00:01
In this problem, you have a diffraction grading of 577 lines per millimeter.
00:08
Here is the diffraction grading.
00:09
D is the separation distance between the slits.
00:14
Now, as you would have seen with double slit interference, there is a path link difference, say, between any adjacent, and that is given by d -sign theta.
00:26
Now, in the case for defraction grading, the principal maximum, the principal one is when delta l, d -sign theta, is, and it's always the same between the delta l is always the same, when it's basically an integer multiple of the wavelength.
00:43
That means they're all in step with each other.
00:46
So when you get out here to this point p where you're looking at, if the top ones are crest at that point, all of them are to give you crest.
00:56
So you get a maximum there.
01:00
They're all in step with each other.
01:02
All in step.
01:03
Number one, number 5 ,000 are all in step.
01:07
So those are the principle maximum.
01:10
So delta l, g sine theta, is equal to m lambda.
01:16
M equals zero, one, two.
01:19
Same formula as the maximum for double slits.
01:21
Even though there's more going on here, that would have to be discussed if you wanted a full analysis of a grading, but we don't need to do that here.
01:29
Now, the only thing left before i talk about some other things, is getting the distance between the slits d.
01:37
Now think about when you did, we're told a frequency, 10 cycles per second.
01:43
If you invert that number, you get the length of a cycle, period.
01:47
Same thing goes on here.
01:49
They have a certain number of lines per millimeter.
01:53
You flip that over and you get the separation between the lines, d.
02:00
So each one, the frequency breaks up the second into pieces, 10 cycles per second.
02:07
This is breaking up a millimeter into certain number of lines.
02:09
It's the exact same idea in terms of inversion to get the spacing, either in time in that case or here in distance.
02:20
So d is going to be one over the 577, and this gives you a number in millimeters.
02:28
And i'll give you a number for what we're going to use.
02:30
This turns out to be 1 .73 times 10 of the minus 3 millimeters, which is 1 .73 times 10 to the minus 6 meters.
02:40
That's what we'll use in a couple of minutes.
02:45
Okay, so let's talk about some other things before that we need.
02:49
For any m, think of m equals 1.
02:54
Data is smallest for smallest lambda.
03:05
So let's say i have some colors, red, green, and blue.
03:09
So here's this, here's m equals zero.
03:12
Here'd be m equals 1.
03:13
You're looking at the screen from behind the gradient from a perspective standpoint.
03:18
So here, red, not red, blue being near the small end, small wavelength, red, violet is smaller, but i'll just use blue, blue, green, red being the longest.
03:35
So the angle from the center line to blue is smaller than it is to the red.
03:39
And likewise, likewise you get that over here, where you get the blue, green, and the red.
03:49
So smallest wavelength corresponds to the smallest angle.
03:53
I'll use that in a minute in a diagram.
03:55
Now, before we do that, we need one other quantity to do the rest of the problem.
03:59
There's our slits.
04:01
Here is our screen, length l away.
04:05
Here's our center line.
04:09
Here is some maximum.
04:12
Say for blue, maybe.
04:13
Doesn't matter.
04:14
Here is the angle theta.
04:17
This is y.
04:19
This is y.
04:25
Tangent theta is y over l.
04:29
Now, be careful here.
04:31
You may be tempted.
04:32
You may have seen in class that you would use, you'd stick this y over l and stick it in the formula for the maxima.
04:45
You can't do that here.
04:47
The only way you can replace sign with tangent if the angle is small, the angles are not small.
04:55
You're going to see me calculate...