# Is there a liquid for which the primary and secondary rainbows overlap for visible wavelengths?

This answer to the question Is there a zone where 'colour' overlaps between double rainbows? raises the possibility that, at suitable wavelengths, the primary and secondary rainbows can overlap, if the wavelength is such that the refractive index goes down to about $n=1.312$ which, according to this page, can happen in the near-IR regime, though dangerously close to the resonances in the IR region.

I have a related question: is this possible to do by changing the liquid instead of the wavelength? That is, I would want a liquid that has a refractive index of around $n_*=1.312$ in the visible range, ideally with a nice linear gradient in the neighbourhood of $n_*$ (and, even better, controllable so that it goes from just below to just above $n_*$ with changes to e.g. the concentration of some dopant or whatever).

For this to work, the liquid would have to be relatively colourless as regards absorption, safe to handle, and easily sprayable. The idea would be to run demonstrations with this in relatively public settings, and the like, so it needs to be reasonably cheaply accessible in reasonable quantities; substances sold for refractive-index matching on per-milliliter bases are somewhat outside of that envelope.

Is this possible?

• The mineral oils tested in ieeexplore.ieee.org/stamp/stamp.jsp?arnumber=1039597 have $n \approx 1.46$ with very low absorption. Apr 24, 2017 at 16:07
• @JonCuster That's way out in the wrong direction. The point is to go from water ($n\approx 1.335$) down by a bit, not up by 10%. Apr 24, 2017 at 16:24
• OK, how about ice? en.wikipedia.org/wiki/List_of_refractive_indices shows a list. I presume ice is shifted down by about the change in volume. Likely the dispersion would remain about the same as well... Apr 24, 2017 at 16:32
• @JonCuster Can you expand on that with a practical suggestion of how you can produce a spray of spherical ice droplets? I thought it went without saying that I looked in Wikipedia and other similar resources and didn't find anything obvious. And, in any case, the question specifically asks for liquids. Apr 24, 2017 at 16:51
• Very well, 2,2,3,4,4,4-Hexafluoro-1-butanol has n=1.312, but one would have to look at the dispersion. Apr 24, 2017 at 17:06

The diameter of the rainbow circle depends on $n_{raindrop}/n_{gas}$, so it might be more convenient to replace the air by a gas or vapor at an adjustable high pressure. Pentane, for example, does not dissolve in water. At 1 bar, pentane vapor has an index of refraction of $n_{pentane.vapor}=1.0017$, so about $1.017$ at 10 bar (I am guessing here that $(n_{pentane.vapor}-1)$ is proportional to the pressure). The temperature should be >120°C, as the vapor pressure of pentane is 10 bar at 120°C. At 10 bar, the raindrops would still be liquid if the temperature is <180°C. So $n_{raindrop}/n_{gas}$ = 1.33/1.017 = 1.308 is possible with pentane at 10 bar.