Uncertainty propagation visible light spectrum

[zdenton]

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:
d=\frac{1}{300000}

l=0.20

\Delta l=0.001

y=0.043

\Delta y=0.005

Homework Equations

\sin\alpha=\frac{\lambda}{d}

\tan\alpha=\frac{y}{l}

The Attempt at a Solution

I combined these equations to end up with:
\lambda=d\times\sin\left(\arctan\left(\frac{y}{l}\right)\right)
The problem is that I don't know how to estimate an uncertainty for this equation. I know that for simple equations like y=q\times r the uncertainty is \Delta y=\left(\frac{\Delta q}{q}+\frac{\Delta r}{r}\right)\times y. 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 \lambda=d\times\sin\left(\arctan\left(\frac{y}{l}\right)\right), it would be greatly appreciated.

 

[Redbelly98]

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)) ?

 

[zdenton]

\frac{x}{x^{2}+1}
Substituting \frac{y}{l} for x gives:
\frac{\frac{y}{\left|y\right|}\times l}{\sqrt{y^{2}+l^{2}}}

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.

 

[diazona]

Hmm... yeah, Wikipedia is being ridiculously detailed about this.

FYI, here's the usual case: if you have a function f(x, y, z) and the uncertainties in the arguments are \delta x, \delta y, and \delta z, then the uncertainty in f is
\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}
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.

 

[zdenton]

OK, thanks for the help so far. I applied the above formula to my equation and received the following result:
\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}}

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

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