What Are the Angles for Different Orders of Diffraction?

AI Thread Summary
The discussion focuses on calculating the angles for different orders of diffraction using a diffraction grating with 6000 lines/cm and monochromatic light from a helium-neon laser with a wavelength of 632.8 nm. Participants clarify that "6000 lines/cm" refers to the number of lines per centimeter, which helps in determining the grating spacing (d). The formula d = λ/N is suggested for calculating the distance between lines, leading to a value of 1.67 x 10^-6 m. The equation sin(θ) = mλ/d is confirmed for finding the diffraction angles. The conversation emphasizes understanding the units and applying the correct formulas to solve the problem.
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Homework Statement


Monochromatic light from a helium-neon laser(lambda=632.8nm) is incident normally on a diffraction grating containing 6000lines/cm.Find angles at which the 1storder,2nd..


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The Attempt at a Solution


here goes my solutions,
given lambda=632.8 X 10^9 ,but 6000lines/cm what does it mean?is that represents d,distance?i have no idea how to start,can you please hint me,thanks...
 
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Imagine a little fence which is constructed with four fence posts per meter. What is the spacing between the posts?

That should help you with the meaning of "6000 lines/cm".
 


heth said:
Imagine a little fence which is constructed with four fence posts per meter. What is the spacing between the posts?

That should help you with the meaning of "6000 lines/cm".

hmm..umm is that mean its actually refers to coverage of an area?so 6000lines/cm X10^-2 change into 6000lines/m and then use d formula d=l/N ?but i can't get the answer..:frown:
 


The units for your grating are "lines per cm". Just like the units for the fence are "posts per m".

No mention of area at all in the problem.

If you work out how to calculate the distance between the posts (which is easier to imagine than lines you can't really see) then you can use the same method to calculate the distance between the lines.

If you can't see how to use that, I'd suggest going back to your textbook and writing out in words what each of the letters in the equation you're trying to apply represents, and thinking about what the values would be for your problem.
 


heth said:
The units for your grating are "lines per cm". Just like the units for the fence are "posts per m".

No mention of area at all in the problem.

If you work out how to calculate the distance between the posts (which is easier to imagine than lines you can't really see) then you can use the same method to calculate the distance between the lines.

If you can't see how to use that, I'd suggest going back to your textbook and writing out in words what each of the letters in the equation you're trying to apply represents, and thinking about what the values would be for your problem.

ok,i got you,i use d=l/N ; l= 10^-2 , N =6000 thus, 10^-2/6000 so my d =1.67 X 10^-6 afterall i shall applied sin@=m X lambda / d ...CORRECT?
 

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