# Which fraction of light is refracted from a source of light under a lake?

I was trying to solve this problem:

"A punctiform source of light is standing inside a lake, at a height h of the surface. f is the fraction of the total of energy emitted that escapes directly from the lake, ignoring the light being absorbed in the water. Given n, the refractive index of water, determine f."

I understand that, since the maximum refraction angle is 90°, there is a maximum incident angle. The next image explains the principle:

http://img441.imageshack.us/img441/9573/lake01b.png

In it, the yellow incident light rays diverge (since the air is less dense than the water), until, at one point, the refractive angle is 90°. Then, the rays stop refracting.

At this point, I applied Snell Law:

n1 * sin i = n2 * sin r
n * sin i = 1 (it's air) * sin 90
sin i = 1/n


Now let's analyze the following triangle:

http://img690.imageshack.us/img690/8421/lake02.png

As you can see, the triangle is formed by: 90 - i, 90 - i and a.

a + 90 - i + 90 - i = 180
a = 2i


The fraction of light that made out of the lake is a over the total circle, that is, 360°. So:

f = 2i/360 = i/180
i = arc sin (1/n)
f = (arc sin (1/n))/180


However, the answer I have for this exercise (and it does seem to be right, because it is from a University*) is $f = \tfrac12 - \tfrac{1}{2n} \sqrt{n^2 - 1}$. And I don't know what I did wrong. It is very important for me to solve this exercise, and I hope someone would have a hint of what I am doing wrong.

*It is a very old test (1969), and there is no resolution anywhere (just the final answer).

Second try, using Solid Angles:

At is the total Area of the light sphere of radius h:

At = 4 * pi * rt²
rt = h
At = 4 * pi * h²


Ap is the partial area of the circle of light that gets out of the water:

rp = h / (tg(90 - i))

tg (90 - i) = sen (90 - i)/ cos (90 - i)
sen (90 - i) = sen 90*cos i - sen i*cos90 = cos i
cos (90 - i) = cos 90*cos i + sen 90*seni = sen i
tg (90 - i) = cos(i)/sen(i) = 1/tg(i)

rp = h*tg(i)
Ap = pi * rp²


So f must be Ap/At:

Ap/At = (pi * rp²) / (4 * pi * rt²)
f = h² * tg²(i) / (4 * h²)
f = tg²(i)/4


Still not there.

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You should consider that this is probably a 3D problem, see en.wikipedia.org/wiki/Solid_angle. In any case, what is v supposed to be? –  Norbert Schuch Apr 1 '13 at 23:37
Got it, v is the square root ... looking at it in 3D indeed gives the correct solution. (I edited the post to contain the right formula; please correct me if my guess was wrong.) –  Norbert Schuch Apr 1 '13 at 23:44
I took a look at Solid Angles. I'd never heard of it, but I studied for a while, and I added what I came up with in the post. Still not right, but I think it is progress. –  Luan Nico Apr 2 '13 at 0:18
According to http://en.wikipedia.org/wiki/Solid_angle, the solid angle given by a cone with angle $2\theta$ is $\Omega=2\pi(1-\cos(\theta))$, i.e., it covers a fraction $\Omega/4\pi = (1-\cos\theta)/2$ of the sphere. This is exactly the fraction of the light leaving the water, where $\theta$ is just the angle you originally called $i$, i.e., $\theta=\sin^{-1}(1/n)$. Thus, $$f=(1-\cos(\sin^{-1}(1/n)))/2 = \tfrac12-\tfrac1{2n}\sqrt{n^2-1}\ .$$