Transmission grating Nλ=dsinθ

  • #1
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Homework Statement


This is a question we have been doing at school. The answer is C. I think A is correct as well.
upload_2018-6-22_14-50-51.png


Homework Equations


I have used nλ=dsinθ and put in test values.
I guess I should be able to do this by just 'inspecting the question' but the sin confused me. So I put in some test values as below.

The Attempt at a Solution


When I used 600nm for wavelength and 100,000 lines per meter the first order and second orders are at 3.4398 degrees 6.8921degrees. Doubling N then gives 6.8921 and 13.8865 degrees. This suggests that a is right... so what am I doing wrong please

 

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Answers and Replies

  • #2
kuruman
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I think you are confusing the number of lines ##N## with the order of a given line. Suppose you were to write the grating equation as ##d \sin \theta = m \lambda## where ##m## is the order of the line (here ##m=2##). The separation between lines, ##d##, is halved when the number of lines ##N## is doubled.
 
  • #3
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Thanks for your reply. I dont think i am confusing N and n. When N is doubled, d is halves like you say. Small n (or m if you like) is the feinge number. So ...

For fixed N then theta for n=1 seems to be half what it for n=2.... ie a is right.... which it isnt according to the mark scheme
 
  • #4
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But... i have sorted it now .... a is wrong because the angle doesn't necessarily double as you go from n=1 to n=2 - at small angles it approximately does, at larger angles (like in the diagram) it doesn't
 
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kuruman
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So which answer do you think is correct?
 
  • #6
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C -according to the mark scheme!
 
  • #7
kuruman
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(c) is indeed the correct answer. Do you see why?
 

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