Here is a problem that I am working on, which is the applying the concepts of diffraction to the setting of the sun:
Air has a small, usually negligible index of refraction. It is 1.0002926. This causes the Sun to actually be below the horizon when it appears to be just on the verge of sinking below it.
Suppose you are on the sea-shore watching the Sun apparently sinking into the ocean. When only its upper tip is still visible, by what fraction of the diameter of the Sun is that tip actually already below the surface?
As an approximation, take the earth's atmosphere as being of uniform density out to a thickness of 8.600 km, beyond which there is no atmosphere. This means that, with the Earth's radius being 6400. km, your line of sight due West along the ocean surface to the horizon will intersect this "upper surface" of the atmosphere at about 331.9 km from your eye. (The diameter of the sun subtends 0.5000 degrees at your eye).
This is how I've attempted to model the problem, but I'm not sure if it's correct…
I'm not really sure how to approach this problem…. Can anyone please help me?
Best Answer
Here is a picture that should help you approaching the problem:
(Needless to say, nothing is drawn to scale)
First, evaluate $\alpha$ (see picture). Then, derive $i$:
$i = 180 - 90 - \alpha$
Then, substitute into Snell's law and derive $i'$:
$i' = arcsin(n sin(i))$ where $n=1.0002926$
Then derive the angle $j$ (see the picture).
Don't forget to convert it to Sun diameters, that is, multiply with $2$, provided that the Sun diameter is $0.5$ degrees.
I get $0.34308205134$ degrees, that is, $68.62$% of the diametre.
Most important, when you have understood it, try to do it by yourself again.
Good luck!