Bertrik
Newly Enlightened
I'm not doubting that the wavelength dependent reflection effect exists, but I doubt if it has significant effect with flashlights. The important parameter is the particle size to wavelength ratio. The sky is blue because of scattering on air molecules (much smaller than the light wavelength), clouds are white because of scattering on water droplets (much larger than the light wavelength).Well scientifically you don't need much evidence because it's already been widely accepted as the general theory (you can google it) that shorter wavelengths (higher frequency) have a higher probability of "bouncing" or "reflecting" off of something such as nitrogen and oxygen molecules in the atmosphere. How you apply that theory to anything else will surely follow the same logic. (A cloud is quite thick, which is why it casts a shadow onto the ground. It's not white because it "scatters" white light. Clouds are white because they reflect almost every spectra of ambient light, only absorbing a few colors.)
The atmosphere is much less dense than clouds, and a very small percentage of light "bounces" off of the molecules, and since shorter wavelengths of light (blue) have a higher probability of bouncing, the sky appears blue and the sun appears more yellowish. It's also the reason why the ocean appears blue. It's also the reason why when during sunset, the sky takes on more of a reddish color than mid-day since the light has to go through more and more atmosphere to reach you, reflecting more and more blue as it reaches your eye, appearing redder.
Anyway, the effect of this phenomenon on flashlights at close distances is very small, but still noticeable. LED's usually have a more pronounced "beam" that you can see in the air because the light gets reflected easier than incandescents.
My assumption is that outdoors reflection of flashlight light is dominated by reflection on tiny water droplets and that these droplets are similar in size to droplets in clouds (about 10 um). In that case, rayleigh scattering does *not apply* and all visible wavelengths scatter about equally.