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Uncertainty propagation visible light spectrum

  • Thread starter zdenton
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  • #1
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Homework Statement


I have conducted an experiment which attempts to calculate the range of the visible light spectrum. Basically white light was shined through a diffraction grating (300 lines/mm) and diffraction theory is applied to calculate the wavelength.

So, here are the variables:
[tex]d=\frac{1}{300000}[/tex]

[tex]l=0.20[/tex]

[tex]\Delta l=0.001[/tex]

[tex]y=0.043[/tex]

[tex]\Delta y=0.005[/tex]


Homework Equations


[tex]\sin\alpha=\frac{\lambda}{d}[/tex]

[tex]\tan\alpha=\frac{y}{l}[/tex]


The Attempt at a Solution


I combined these equations to end up with:
[tex]\lambda=d\times\sin\left(\arctan\left(\frac{y}{l}\right)\right)[/tex]
The problem is that I don't know how to estimate an uncertainty for this equation. I know that for simple equations like [tex]y=q\times r[/tex] the uncertainty is [tex]\Delta y=\left(\frac{\Delta q}{q}+\frac{\Delta r}{r}\right)\times y[/tex]. Unfortunately I don't know how to apply this to a more complex equation. If anyone could lead me in the right direction as to an equation which would give the uncertainty for [tex]\lambda=d\times\sin\left(\arctan\left(\frac{y}{l}\right)\right)[/tex], it would be greatly appreciated.
 

Answers and Replies

  • #2
Redbelly98
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It can be done using calculus, if you've had calculus. But first I would get rid of the trig functions.

What's an equivalent expression for sin(arctan(x)) ?
 
  • #3
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[tex]\frac{x}{x^{2}+1}[/tex]
Substituting [tex]\frac{y}{l}[/tex] for [tex]x[/tex] gives:
[tex]\frac{\frac{y}{\left|y\right|}\times l}{\sqrt{y^{2}+l^{2}}}[/tex]

I hadn't thought of doing this, so it seems to be a step in the right direction. I have done limited calculus, I'm just finishing the first year of IB Math HL so we're starting on integration right now. I looked briefly at the wikipedia page for error propagation and didn't really understand it.
 
  • #4
diazona
Homework Helper
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Hmm... yeah, Wikipedia is being ridiculously detailed about this.

FYI, here's the usual case: if you have a function [tex]f(x, y, z)[/tex] and the uncertainties in the arguments are [tex]\delta x[/tex], [tex]\delta y[/tex], and [tex]\delta z[/tex], then the uncertainty in [tex]f[/tex] is
[tex]\delta f = \sqrt{\left(\frac{\partial f}{\partial x}\delta x\right)^2 + \left(\frac{\partial f}{\partial y}\delta y\right)^2 + \left(\frac{\partial f}{\partial z}\delta z\right)^2}[/tex]
Of course, there are some conditions on that formula, i.e. small, independent (uncorrelated) uncertainties and Gaussian distributions, but probably 99% of the time that formula is good enough.
 
  • #5
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OK, thanks for the help so far. I applied the above formula to my equation and received the following result:
[tex]\delta f = \sqrt{{\delta l}^{2}\,{\left( -\frac{d\,\left| l\right| \,y}{{l}^{2}\,\sqrt{{y}^{2}+{l}^{2}}}+\frac{d\,y}{\left| l\right| \,\sqrt{{y}^{2}+{l}^{2}}}-\frac{d\,\left| l\right| \,y}{{\left( {y}^{2}+{l}^{2}\right) }^{\frac{3}{2}}}\right) }^{2}+{\delta y}^{2}\,{\left( \frac{d\,\left| l\right| }{l\,\sqrt{{y}^{2}+{l}^{2}}}-\frac{d\,\left| l\right| \,{y}^{2}}{l\,{\left( {y}^{2}+{l}^{2}\right) }^{\frac{3}{2}}}\right) }^{2}}[/tex]

Substituting with the variables in my first post returns the result:
[tex]\delta f = 7.79\times 10^{-8}[/tex]

Which is exactly what I received when I tried using an online uncertainty calculator! Thank you so much!
 
Last edited:

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