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Inside the rainbow all the colors of the rainbow are joint together to form white light. But how are all these different color cones formed?

We see only the outside bands of the rainbow colors but the inner parts are joint together to form white light within/below the rainbow, how is that possible?

Are perhaps all the different 'colorwaves' scattered on the tiny waterdrops and giving it a total white color which is coming into our eyes from all directions? And it is not blue, as you could think because of the dominants of blue by N2 and O2, but white because the droplets underneath the rainbow are, just the way they make the clouds white, scattered on H2O??

Is this right or is there another explanation? As you can see it is not bright white but still more white compared to the right side of the rainbow....

enter image description here

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  • $\begingroup$ Well you might be interested in this en.wikipedia.org/wiki/Fog_bow It's something to do with droplet size and viewing angle. Maybe someone can bring the equations. $\endgroup$ – JMLCarter Dec 31 '16 at 0:23
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    $\begingroup$ I think you may be asking this question: physics.stackexchange.com/questions/88265/… $\endgroup$ – user2309840 Dec 31 '16 at 3:45
  • $\begingroup$ Well that question and the answers comes really close to my question, but actually just stops where my question begin. $\endgroup$ – Marijn Dec 31 '16 at 10:00
  • $\begingroup$ This is quite late, but the answer is that you're looking at a blurred supernumerary bow. These are "echoes" of the rainbow due to the wave nature of light, which have gotten blurred together here to make white. $\endgroup$ – knzhou Jan 7 at 15:52
  • $\begingroup$ Related: en.wikipedia.org/wiki/Alexander%27s_band $\endgroup$ – PM 2Ring Jan 7 at 16:05
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"...how are all these different color cones formed?"

Say a ray of light hits a raindrop with and angle of incidence of A. The refraction angle is B=arcsin(sin(A)/n). The actual values aren't important yet; only that n>1, so we know that B<A, and the amount of deflection is the positive angle A-B.

The light that enters the drop (yes, some will reflect without entering) will reach the back of the drop. The angle of incidence there will be this same angle B. The deflection angle of the portion that reflects will be 180-2B, and the portion that exits the drop will be A-B, both in the same direction as the original deflection. And this repeats every time the remaining light hits the surface.

As a result, light that exits the drop after one internal reflection deflects a total of 180+2A-4B. But since you are looking 180 degrees away from the original ray already, it is easier to use 4B-2A. Here's a plot:enter image description here

The brightness of each color, at any particular deflection angle, is inversely proportional to the slope of these curves. Which is zero at the maxima shown. Think about it. :) At lower angles, each color has about the same brightness, and so appears white: enter image description here

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  • $\begingroup$ Note that the brightness of the color is also affected by the fact that relatively few rays will hit the droplet at zero incidence (they have to be aligned perfectly to do that), and very few will hit the droplet at grazing incidence (the cross-section is infinitesimal.) If you take into account the cross-sectional effects, the overall intensity is multiplied by $(\sin A) (\cos A) = 2 \sin (2A)$. But that said, this doesn't affect the main conclusion: inside the rainbow arc, all colors are reflected approximately equally. $\endgroup$ – Michael Seifert Jan 7 at 16:58
  • $\begingroup$ Yeah, I have another diagram that replaces "angle of incidence" with "distance from center." Using the angle makes the explanation simpler, since it is the angles that drive the process. And as you say, the conclusion is the same. $\endgroup$ – JeffJo Jan 9 at 13:13

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