I'm aware of a similar question being asked in Do rainbow shows spectral lines?:

The response to this question is that the body producing the light is not the water droplet that merely diffracts it but rather the Sun, which acts as a black body and thus produces a complete spectrum.

However, it is not entirely true that the sun is a black body: like all stars, it has an spectral class with absorption lines.

I am wondering why these lines are not observable in the rainbow or when decompossing sunlight with a prism in otherwise uncontrolled conditions. Are they simply too narrow to be seen without optical instruments? Is it a product of the light source not being coherent? How is this consistent with the possibility of obtaining spectra from distant stars in which the spectral lines are visible?

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    $\begingroup$ Can you see the spectral line in sunlight you put through a prism directly? I took a detailed solar spectrum in grad school, but it required a rather more powerful spectrometer than a prism. $\endgroup$ Commented Nov 3, 2019 at 15:38
  • $\begingroup$ Certainly not with any of the prisms that i have, my question is precisely why the prism is not powerful enough on its own. $\endgroup$
    – Mario
    Commented Nov 3, 2019 at 18:41
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    $\begingroup$ A rainbow is a very different phenomenon than a spectrum from a prism. In a rainbow the light frequencies are spread around in a very specific way that has to do with the physics of how light is refracted inside a drop of water. For this reason I doubt it would be possible ever to observe a spectral line in a rainbow, no matter how ideal the conditions. This is not at all obvious, however, and it's a good question. $\endgroup$
    – N. Virgo
    Commented Nov 4, 2019 at 1:22
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    $\begingroup$ I'm afraid I'm unaware of how it is different, is there a reference in which i can read further into this? $\endgroup$
    – Mario
    Commented Nov 4, 2019 at 11:08
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    $\begingroup$ if you take the sunlight and send it through a narrow slit first before it goes to the prism you will have absorption lines present. Without the slit the rays interfere with eachother $\endgroup$
    – ChemEng
    Commented Nov 5, 2019 at 3:23

3 Answers 3


First you must understand how a (primary) rainbow is formed. Here is a summary with some details that are not in the linked article:

Whenever a beam of light encounters an air-water boundary, it is either reflected or refracted. To form a primary rainbow, we must first have enough small water droplets in the air, as these are close to spherical. Parallel light beams that reach such a water droplet must be refracted once as they enter, reflected off the inner surface of the droplet once, and refracted once more as they exit:

(image from the linked article)

There are 3 important points to note here:

  1. Water droplets in the air are never perfectly spherical. That is one source of fuzziness.

  2. Parallel beams entering the same water droplet can exit at different angles! This is the major reason why rainbows can never give sharp spectra even if you have perfectly spherical water droplets (say in outer space). Why then do we still see the rainbow? There are three reasons, which together result in the rainbow being seen roughly at 42°:

    • Different incident light beams will have different amount reflected/refracted. In particular, the bottommost incident beam (in the diagram) will mostly pass through without being reflected at the back of the droplet, and the topmost incident beam will mostly be reflected rather than enter the droplet.

    • The light beams emerging from the droplet after the above process are 'denser' along the so-called caustic ray, because the emergent angle does not vary monotonically with the distance of the incident beam from the central axis, and it reaches a maximum for the caustic ray, around which the emergent angle varies less.

    • The incident light beams further from the central axis undergo greater refraction, hence resulting in greater separation of different wavelengths. In contrast, the rainbow rays from incident beams close to the central axis largely overlap one another and wash out. (See this webpage for an image illustrating this.)

  3. Light beams may encounter more than one droplet! This is another major reason why we cannot expect a sharp spectrum from a (natural) rainbow.

  4. Even if we assume that the water droplet is a point, the light beams from the sun will not be perfectly parallel. In fact, the sun subtends an angle of about 0.5° to an observer on Earth, so this leads to roughly that same amount of spreading of the rainbow as compared to one generated by a point light source. Still, this is a much less significant effect than point 2.

(jkien's answer is incorrect but inexplicably has lots of upvotes.)

