What Are the Angles for Different Orders of Diffraction?

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Homework Help Overview

The discussion revolves around a problem involving diffraction grating and the calculation of angles for different orders of diffraction using monochromatic light from a helium-neon laser. The specific parameters include a wavelength of 632.8 nm and a grating with 6000 lines per cm.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the meaning of "6000 lines/cm" and its relation to the distance between lines on the grating. There is an attempt to clarify how to calculate the distance 'd' using the formula d = λ/N, where λ is the wavelength and N is the number of lines.

Discussion Status

Some participants have provided guidance on interpreting the units of the grating and suggested methods for calculating the necessary distances. There is an ongoing exploration of how to apply the relevant equations, with some participants expressing uncertainty about their calculations.

Contextual Notes

Participants are navigating the definitions and relationships between the variables involved in the diffraction problem, with some confusion about the units and their implications for the calculations.

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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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