00:01
So the first thing we'll do in this problem is draw this setup.
00:05
We have an eye drop with an index of a fraction of 1 .45, air with an index of a fraction of 1, and the cornea with an index of a fraction of 1 .38.
00:19
So because the index of a fraction of the cornea is less than the index of a fraction of the drop, we have our two rays coming here.
00:27
This ray here is not going to have any sort of phase shift.
00:31
Meanwhile, because the drops index, is greater than air's index, is going to be a phase shift of pi for this first ray.
00:41
So in order to have constructive interference at 600 nanometers of red light, we're going to have to use the typical destructive interference equation in order to cause a phase shift of pi in this second ray.
00:58
So to find the minimum thickness, we let m be equal to zero, and this gives us t is equal to lambda over four, which is equal to using 600 for red light, this gives us 150 nanometers, which is the answer to part a.
01:14
Now, to see what else is reinforced, we can just move m up to one, and now plugging in 150 for our thickness, we have 300 is equal to three halves lambda, which implies that lambda is equal to 600 divided by 3, which is 200 nanometers, which is out of the spectrum of visible.
01:36
Light.
01:38
And now we can use the opposite condition, the destructive condition in our case now, to see what is canceled...