00:01
The first step is to use the total internal reflection info that they give us in order to figure out the index of refraction of the liquid.
00:09
And so we know that the index of refraction of the liquid times sine of 42 .5 degrees.
00:24
All right.
00:25
This is the critical angle is equal to in air times sign of 90.
00:35
90 degrees because when we've reached the critical angle, the angle between the normal is 9 degrees in the air.
00:43
And so a sign of 90, that's just one.
00:46
In air, that's just 1 2.
00:48
And so this right -hand side is just 1.
00:52
And so this allows us to sulfur the index of a fraction of the liquid.
00:56
It's just 1 over the sign of 42 .5 degrees.
01:02
When you do this division, you get 1 .48.
01:08
Now that we know the index of a fraction of the liquid, we can actually start to solve the problem.
01:15
Part a gives us that the index and the material we start with is 1 .48, since we're starting in the liquid, and we're moving to the air.
01:25
So that means nb is 1 .0...