Having little experience teaching uncertainties and error propagation to high school students I am learning a lot of new things this year. One thing in the textbook we use states that the uncertainty in Sinx is Cosx * absolute uncertainty in x expressed as radians. That's fine with me.

Now, if I then used Snell's law to find refractive index (Sini/Sinr) how do I propagate those uncertainties? Can I add them?


2 Answers 2


Assuming uncorrelated errors (which is not necessarily the case here, but is fair enough for high school), the error $\delta f$ in the function $f(\theta_i,\theta_r) = \frac{\sin(\theta_i)}{\sin(\theta_r)}$ due to uncertanties $\delta \theta_i$ and $\delta \theta_r$ is

$$\delta f = \sqrt{\left(\frac{\partial f}{\partial \theta_i}\delta\theta_i\right)^2+\left(\frac{\partial f}{\partial \theta_r}\delta\theta_r\right)^2}$$

in this case, $$\frac{\partial f}{\partial \theta_i} = \frac{\cos(\theta_i)}{\sin(\theta_r)}$$ and $$\frac{\partial f}{\partial \theta_r} = -\frac{\sin(\theta_i)}{\sin^2(\theta_r)} \cos(\theta_r)$$ This can be simplified somewhat if you note that

$$ \frac{\partial f}{\partial \theta_i} = f(\theta_i,\theta_r) \cdot \cot(\theta_i)$$ and $$\frac{\partial f}{\partial \theta_r} = -f(\theta_i,\theta_r) \cdot \cot(\theta_r)$$

Putting all of that together and doing a bit of algebra, $$\frac{\delta f}{f} = \sqrt{\delta\theta_i^2 \cdot \cot^2(\theta_i) + \delta\theta_r^2 \cdot \cot^2(\theta_r)}$$

This quantity is the relative uncertainty in $f$, and can be interpreted as a percent. If $\frac{\delta f}{f} = 0.1$, then your uncertainty in $f$ is 10% of its calculated value. That is, if you measure $f$ to be 1.5, then you would report your measurement as $f = 1.5 \pm 0.15$.


The two most common types of multivariate error propagation are addition and multiplication (and division, though this turns out to be the same as multiplication). For addition, we have $$\delta(a+b)=\sqrt{(\delta a)^2+(\delta b)^2},$$ for multiplication we have $$\delta(ab)=ab\sqrt{\left(\frac{\delta a}{a}\right)^2+\left(\frac{\delta b}{b}\right)^2},$$ and similarly for division we have $$\delta(a/b)=(a/b)\sqrt{\left(\frac{\delta a}{a}\right)^2+\left(\frac{\delta b}{b}\right)^2}.$$

Applying this to your question of $\delta(\sin(i)/\sin(r))$, we have $$\delta(\sin(i)/\sin(r))=(\sin(i)/\sin(r))\sqrt{\left(\frac{\delta \sin(i)}{\sin(i)}\right)^2+\left(\frac{\delta \sin(r)}{\sin(r)}\right)^2}.$$ And of course you already know the expression for $\delta(\sin(\theta))$.

In general, if you have two independent measurements $a$ and $b$ with associated uncertainties, and you have some calculated value $f(a,b)$, then the general form is $$\delta f=\sqrt{\left(\frac{\partial f}{\partial a}\delta a\right)^2+\left(\frac{\partial f}{\partial b}\delta b\right)^2}.$$ The reason for this can be loosely seen by considering the uncertainty for each dimension independently, which will be $\delta f_a=\frac{\partial f}{\partial a}\delta a$. We then keep in mind that these measurement are made independent of one another, meaning the uncertainty of one has no influence on the other. As such, the way to add them is geometrically at right angles (this is hand-wavey, but its the best I can do). With this, Pythagorean theorem gets you the rest of the way.

  • $\begingroup$ Thank you. That helps. Forgive one small question, the lowercase delta there reads as 'uncertainty', yes? $\endgroup$ Sep 20, 2019 at 1:52
  • $\begingroup$ Yes, that's correct. $\endgroup$
    – Bronicki
    Sep 20, 2019 at 2:24

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