Optics 与虹，霓，Alexander 暗带

作者: optic_css | 来源:发表于2019-01-09 00:24 被阅读0次

Optics 与虹，霓，Alexander 暗带
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无题
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Optics of rainbow___ breif intro

Rainbow, as we all know, is the phenomenon which well reflect the refraction and scattering properties of light, this article will calculate briefly about its mechanism, in a much simple way.

As the ray of light shown above, the sunlight shines in parallel to the water droplet in the air with the angle $\alpha$ .

From the Snell law of refraction.

$\frac{sin\alpha}{sin\beta } =n$ ,

besides, notice that $\theta$ satisfied

$\frac{\theta}{2}=\beta -(\alpha -\beta ) =2\beta -\alpha$

$\implies \theta =4\beta -2\alpha$ ,

hence, obtain the relationship between $\theta$ and $\alpha$ as

$\theta =4arcsin\frac{sin\alpha}{n} -2\alpha$ ,

The figure above tells us there exist a maximum in the function, which will be treated as one side of the rainbow, it will be at $\theta \approx 42.30 degrees$ when $n=1.33$ , of course, the red one. When it comes to the purple one, the same, $\theta \approx 40 degrees$ .

From the fermat's principle, different colors of light will converge near the extremum area, that's why we can observe the rainbow.

The other confusion of the rainbow is the existence both of the neon and the rainbow, and there is always a "Alexander dark belt" between them.

The reason, however, is easy to understand, when the sunlight goes into the eyes after more times of refractions and reflections (almost of them are 2 and 3 times), the odd times and even times will give the different curve of $\theta$ , and the curve which have the minimum will give another type of "rainbow" named "neon", they will have band of light with reverse order, as expected.

In the neon, the red light will have the extremum value at $52degrees$ , purple on $54degrees$

It's clear that the "Alexander dark belt" is just the angle between $42$ and $52$ .

Finally, the rainbow is the light belt we can see only at the specific angle $\theta$ , and the shape of it is no more than the cross section of the cone. (As the figure shown below)