Uncertainty problem

  • #1
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Homework Statement


A common method for finding wavelength is to use diffraction grating. The relationship between wavelength λ and the angle of max intensity θ for first order interference is λ = d*sinθ where d is the spacing between lines on the grating, which is the inverse of the grating line density N (d=1/N). This data was taken on a laser of unknown wavelength:
N = 602 ± 2 lines/mm
θ = 22.4°± 0.4°

Homework Equations


Δλ = √[ ( Δd*∂λ/(∂d) )2 + Δθ*∂λ/(∂θ) )2 ]
σx = (w/2)/√6 ~ pdf uncertainty for analog measurements // I wouldn't think this is needed, since it's for measurements with calibration uncertainty...right?

The Attempt at a Solution


Well, the above equation is part of the attempt.
The partial with respect to distance (d) is sinθ.
The partial with respect to θ is d*cosθ.
∴ Δλ = √[ ( 2*sin(22.4°) )2 + 0.4*cos(22.4°)/602 )2 ] = 0.762

λ = 6.33 ± 0.762 mm
 

Answers and Replies

  • #2
BvU
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Hello adamaero, :welcome:

Is 2 the uncertainty in d ? Would that be in meters, or in millimeters ? :smile:
 
  • #3
gneill
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Hi adamaero, Welcome to Physics Forums!

I don't see where the uncertainty in N appears in your calculations. You aren't given an uncertainty for d, yet it looks like you've used the value 2 for it ???

Write out the actual function that you need to differentiate by first substituting in the known relationship between N and d. THen you can take the partial derivatives with respect to the variables for which you have uncertainties.

Another thing, in your final results you give the wavelength and its uncertainty the units mm. This is unlikely for a laser light wavelength. I'd expect something in the nm (nanometer) range. So check your calculation's orders of magnitudes.


Edit: Whoops! BvU beat me to it!
 
  • #4
BvU
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Second gneill re order of magnitude. Easily fixed.

For error propagation you can also do a rough check: relative errors add up quadratically in products and quotients; absolute errors in sums and differences. So here 1/N has a 0.3% error, just like N itself. ##\sin\theta## is about 2%, so that's way bigger and you can pretty much ignore the 0.3% when adding in quadrature.
The 'check' then already gives the result to adequate precision

Google "He Ne laser wavelength" to find an uncanny precise agreement with your result !
 
  • #5
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I don't see where the uncertainty in N appears in your calculations. You aren't given an uncertainty for d, yet it looks like you've used the value 2 for it ???
I thought I could just do the inverse, and it would be the same. Why not?

Write out the actual function that you need to differentiate by first substituting in the known relationship between N and d. THen you can take the partial derivatives with respect to the variables for which you have uncertainties.
Δλ = √[ ( ΔN*∂λ/(∂N) )2 + Δθ*∂λ/(∂θ) )2 ]
∂λ/(∂N) = -sin(θ)*N^(-2)
∂λ/(∂θ) = cos(θ)/602

θ = 0.39095 ± 0.00698
Why does this problem require that I switch to radians?

ΔN = 0.5
Δθ = 0.00698

∴ Δλ = √[ ( -.5*sin(0.39095 rad)/6022 )2 + 0.00698*cos(0.39095 rad)/602 )2 ] = 1.334*10^(-4)
which isn't in nm--where did I go wrong? I think it has something to do with bolded partial.
Another student got 633±11 nm.
 
Last edited:
  • #6
BvU
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it has something to do with bolded partial
No. It has to do with units.
 
  • #7
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No. It has to do with units.
I converted to radians. Can you be more specific please?
 
  • #8
gneill
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I thought I could just do the inverse, and it would be the same. Why not?

Try using the derivative method to find the uncertainly for f(N) = 1/N. Compare with 1/ΔN.

Why does this problem require that I switch to radians?

Radians are the "natural" units for angles. The trig functions are based on length ratios on the unit circle and the angles can be identified with arc lengths along the perimeter of the unit circle.
 
  • #9
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Try using the derivative method to find the uncertainly for f(N) = 1/N. Compare with 1/ΔN.
1/N = -1/N2
1/ΔN...what? Why would I take the derivative of this? I don't know what to do with the delta.
 
  • #10
gneill
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1/N = -1/N2
1/ΔN...what? Why would I take the derivative of this? I don't know what to do with the delta.
No. You're saying that your used 1/ΔN for the uncertainty in d. I'm suggesting that you compare the value of 1/ΔN with what you obtain using the "proper" method.
 
  • #11
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1/ΔN = 1/2
Do you mean this is incorrect?

1/N = -1/N2
I don't know what you're trying to point out.
 
  • #12
gneill
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1/ΔN = 1/2
Do you mean this is incorrect?
Yes. It is incorrect.
1/N = -1/N2
I don't know what you're trying to point out.
I'm trying to point out why it's incorrect. Suppose your function is f(N) = 1/N. You are given values N = 602, ΔN = 2. What's the uncertainty in f(602)?

You're using the derivative method to find the uncertainty in your function λ = d sin(θ). I'm suggesting that you try the same method on d = 1/N separately to check your assumption about the uncertainty in d being 1/ΔN.
 
  • #13
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I just did trial by error until I got that other students answer. So I guess ΔN = 2. That makes sense.
Why couldn't anyone just say that?
I suppose I was overthinking it.
 
  • #15
gneill
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I just did trial by error until I got that other students answer. So I guess ΔN = 2. That makes sense.
Why couldn't anyone just say that?
I suppose I was overthinking it.

You were given ΔN = 2 in the problem statement:

N = 602 ± 2 lines/mm
θ = 22.4°± 0.4°
 
  • #16
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You were given ΔN = 2 in the problem statement:
I know, that's what I'm saying.
I had that part correct in the initial (wrong) solution. Why couldn't anyone just say that when I was overthinking that part, and that I had it right in the beginning?
 
  • #17
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Is everyone reading the posts too fast or something?
Simply saying something along the lines that the initial partial equation was only part correct, and that the degrees needed to be in radians would have helped the most.
I suppose you did, sort of, but it just seemed like a round about way of saying so. I assume everyone's trying to get me to figure it out myself, but...I really didn't.
That's the crux of it. It's hard helping people. Giving too much help is worse than giving not enough. Get what I mean?
 
  • #18
gneill
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Hi adamaero, I agree that it can be tricky to find the best way to help someone without giving too much help in the process. It can sometimes be difficult to judge precisely what is tripping up questioner, and the best way to shed light on the issue so that the "aha!" of understanding takes place.

In this case it seemed that your problem lay with assuming that if d = 1/N and the uncertainty in N is ΔN = 2 (lines per mm), then the uncertainty in d would be 1/ΔN, or 1/2. I was trying to lead you to the conclusion that the uncertainty in 1/N is not 1/ΔN without just declaring it. I think it's important that students be able to confirm assumptions with a bit of investigation. That's why I suggested you apply the basic uncertainty process to the function f(N) = 1/N with the given data, to see if the uncertainty in 1/N matches your assumption of 1/ΔN.

Your competence with applying partial differentiation and the error analysis formulas led me to believe that you would see the issue quickly. If I misjudged I apologize. I didn't plan on derailing your progress with the digression.
 
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