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:


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:


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.

  • 1
    $\begingroup$ 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? $\endgroup$ Commented Apr 1, 2013 at 23:37
  • $\begingroup$ 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.) $\endgroup$ Commented Apr 1, 2013 at 23:44
  • $\begingroup$ 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. $\endgroup$
    – Luan Nico
    Commented Apr 2, 2013 at 0:18

1 Answer 1


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, $$ f=(1-\cos(\sin^{-1}(1/n)))/2 = \tfrac12-\tfrac1{2n}\sqrt{n^2-1}\ . $$

  • $\begingroup$ Thanks a lot! I'd upvote you, but I don't have enough reputation here yet. $\endgroup$
    – Luan Nico
    Commented Apr 2, 2013 at 0:57
  • 2
    $\begingroup$ @LuanNico Although this is obviously the solution that was sought, since it gave the "correct" answer, I believe that this is not really the correct solution to the problem as it is formulated. This solution assumes that no light is reflected for the angles below total internal reflection. That is not true. The correct solution would require integrating expressions for transmittance derived from the Fresnel equations over the solid angle of the cone. $\endgroup$
    – jkej
    Commented Apr 2, 2013 at 15:12

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