Finding Maximum Number of Interference Maxima?

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Homework Statement


Light of wavelength λ = 535 nm shines through two narrow slits which are 670 μm apart. What is the maximum number of interference maxima which could conceivably be observed (assuming that diffraction minima do not extinguish them and the screen is arbitrarily large)?Your answer should be an integer. There is no sig-fig requirement for your answer

  1. Hint: What is the maximum angle allowed? Did you remember to count the maxima below the center?

Homework Equations


When I tried to answer the first time, the question gave me the a hint to look for the max angle allowed, so I think I'll need to use
dsinθ = mλ, where d=slit distance = 670 μm = 6.7 x 10-7m and λ=535nm = 5.35 x 10-7m
I also think I'll need y=mdL/λ, where y=distance between maxima, m= maxima integer, d=slit distance, L=distance between slits and screen (not given) and λ= wavelength, so maybe not this equation (because L isn't given?)
I thought about using wsinθ=mλ (where w=slit width), but because the question states to assume diffraction minima do not extinguish the maxima, I didn't think it was necessary to factor this in as well.


The Attempt at a Solution


Honestly I'm not even sure where to begin with this. I tried solving for the maximum angle as the question gave me feedback, and I got 52.98°. But I don't know if this is even relevant or what. For an earlier question I did m= d/λ and got the correct answer, but for some reason this does not work here.

Thanks in advance for any help!
 

Answers and Replies

  • #2
ehild
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What can be the diffraction angle maximum ? Can it exceed 90°?
 
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If it's 90°, wouldn't the equation be dsin(90°) = mλ? And sin(90°) = 1
Rearranging for m: m=d/λ= (6.7 x 10-7m)/(5.35 x 10-7m) = 1.26. Is this the number of maxima produced? Or just for one side of the central maxima?
So I multiplied by 2 (for either side) and got 2, but that's wrong. So I'm getting confused somewhere.
 
  • #4
ehild
Homework Helper
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If it's 90°, wouldn't the equation be dsin(90°) = mλ? And sin(90°) = 1
Rearranging for m: m=d/λ= (6.7 x 10-7m)/(5.35 x 10-7m) = 1.26. Is this the number of maxima produced? Or just for one side of the central maxima?
So I multiplied by 2 (for either side) and got 2, but that's wrong. So I'm getting confused somewhere.
Yes, m would be 1,26, but it must be integer. You need to tale the integer part. It is 1, so the orders of ±1 are observed, one maximum at both sides of the central maximum. The question was What is the maximum number of interference maxima which could conceivably be observed? . You observe the central maximum, too.
 
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