Transmission Grating: Determining the Correct Order of Diffraction

AI Thread Summary
The discussion centers on determining the correct order of diffraction for a transmission grating problem. The user initially believes both options A and C could be correct but ultimately concludes that C is the right answer according to the mark scheme. The confusion arises from the relationship between the number of lines (N) and the order of diffraction (n), specifically how the angle changes with different orders. The user clarifies that while doubling N halves the line separation, the angles for different orders do not simply double, especially at larger angles. The final consensus confirms that C is indeed the correct answer based on the grating equation.
maxelcat
Messages
33
Reaction score
4

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

 

Attachments

  • upload_2018-6-22_14-50-51.png
    upload_2018-6-22_14-50-51.png
    12.9 KB · Views: 1,940
Last edited by a moderator:
Physics news on Phys.org
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.
 
Thanks for your reply. I don't 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 isn't according to the mark scheme
 
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
 
So which answer do you think is correct?
 
C -according to the mark scheme!
 
(c) is indeed the correct answer. Do you see why?
 

Similar threads

Back
Top