# Calculating uncertainty with a set of angles

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1. Jan 11, 2017

### zexxa

1. The problem statement, all variables and given/known data
Hi there, the issue I'm having right now is finding the right way to calculate uncertainties when trigonometric terms come in play.

The question goes like this, we're given 2 sets of 5 angles in degrees, the incident angle (10.0, 20.0, 30.0, 40.0, 50.0) and the other is the refracted angle (6.9, 13.6, 20.0, 25.4, 31.1). The measurements were taken with an analogue protractor with a ±0.1° resolution.

The refractive index n is to be calculated with it's standard uncertainty.

2. Relevant equations

$$n = \frac{sin(α)} {sin(β)}$$$$dn = \sqrt { (\frac {δn} {δα} dα)^2 + (\frac {δn} {δβ} dβ)^2}$$

3. The attempt at a solution
I first tabulated the data and found n for each α,β entry and averaged all 5 values of n which was 1.466461.
The standard deviation of n is 0.024073, which then I used to find its Type A uncertainty with the equation $$u=\frac{σ}{\sqrt 5}$$
When I tried to find the Type B uncertainty however, I hit a wall. This was rest of my work:
$$\frac {δn}{δα}=\frac {cos(α)} {sin(β)}$$$$\frac {δn}{δβ}=\frac {-sin(α) cos(β)} {sin(β)^2}$$$$∴dn= \sqrt {n^2 (cot(β)^2 dβ^2 -dα^2) + \frac{dα^2}{sin(β)^2}}$$
I do not know what I should substitute in place of β when calculating since the value of β is a dependent variable in the experiment and have been at a loss for a few days.

Last edited: Jan 11, 2017
2. Jan 11, 2017

### BvU

Hello zexxa,

With quotients and products it's always easier to work with relative errors, like here $${\delta n\over n} = {\delta\alpha\over\tan\alpha} - {\delta\beta\over\tan\beta}$$

I copied your data to a spreadsheet and found $n = 1.4687$ as average, not your 1.466. And sigma 0.0217, not 0.0241
A typo ?

3. Jan 11, 2017

### zexxa

Hi BvU

Oh dear I put in the wrong numbers in my excel spreadsheet, your values there are right.
To be honest I don't see how you arrived at $${\delta n\over n} = {\delta\alpha\over\tan\alpha} - {\delta\beta\over\tan\beta}$$
I've been deriving the uncertainties using the second equation under "Relevant Equations" and can't seem to resolve my answer to match yours. But regarding your relative error, the same issue will still arise - what would you substitute in α and β, since the readings are not similar?

4. Jan 11, 2017

### Cutter Ketch

your derivatives are correct. I didn't check your algebra, but I presume your resulting expression for error is correct. For β you use the measured value for a single measurement. This gives you an error bar FOR THAT SINGLE MEASUREMENT. Each measurement will have a different error. Putting the measurements together to make a best estimate of n from all the measurements would then be a weighted average instead of a simple average giving more weight to the better measurements.

This makes sense because you can see how much better you can measure n when the incident angle is 45 than when it is 1. Those two measurements clearly have different error and shouldn't be given the same importance.

5. Jan 11, 2017

### BvU

Good we found that
$$\frac {δn}{δα}=\frac {cos(α)} {sin(β)}\ \ \Rightarrow\ \ \frac {δn}{n} = \frac {\cos(α)} {\sin(β)} {\sin\beta\over\sin\alpha}{δα}$$and
$$\frac {δn}{δβ}=\frac {-sin(α) cos(β)} {sin(β)^2}\ \ \Rightarrow\ \ \frac {δn}{n} = \frac {-\sin\alpha \cos(\beta)} {\sin^2(β)} {\sin\beta\over\sin\alpha}{δ\beta}$$
If we assume the measurements of $\alpha$ and $\beta$ are independent we can add these constributions in quadrature.

(you want to use \cos and \sin in tex and don't need the brackets)

You want to insert the actual $n, \ \alpha$ and $\beta$ for each measurement and sum over the measurements.

The post by Cutter is a refinement in the sense that each measurement in fact has a different weight due to the absolute error in the angle measurement. So for all summation terms a weight factor enters.

6. Jan 11, 2017

### zexxa

Thanks very much @BvU and @Cutter!
Cleared up a lot of misconceptions and I understand how it works now!

P.S. Thanks for the Tex tips, still not used to it yet but I'll get the hang of it :)

7. Jan 11, 2017

### Cutter Ketch

You are most welcome