00:01
We're asked to determine the maximum angle, which they call theta, for which the light rays incident on the end of the light pipe and figure p22 .38, are subject to total internal reflection.
00:13
So assume that the pipe has an index of refraction in sub 2 of 1 .36 coming out of air, which has an indexed refraction of 1.
00:22
Okay, so we're going to apply snell's law to point a, and at point a, light travels from pipe to air.
00:32
Thus, the index of reflection in 1 is equal to 1 .36, and the index of refraction in 2 here is equal to 1.
00:40
So we are actually going to reverse these because of the direction of travel.
00:46
So we are going to call n2 .1 .0 and in 1 .36.
00:55
Therefore, the critical angle, sine theta c, is equal to n1, or in 1.
01:05
2 divided by n1, so therefore the critical angle theta c is equal to the inverse sign of the ratio n2 to n1.
01:24
Using that, we've now found the critical angle, which is 47 .3 degrees.
01:37
So from the figure, the angle of refraction can be calculated as follows, just using the fact that the total inside angles of a triangle have to add up to 180 degrees.
01:50
We have theta c plus theta r is equal to 90 degrees or theta r just plugging the value for the critical angle that we just found is equal to 42 .7 degrees...