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Wave optics and diffraction gratings

 
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Jun18-10, 11:15 PM   #1
 

Wave optics and diffraction gratings


1. The problem statement, all variables and given/known data

I'm really struggling with this one.

"A diffraction grating with 600 lines/mm is illuminated with light of wavelength 500 nm. A very wide viewing screen is 2.0 m behind the grating."

a) What is the distance between the two m=1 fringes?
b) How many bright fringes can be seen on the screen?

2. Relevant equations

d sinθ= mλ where d is the distance between the slits in the grating and m is the diffraction order of the fringes
y=Ltanθ where y is the distance between the fringes and L is the distance to the viewing screen

3. The attempt at a solution

I finally got the first part. I found d (d=1/N where is slits/mm) and then using d sinθ= mλ, I solved for θ for both fringes and then plugged this into y=L tanθ. I found the difference between these two y values (0.629 m each way) to get 1.3 m, which the book says is correct.

But now I have no idea how to solve the second part. How can I know how many fringes there are on the screen if I don't know the width of the screen?
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
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Jun19-10, 02:06 AM   #2

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What do you think, will that ray reach the screen which is diffracted at 90 degrees or more?

ehild
Jun20-10, 04:11 PM   #3
 
I tried that, though. The first fringe was at 17.45 degrees, so I figured you could go out to only 90 degrees in each direction; this means you could fit five fringes in either direction, plus the one in the middle, so 11 total. Doesn't that make sense?

But the book says there are supposed to be only seven fringes...
Jun20-10, 04:33 PM   #4

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Wave optics and diffraction gratings


You know that d sinθ= mλ. Plugging in the data, sinθ =0.3 m. sin(90°)=1, so 0.3 m <1. It can be 0, 0.3, 0.6, 0.9 and also -0.3, -0.6, -0.9.

ehild
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