  • $\begingroup$ Point 3 is most likely a negligible effect in most circumstances. Point 2 is certainly the most important point, but this description seems a bit convoluted and still it doesn't give any quantitative information about what kind of spectral resolution is achieved by the rainbow. Also, since this answer totally ignores the blurring caused by the fact that the light from the sun is not perfectly parallel, it is basically equally incorrect to jkien's answer. Taken together, these two answers cover the most important reasons that the rainbow cannot resolve the narrow spectral lines of the Sun. $\endgroup$
    – jkej
    Commented Nov 4, 2019 at 15:29
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    $\begingroup$ @jkej: Point 3 is not negligible at all; I'm not sure why you think it is. As for your inquiry about "spectral resolution", the point is that parallel incident beams of the same wavelength do not remain parallel, and the possible emergent angles are not even confined to a narrow band but cover the entire range from 0° to about 42°. The distribution has a peak at around 42°, but it smoothly varies with emergent angle (see the graph in the linked article). Thus arguably "resolution" is not well-defined. But I did forget to say that the incident sunlight are not quite parallel beams; I'll add it. $\endgroup$
    – user21820
    Commented Nov 4, 2019 at 15:40
  • $\begingroup$ The fact that all emergent angles from 0° to 42° are possible doesn't mean that you can't define a a spectral resolution based on the very clear peak at ~42°. If that was the case you couldn't talk about resolution of any dispersive instrument because you always have some form of tails on lineshape functions. One common way of defining resolution is the FWHM of the peak. According to this site the FWHM of the 1st order peak of the rainbow seems to be roughly 0.3-0.4°, which would be similar to the divergence of the incoming Sun light. $\endgroup$
    – jkej
    Commented Nov 4, 2019 at 16:09
  • $\begingroup$ Regarding point 3, it would probably require some form of radiative transfer modeling to prove definitively that multiple scattering is negligible in rainbows, but I have never seen it mentioned as important before. On the other hand, you don't mention the effect of multiple internal reflections in the droplet, and this is certainly a more important effect as evidenced by the fact that we can see the secondary rainbow in favorable conditions. $\endgroup$
    – jkej
    Commented Nov 4, 2019 at 16:17
  • $\begingroup$ @jkej: I know very well that you can define spectral resolution based on the distribution, but the FWHM is clearly not a good measure for this purpose because of the heavy tail (see the graph in the webpage you linked to). In contrast, the sun (as observed from the Earth) has a sharp boundary, so the 0.5° spread has a sharp profile. I also know about the secondary bow, but that is offset by a reasonably large angle so the effect on the primary bow is negligible. And I'm taking feedback into account, so I see no reason for the downvote. $\endgroup$
    – user21820
    Commented Nov 4, 2019 at 16:24

Unfortunately, rainbows produced by the Sun are blurred because the Sun is not a point source. The apparent diameter of the Sun is 0.5°, while the width of the rainbow spectrum is about 2°.

However, be patient and wait until the next bright supernova appears at the night sky. It should produce brighter rainbows than the Sun, with purer, saturated colors, and perhaps with the Fraunhofer lines of the supernova visible to the naked eye.

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    $\begingroup$ Have rainbows from supernovae been observed, or is it a hypothetical situation? $\endgroup$
    – svavil
    Commented Nov 4, 2019 at 0:34
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    $\begingroup$ -1 I'm afraid - this answer is incorrect. A rainbow is a very different phenomenon to a spectrum from a prism. The reason for the 'blurring' of a rainbow compared to a spectrum is not to do with the sun being a non-point source, but rather it has to do with the physics of how a rainbow is produced. A bright supernova could indeed produce a rainbow - that's a lovely thought - but it will not look much different from one produced by the Sun. $\endgroup$
    – N. Virgo
    Commented Nov 4, 2019 at 1:19
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    $\begingroup$ @Nathaniel how is it a "very different phenomenon"? Both processes disperse a light source by refracting light, changing its angle leaving the medium in a wavelength-dependent fashion. And I would in fact expect a rainbow from a point source to look noticeably different, not being convolved from the disc of the sun. $\endgroup$
    – MooseBoys
    Commented Nov 4, 2019 at 5:31
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    $\begingroup$ @MooseBoys The water droplets causing the rainbow disperse light in all directions. The ~42° is simply a sharp peak, but with finite width. You can see the lineshape functions (for the 1st order) here. The resolving power of an ideal prism on the other hand is diffraction limited. But you are right that the rainbow from a point source would probably be at least somewhat noticeably different, because the lineshape functions are narrower than the ~0.5° of the sun. $\endgroup$
    – jkej
    Commented Nov 4, 2019 at 9:43
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    $\begingroup$ I just realized that the website I linked to above also has a simulation of what a supernova rainbow may look like. The difference seems to be mostly noticeable for the inner rings, at least for a casual observer. $\endgroup$
    – jkej
    Commented Nov 4, 2019 at 9:52

As you point out, a sufficiently sophisticated spectrum device will allow measurement of spectral lines in the spectrum of the Sun.

For instance (history of chemistry), the existence of the element Helium was first noticed by the fact that a spectrum of the Sun includes lines not accounted for by the spectra of the elements known at the time.

While there is sufficient separation of colors to created noticable color effect, in actual fact there is a lot of overlap. The amount of overlap is such that any details get averaged out.

A well manufactured prism will not have the kind of overlapping that you have with a rainbow, so I do expect good separation of the individual colors of the light.
I don't know what level of spectroscopy is possible with prisms.
I'm not sure, but it may well be that with a prism there is not enough dispersion of the light to obtain sufficient detail.

To my knowledge: as soon as the technology became available spectrosopy moved to diffraction gratings. The closer the spacing of the lines of the diffraction grating the bigger the separation of the spectral lines.

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    $\begingroup$ It just needs an entrance slit. $\endgroup$
    – user137289
    Commented Nov 3, 2019 at 21:41
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    $\begingroup$ Could you clarify why rainbows have a lot of overlap? It doesn't seem obvious reading the comments from the other answer. $\endgroup$
    – Calimo
    Commented Nov 4, 2019 at 7:49

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