Diffraction grating and Dirac comb

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SarahLou
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Homework Statement


I need to measure (with the ruler) the width the depicted sinc envelope and the period of the depicted Dirac comb light pattern.

And from the above I need to calculate the width of one slit a (i.e. aperture width) of the grating, and the period of the grating dx (i.e. distance between each slit). I have no idea where to start :(
diffraction.jpg

Homework Equations

The Attempt at a Solution


I suck at physics and spent few hours reading everything I could find on the internet on convolution, diffraction and Fourier theorem. I still have no clue what I need to measure!
 
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distance between each point? I really don't know :/
 
I forgot to mention the distance between the slit and the points on the paper was 1000mm and the wavelength was 633nm. I tried to calculate width of one slit a = 9Lλ/0.31 = 9(1000)0.000633/0.31= 18.3mm
 
SarahLou said:
I forgot to mention the distance between the slit and the points on the paper was 1000mm and the wavelength was 633nm. I tried to calculate width of one slit a = 9Lλ/0.31 = 9(1000)0.000633/0.31= 18.3mm
You mention two distances. Which one was 1000 mm? Do you realize that 1000 mm is 1 meter? I doubt the slits or the points on the paper were 1 meter apart.
 
vela said:
You mention two distances. Which one was 1000 mm? Do you realize that 1000 mm is 1 meter? I doubt the slits or the points on the paper were 1 meter apart.

There was 1 metre distance between the slit frame and the photographic paper that recorded the diffraction. 1 metre allowed the points to be perfectly focused. Six figure distraction is corresponding to the last image on the right I think, and I'm guessing the middle one above it corresponds to the slits and the distance between them?
 
Ah, okay, I totally misread what you wrote. I took it to mean the distance between the points was 1 m or the distance between the slits was 1 m whereas you meant the distance from the slits to the screen is 1 m.

You're right that the diffraction pattern corresponds to the figure on the bottom right, so by measuring the distance between the points, you can get a measure of what ##\delta \omega## is. Since ##\delta x## and ##\delta \omega## are related, you can, in principle, be able to figure out what ##\delta x## is. Note, however, that you can't actually measure what ##\delta \omega## is with the ruler; you're only getting a quantity that's proportional to ##\delta \omega##. If you don't see why that's the case, consider the units of ##\delta \omega##. So part of your task is to figure out how ##\delta \omega## is related to the linear distance between the points on the screen.

Do you see what you would measure on the diffraction pattern to figure out what ##a## is?