sophiecentaur said:
The rainbow is an effect due to the refractive index of water and not the air around the droplets. In a vacuum (nonsense I know because the drops would evaporate but . . . . .) a rainbow would still form and the geometry would change very very little.
Actually, the rainbow is a poor example. Because the spherical water drops, unlike prisms (or atmosphere!), do not refract a single wavelength to a single line. As can be seen in monochromatic red rainbows - these still and more conspicuously have the contrast between bright interior and dark exterior of the first rainbow (and dark interior and bright exterior of the second).
Consider the Cauchy formula:
n=A+B/λ
2
Consider the definition of Abbe number:
V
d=(n
d-1)/(n
F-n
C)
The indices refer to Fraunhofer lines.
Fraunhofer F is at 486 nm; C is at 656 nm and d at 589 nm.
What is the factor 1/λ
2 for these lines, in terms of inverse square μm?
For Fraunhofer C, it is 1000
2/656
2=2,32
For Fraunhofer F, it is 1000
2/486
2=4,23
What is visible light?
From my university spectroscopy lesson back in 1990s, I distinctly recall seeing spectral lines to the violet of 400 nm (don´t now recall exact number or substance). If we call Fraunhofer K visible, at 393 nm, then it is at the position of 1000
2/393
2=6,47. On the red edge, let´s assume Fraunhofer A is visible at 759 nm. Then it is at the position of 1000
2/759
2=1,74
For a refractor which is close to Cauchy formula, the relative width of spectrum from Fraunhofer K to Fraunhofer F, 393 to 486 nm, the violets and blues, is 6,47-4,23=2,24 units of 1/μm
2; the relative width of spectrum from Fraunhofer F to A, 486 to 759 nm, covering all of greens, yellows, oranges, reds is 4,23-1,74=2,49 units.
Note that this is relative width of parts of spectrum. The geometry of the refractor and the Cauchy formula coefficients A and B cancel out because they are shared by all parts of spectrum
I admit that I was mistaken that "most" of the rainbow width is in "blue", unless blue is counted redwards of 486 nm. It would be actually true if blue extends, to, say, round 500 nm. But my point is that Cauchy formula compresses the red part of spectrum and stretches out the blue/violet part.
"Rainbow" is problematic because as I mentioned, the refracted light of rainbow has the long tail to the interior. Which means that the extreme violet colour of a rainbow, inherently dim for eye, is overlapped by the bright background of other colours.
The other point about the width of spectrum is that the violets and blues are most strongly attenuated. Which means that, for example, at the end of the rainbow the inner violet/blue colours of rainbow vanish. In case of strong attenuation, the whole rainbow may be monochromatic red.
But atmospheric refraction?
The atmospheric refraction is strongest near horizon and much weaker well above the horizon.
But near horizon where the atmospheric rainbow should be widest, the blues which make up most of the rainbow width are attenuated away. Well above the horizon, where the blue does get transmitted, the whole rainbow is narrower. Either case, too narrow to resolve by naked eye.
