
    – user1993
    Apr 2, 2018 at 0:34
  • $\begingroup$ Contrast would be much lower. Many other structures in a cell will also scatter light at the same wavelength. One needs very intense pigments to see something. That is the stuff used in traditional microscopy stains, but usually those kill the cell. $\endgroup$
    – user137289
    Apr 2, 2018 at 6:20

1 Answer 1


The quote from the paper seems rather sloppy to me.

Assume a substance that exhibits no scattering but only fluorescence. If and only If the exciting wavelength is short enough, you'll get a monochromatic "response" from the substance, always the same color, independent from the exciting light's wavelenght.

Nonfluorescent substances (e.g. the blue chair) will always scatter light of the same wavelength as the incoming light. The chair appears blue because it scatters more blue than e.g. red light.

"Normal" substances will scatter many wavelengths, even if they appear blue. You can check this with a laser pointer. You'll basically always see a red dot from a red laser pointer, even when pointed at a blue chair because some red light will be scattered into your eyes.

If you start with monochromatic blue light and scan through the whole spectrum, the scattering will vary in intensity, but $\lambda_\text{in}=\lambda_\text{scatter}$.

  • $\begingroup$ I am trying to understand the two statements in your penultimate paragraph. So if red and blue lights were to fall on a "blue" chair, then it will scatter blue back at the same blue wavelength, whereas it will absorb the red, so there is no red to scatter? In this case, if I shine only red light on a "blue" chair, it should absorb almost all of the light and thus appear black? $\endgroup$
    – user1993
    Apr 3, 2018 at 5:09
  • 1
    $\begingroup$ Your typical blue chair will scatter some red light, but you're on the right track. $\endgroup$
    – Jasper
    Apr 3, 2018 at 6:51

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