00:01
Okay, so here we start with the intensity equation for some angle theta goes like the peak intensity times the cost squared of pi d sine theta over lambda.
00:27
Okay, so we're looking for when the intensity is going to be i not over 2.
00:34
So we want to solve for when i theta equals i not over 2.
00:39
This will give us the half angle.
00:42
Okay, so this will happen at what's called i theta over 2 when the intensity is half.
00:52
So if we look and we have i not over 2 equals i not, so cose squared, i, d, sine theta, 1ā2, i'll call it.
01:13
It's the half angle over lambda so of course we can divide by i not and we get one half has to equal the co squared of pi d sine theta one half over lambda okay so we could take the um the square root and then we can look at the coast inverse so we take the square root here we get one over root two the same thing as root 2 over 2.
01:59
And this is going to equal the cosine, pi, d, sine, theta, 1 -half over lambda.
02:13
I'm going to go to a new page here.
02:16
We know the cos inverse of root 2 over 2 is pi over 4.
02:28
So pi over 4 will equal pi, p.
02:34
D.
02:36
Sign...