00:01
Shown here is the given figure from the problem.
00:03
Question a asked, at what point does the beam first exit? the beam relaxates it if total internal reflection does not occur.
00:30
And note that total internal reflection occurs when the critical angle is less than the incident angle.
00:46
Or when the incident angle is exceeds the critical angle.
00:50
So to solve this problem, we have to determine the critical angle first, which is this formula.
01:01
Sign inverse of the index of refraction, the lower index of refraction over the higher.
01:10
So since this is an interface with air, then this is the lower index fraction since the index of fraction is of crown glass is higher than air.
01:28
Then so we can find that the critical angle is this note that the fraction of cranglass is 1 .523.
01:39
So the critical angle is 41 degrees.
01:42
Now let's start with point a.
01:46
We can see that the critical angle is less than the incident angle at point a.
01:57
Therefore, tir this occurs here.
02:05
So all of the light is reflected.
02:10
Since the incident angle exceeds the critical angle.
02:13
So 60 degrees is critical 41.
02:16
Now at point we can see that critical angle is 41 degrees again and the incident angle how do we know the incident angle at point b? we will use the geometry.
02:34
So this is our right triangle...