00:02
Now, i know that the resolving power is n times m, whereas to use m equals 1, and n is going to be 5 ,000 lines per centimeter times 3 .5 centimeters.
00:26
So our answer is just going to be 5 ,000 times 3 .5.
00:34
Put that in a calculator.
00:37
Of course, i could probably do that in my head.
00:39
But anyway, the resolving power is 17 ,500.
00:47
Let's move one to b.
00:58
All right.
00:59
As i look in section 36 .5, i see that in order to resolve, the sodium doublet, you need to have a resolving power of 1 ,000.
01:15
17 ,500 is greater than 1 ,000.
01:21
So, yes, you can resolve that.
01:24
See, we've got here.
01:29
Okay, so it says that we're going to use the second order, and we're resolving spectral lines that are very close to 587.
01:46
8 .002 times 10 to the negative 9th power meter spectral line of iron.
02:07
Alright, so resolving power is lambda over delta lambda.
02:30
Longer than the iron line.
02:33
So that would be, i'm going to put wavelengths of i for wavelength of iron.
02:45
Okay.
02:48
What is the shortest wavelength you can distinguish from the iron line? okay.
03:00
Which i think i could really, could really just change this to i here...