00:01
So once again, welcome to a new problem.
00:05
We still have a diffraction grading.
00:07
You have a diffraction grading.
00:09
And our goal in this problem is to find our d, which is the, so we have to find two things.
00:18
First thing we want to find is the slit separation.
00:26
So slit separation is this gap right here.
00:30
That's d.
00:32
That's the slit separation.
00:35
And the next thing we want to find out in this problem is a, which is the slit with, these are two items.
00:43
We have a mixture, so light is coming in with two wavelengths.
00:50
The faster one lambda 1 is 500 nanometers.
00:54
The second one lambda 2 is 600 nanometers, and they're moving perpendicular to the grading, the grading itself, the grading itself.
01:07
This distance, l is the length of the grading and the number of lines n is the number of lines coming in.
01:32
So assume this is lambda 1, 2, it's a mixture, it's a unique wavelength, which is a mixture.
01:42
And our goal is to find d and a.
01:45
And on top of that we also want to find the biggest order of our maxima you get from the grading.
02:06
So those are things you're dealing with.
02:09
One of the assumptions is that if the light is coming in like that, it's going to disperse that way with an angle of theta.
02:18
Theta is assumed to be less than or equal to 30.
02:24
So the first and second max of the wavelengths of lumbdas show up when this is true.
02:42
So when theta is less than equal to 30, that's when the lumbdas show up.
02:50
So we're just going to go ahead and jump right into the problem.
02:54
And the first part we're seeing is d sine theta equals to.
02:58
To m lambda.
03:00
Remember we had two wavelengths.
03:02
Lambda 1 is 500 nanometers and lambda 2 is 600 nanometers.
03:09
Those are the two wavelengths we had.
03:11
The assumption is that the angle is less than 30.
03:16
So it follows that if this is true, we can divide both sides by d and so sine theta equals to m lambda over d.
03:26
Or we can flip it and say m lambda over d.
03:30
Equals to sine theta but sine theta is less than sine 30 so it follows that m lambda all of d has to be less than equal to sine of 30 so this is lambda 1 2 if you can recall it's a mixed it's a mixed wavelength multiply both size by d we get to see that m lambda the mixed wavelength is less than or equal to d sine 30 alternatively you could see that as d sine 30 being greater than or equal to m times the mixed wavelength.
04:13
If you flip it, you get one over d is less than equal to sign 30, all of m lambda 1, 2, which is mixed wavelength.
04:28
Recall the relationship between the number of lines, n, and the length, length together with the separation, meaning that the number of lines is inversely proportional to the separation.
04:47
If you increase the separation, then you're decreasing, or rather if you increase the number of lines, you're decreasing the separation.
04:55
And so looking at the equation, we can see that there's an inverse relationship.
05:01
The resolving power, resolving power is therefore proportional to the number of lines.
05:14
And so we want to maximize this dispersion or the resolving power.
05:21
We want to maximize that.
05:25
So if we maximize n, then 1 over d will also be a maxima.
05:34
And for that to happen, you see that since one of the d is a maxima and we had one of the d is less than or equal to sign 30m lambda 1, 2.
05:51
So we had a choice.
05:52
Do we take less than or do we take the equal to? we take the equal to because we want to maximize 1 over d.
05:59
Why is the 1 over d maximized? is because we want to maximize the resolving power.
06:06
That's one of the goals of the problem.
06:08
We want to maximize the resolving power.
06:10
So then one over d has to equal m lambda 1 -2 over 5 -1 -2 over sign of 30.
06:18
So we take the equal one and we ignore them less than because our goal is to maximize.
06:23
So we can now comfortably solve for d and see that d is sine 30 over m.
06:29
The mixed lambda at maximum, at maxima, m equals to 2, the order is 2, and lambda 2 is 600 nanometers.
06:42
So that's when the max separation happens.
06:48
These two conditions are made...