# Application of diffraction problem!

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... • This seems to me to be a literal copy of a homework question. If so I think you should add the "homework" tag and I think in any case you should show what effort you have made. – user16228 Nov 26 '12 at 1:16
• I have a picture that I sketched, but it didn't let me post it because I'm a new member...but other than that, I have no idea how to approach this problem. – Vanessa Nov 26 '12 at 1:18
• OK, well maybe you could describe the picture you have drawn in your question . I can't really help you now with this question, it's about 2 AM here and I am about to go to bed. I'm sure there are plenty of other people here who will give you good help though. – user16228 Nov 26 '12 at 1:23
• Alright...I'm actually going to try and recreate it in paint and find a way to upload it... – Vanessa Nov 26 '12 at 1:25
• This is how I'm imagining the situation, but I'm not sure if it's correct: – Vanessa Nov 26 '12 at 1:28 (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!

• Thanks so much for your answer! I was able to work it out myself, but I didn't need to multiply by 2...is it possible that you could explain that step? – Vanessa Nov 27 '12 at 3:00
• Because they ask "By what fraction of the diameter of the Sun...?". Since that diameter is half a degree, you divide your solution in degrees by 0.5, that is, multiply by 2. Then you get 0.68... but I multiplied by 100 to see it as a percentage. – Eduardo Guerras Valera Nov 27 '12 at 9:18

## protected by Qmechanic♦Dec 4 '13 at 20:16

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