# Too small error on the calculus of wavelenght

I have this function: $$\lambda=d \sin(\arctan(\frac{x}{z}))$$ and I want to find its absolute error. $d$ is a constant ($10^{-6}$), $x =(0.716 \pm 0.001)$ m, and $z=(1.000 \pm 0.001)$ m.

For the error of $\lambda$ I have used $$\Delta\lambda=\sqrt{\left(\frac{\partial\lambda}{\partial x}\Delta x\right)^2 + \left(\frac{\partial\lambda}{\partial z}\Delta z\right)^2}\\ \Delta\lambda=d\sqrt{\left(\frac{z\Delta x}{(z^2+x^2)(\sqrt{\frac{x^2}{z^2}+1})}\right)^2 + \left(\frac{x\Delta z}{(z^2+x^2)(\sqrt{\frac{x^2}{z^2}+1})}\right)^2}$$

And I have obtained $6.6 \cdot 10^{-10}$ m. I expect a much larger error. Is this wrong?

• You also have to put the error in perspective, because the relative error ($\frac{\Delta\lambda}{\lambda}=0.001$) is similar to that of $x$ and $z$. – fibonatic Dec 17 '13 at 19:51

You have $\lambda=d\frac x{\sqrt{x^2+z^2}}$ It looks OK to me. The errors in each $x,z$ are about one part in $1000$ and your final error is about that. Since the derivatives are a bit less than $1$, that seems sensible.
• many thanks for your answer! Just the last question: if I use a diffraction grating to measure $\lambda$, characterized by a pace of $10^{-6}$m, I obtain $\lambda=582,2$ nm. If I use another diffraction grating characterized by a pace of $2*10^{-6}$ m, I obtain $\lambda$=635,3 nm. The expected value is $\lambda=632,8$ nm. So the second value is Ok, but which could be the causes of this big difference respect with the first value that I have found? Thanks again! – sunrise Dec 17 '13 at 20:51
• I have no idea. Note that even the second one is off by $2.5$ nm, which is four times larger than you calculate above. – Ross Millikan Dec 17 '13 at 20:54