Talk:Rainbow

Re explanation

"The light at the back of the raindrop does not undergo total internal reflection, and some light does emerge from the back. However, light coming out the back of the raindrop does not create a rainbow between the observer and the sun because spectra emitted from the back of the raindrop do not have a maximum of intensity, as the other visible rainbows do, and thus the colours blend together rather than forming a rainbow"

Am I right in thinking that this needs to be edited?

...light coming out of the back of the rainbow...

...travels between the raindrop and infinity and so can never create a rainbow between the observer and the sun, not even the rainbow can do this because it is created between the observer and infinity ie the sun is behind the observer.

I think what the original text is getting at is that light travelling out of the back of the raindrop is diffuse and cannot create further rainbows beyond the point of maximum intensity which I think must be at 42° between the sun and the observer as it is reflected at the appropriate distance from the observer depending on the hight of the sun.

So maybe it should read "light coming out the back of the raindrop does not create a rainbow in relation to the observer and the sun..."

but then I'm not a scientist I'm a wiki reader so your thoughts please, --Indipage (talk) 20:03, 27 June 2014 (UTC)

What is meant by those sentences might be too complicated for Wikipedia. The diagram at the right might illustrate the word 'maximum'. Light rays may enter the water droplet anywhere between the center (eccentricity = 0) and the edge (eccentricity = 1). The diagram shows the exit angle after 0, 1, 2, and 3 internal reflections. The maxima of these curves correspond to (1) the primary rainbow at 42°, (2) the secundary rainbow at -50°, and (3) the tertiary rainbow at -137°. As curve 0 has no maximum, it does not generate a rainbow, merely 'zero order glow'. Ceinturion (talk) 18:00, 9 July 2014 (UTC)

Formula for the 42° angle

The todo-list at the holy top of this Talk page was added in 2007, and it has been ignored ever since. It is possible to add the formula for the angle, 4 arcsin(q/n) - 2 arcsin(q), where q = √((4-n²)/3), but formulas scare away readers. Or is it time to remove this item from the todo-list (or the entire todo-list)? Ceinturion (talk) 22:00, 2 September 2014 (UTC)

The explanation is actually not difficult. For a quantitative calculation we need the entire angle change as a function of incidence angle. Is the total angular change not dependent on the angle of incidence many rays are going in the same direction, that means the intensity is highest at that angle of incidence. 84.118.81.7 (talk) 18:25, 16 November 2014 (UTC)

If $\theta _{i}$ is the angle of incidence for refraction then the total change of direction to the incident beam after k reflections in the droplet is

$tot=2\left(\theta _{i}-asin(sin(\theta _{i})/n)\right)+k\,\left(\pi -2\,asin(sin(\theta _{i})/n)\right)$

The condition for maximum intensity is, that the total angular change tot does not depend on the angle of incidence $\theta _{i}$.

${\frac {d(tot)}{d\theta _{i}}}=2\,\left(1-{\frac {(k+1)\,cos(\theta _{i})}{n\,{\sqrt {1-\left(sin(\theta _{i})/n\right)^{2}}}}}\right)=0$

The sine function of the incident angle at maximum intensity is

$sin(\theta _{i})={\sqrt {\frac {\left(k+1\right)^{2}-n^{2}}{\left(k+1\right)^{2}-1}}}$ — Preceding unsigned comment added by 84.118.81.7 (talk) 15:33, 3 November 2014 (UTC)

Your statement that "q is the incident angle" should be "arcsin(q) is the incident angle" if you are using my q. Anyways, the intention of my comment was not to discuss lots of formulas. Ceinturion (talk) 20:08, 3 November 2014 (UTC)

Ok, I better choose $\theta _{i}=asin(q)$. With k = 1 it's your formular. It might be derived also from ${\frac {d(tot)}{dq}}=0$.84.118.81.7 (talk) 23:07, 3 November 2014 (UTC)

Dispersion needs mention in the first line.

204.214.145.6 (talk) 22:42, 1 October 2014 (UTC) The word "Dispersion" should be included in the first line of the Article, along with Reflection and Refraction. First, there is the Refraction that occurs when light enters the drop. Next is the Dispersion of different wavelenghts (colours), as they get refracted at different angles. Then comes the Total Internal Reflection, and finally the refraction that occurs when light comes out of the drop. Hence, I would prefer to put Refraction, Dispersion and Reflection in that order. Raghavendra P Umarji 204.214.145.6 (talk) 22:42, 1 October 2014 (UTC)

Violet≠Purple --> MinutePhysics (youtube) video

i just stumbled across a video from MinutePhysics all about misconceptions over the colours in a rainbow. ≈Sensorsweep (talk) 02:05, 29 October 2014 (UTC)

Order of colors

The picture "White light separates into different colours" puzzles me. The red light is depicted there "on the bottom". If the picture were correct, the red color would lie in the inner part of a primary rainbow, but it lies in the outer part. Is the picture wrong? In any case this needs a clarification... — Preceding unsigned comment added by 70.114.197.37 (talk) 01:17, 3 November 2014 (UTC)

The figure is correct. The red light is directed more down than the blue, which means you will see it from water droplets that are higher up in the sky. I agree that the reason for this inversion in the order of the colors is not well explained in the article. Ulflund (talk) 19:44, 3 November 2014 (UTC)