Finding the largest angle from the central maximum (single slit diffraction)

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In a single slit diffraction problem, a slit width of 1400 nm and a light wavelength of 490 nm are given. The largest angle from the central maximum at which intensity is zero is sought, with the correct answer being approximately 44 degrees. The formula used is a sin(θ_n) = nλ, where 'a' is the slit width. To find the largest angle, one must determine the maximum value of 'n' that keeps sin(θ) less than or equal to 1, which corresponds to the largest angle being 90 degrees. The solution involves trial and error to find the appropriate value of 'n' until the largest angle that fits the criteria is identified.
HenryHH
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Homework Statement



A single slit, 1400 nm wide, forms a diffraction pattern when illuminated by monochromatic light of 490 nm wavelength. The largest angle from the central maximum at which the intensity is zero is closest to:

A) 44° B) 38° C) 35°


d = 1400 nm, lambda = 490 nm.


Homework Equations




sin(theta) = wavelength/d


The Attempt at a Solution



Plugging in variables, I have sin(theta) = 490/1400 = .35. To get theta, I multiplied .35 by sin inverse (sin^-1). I keep getting approximately 20 degrees, but the correct answer is A.) 44 degrees. I don't understand what I'm doing wrong, unless there's some extra step I should be doing... ?
 
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You have found the first minimum, or the one closest to the central maximum. You are being asked to find the minimum farthest from the central maximum.
 
Thanks. Will I need to use a different formula since I don't have any other variables?
 
No, it is the same formula:

a \text{sin}(\theta_n) = n \lambda

It is just more generalized, along you to find the nth minimum.
 
Thanks. This is probably a dumb question, but how would I figure out what to plug-in for n? I now know that I am finding the minimum farthest from the central maximum, but I don't know what number would correspond with that, if that statement makes any sense.
 
You will have to do this by trial and error. Figure out what the maximum meaningful diffraction angle is in a single slit experiment, and then find the biggest value of n that results in an angle less than that maximum angle.
 
I'm kind of confused... when you say trial and error, will I just be plugging in random numbers for n? Also, just to be sure, is a in the formula you posted the same as d (the slit width)?

My confusion with doing it the trial and error way deals with the fact that I won't know when I've gotten the correct answer. The correct answer is 44 degrees, but I don't know how they determined that 44 degrees was the largest angle versus any other angle...
 
HenryHH said:
I'm kind of confused... when you say trial and error, will I just be plugging in random numbers for n? Also, just to be sure, is a in the formula you posted the same as d (the slit width)?

Yes, they are the same.

My confusion with doing it the trial and error way deals with the fact that I won't know when I've gotten the correct answer. The correct answer is 44 degrees, but I don't know how they determined that 44 degrees was the largest angle versus any other angle...

If you use n = 3, you get \text{sin}(\theta) = 1.05. Is there any angle \theta such that that can be true? The same argument holds for all n > 2.
 
So basically, I'm just plugging in 1,2,3... for n until I get the biggest number for theta that also happens to be an answer choice? In other words, this particular problem wouldn't be solvable if it wasn't a multiple-choice question?
 
  • #10
The general way to solve this is to recognise that the largest angle is 90 degrees so this largest value of Sin∅ is 1
This gives the max value of n to be a/λ which must be rounded down tot find the number of minima
 

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