0
$\begingroup$

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 $\lambda$?

$\endgroup$

2 Answers 2

1
$\begingroup$

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 $\sigma_\lambda = \lambda \cdot \delta\lambda$.

$\endgroup$
0
$\begingroup$

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:

$$Var(X)=\sigma_x^2$$

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 $x\frac{1}{D}$:

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

OR:

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

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.