I am doing an experiment to measure the wavelength of light using a double-slit interference pattern. The general formula is

$$\lambda = \frac{xd}{nD},$$

but I have some observed uncertainties in the measurements. Namely, this becomes

$$\lambda = \frac{d(x \pm \sigma_x)}{n(D \pm \sigma_D)}.$$

Using error propagation rules, I rewrote this as

$$\lambda = \frac{xd}{nD} \pm \frac{d}{n}\left(\frac{\sigma_x}{x} + \frac{\sigma_D}{D}\right).$$

This comes from [incorrectly] merging the rules shown here. The problem is that the division rule uses relative uncertainty, and the constant multiplication rule uses absolute uncertainty. How do I derive the uncertainty in $\lambda$?


2 Answers 2


Multiplication and division, whether a constant is involved or not, both use the fractional uncertainty. You can think of them as two cases of using the binomial approximation, where you don't care about signs. If $a$ has absolute uncertainty $\Delta a$ and fractional uncertainty $\delta a = \frac{\Delta a}a$, then you can make statements like

\begin{align} a \pm \Delta a &= a \cdot\left( 1 \pm \delta a \right) \\ \frac 1{a \pm \Delta a} &= a^{-1} \cdot \left( 1 \pm \delta a \right) ^{-1} = \frac 1a \cdot \left(1 \mp \delta a\right) = \frac 1a \pm \frac{\Delta a}{a^2} \end{align}

Note that the distinction between $\pm$ and $\mp$ isn't important for uncertainties, because independent uncertainties add in quadrature, as mentioned in your reference.

In your case, then, the correct relationship is $$ \delta\lambda = \frac{\sigma_\lambda}\lambda = \sqrt{ \left(\frac{\sigma_x}{x}\right)^2 + \left(\frac{\sigma_D}{D}\right)^2}. $$ and the dimensionful uncertainty in your final wavelength is $\sigma_\lambda = \lambda \cdot \delta\lambda$.


Although the correct answer has been given, it seemed interesting to me to link it to where these rules come from: statistics.

Your $\sigma_x$ and $\sigma_D$ could be the standard deviations of $x$ and $D$. However, in combining variables, not the standard deviation, but the variance is used:


For e.g. the sum of two independent variables, this leads to exactly the rules described in your link:

$$Var(X+Y)= Var(X)+Var(Y) = \sigma_X^2 + \sigma_Y^2\\ \Rightarrow \sigma_{X+Y} = \sqrt{\sigma_X^2 + \sigma_Y^2}$$

For your problem, we need the following rules for the variances of a product of independent random variables and of a random function (see here and here )

\begin{align} Var(XY) &= E(X)^2.Var(Y) + E(Y)^2.Var(X) + Var(X).Var(Y) \tag{1} \\ Var(f(X)) &= E(f'(E(X))^2.Var(X) \tag{2}\end{align}

where E(X) is the mean of X (if you only measured once: your measurement outcome).

If you look at $\frac{x}{D}$ as $x\frac{1}{D}$:

$$Var(\frac{1}{D})=\sigma_{1/D}^2 = \frac{1}{D^4}.Var(D) = \frac{1}{D^4}.\sigma_D^2\\ Var(x\frac{1}{D}) = \sigma_{x/D}^2 = \frac{x^2}{D^4}\sigma_D^2 + \frac{1}{D^2}\sigma_x^2+\frac{1}{D^4}\sigma_D^2\sigma_x^2$$


$$\sigma_{x/D} = \frac{x}{D}\sqrt{\left(\frac{\sigma_x}{x}\right)^2+\left(\frac{\sigma_D}{D}\right)^2+\left(\frac{\sigma_x}{x}\right)^2\left(\frac{\sigma_D}{D}\right)^2}$$

Here you see that Rob's answer is an approximation of the more correct one above (although this is an approximation as well, because equation 2 is itself an approximation.


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