2
$\begingroup$

This is from John Taylor- "An Introduction to Error Analysis"
The passage asks us to calculate the uncertainty of refractive index n using Snell's Law
$$\sin i=n \sin r$$
The fractional uncertainty in $n$ is the quadratic sum of those in $\sin i$ and $\sin r$:
$$\frac{\delta n}{n}=\sqrt{\left(\frac{\delta \sin i}{\sin i} \right)^2+\left(\frac{\delta \sin r}{\sin r}\right)^2}$$
The calculation is then carried out as shown in the table below
enter image description here
And then in the last paragraph, the author mentioned that

"If you repeat an experiment like this one several times, the error calculations can become tedious if you do them for each repetition. You should recognize, however, that you almost never need to do the error calculations for all the repetitions; if you find the uncertainties in $n$ corresponding to the smallest and largest values of $i$ (and possibly a few intermediate values), then these uncertainties suffice for most purposes."

I don't understand why don't we need to do the error calculations for all the repetitions, since the $\frac{\delta n}{n}$ values look quite different for each repetition in the table? And what does he mean by "if you find the uncertainties in $n$ corresponding to the smallest and largest values of $i$"?

Really appreciate the help!

$\endgroup$
0
$\begingroup$

I think what the author means is that if you had values $i=30^\circ$ and $i=19^\circ$ then you could make an estimate of the error as being about $7\%$ as midway between $9\%$ and $5\%$.

The mention of intermediate values is there to make the interpolation of the error of intermediate values more accurate.


Note that in this example you could make the error calculation easier by noting that $\delta(\sin i)\approx \delta i \cos i $ and so $\dfrac{\delta ( \sin i)}{\sin i }\approx \delta i \cot i$ where $\delta i$ is in radians.

$\endgroup$
1
$\begingroup$

There is no reason you shouldn't do the error calculation for each measurement - especially if you enter the measurements into a spreadsheet.

However, I think all the book is getting at is that there is a clear relationship between the value of the angle and the size of its contribution to the error in $n$.

If $y = \sin \theta$, and there is a (small) error in $\theta$, then it is the case that $$\frac{\Delta y}{y} \simeq \frac{\cos \theta\ \Delta\theta}{\sin \theta} = \frac{\Delta\theta}{\tan \theta}$$

Thus as $\theta$ increases, the percentage error decreases, as you see in the table.

Or, wrapping up the whole expression for $n$ and assuming $i$ and $r$ are measured with the same precision $\Delta \theta$ $$ \frac{\Delta n}{n}\simeq \Delta\theta \left( \frac{\cos^2 i}{\sin^2 i} + \frac{\cos^2 r}{\sin^2 r}\right)^{1/2}$$

A bit of trigonometry and use of Snell's law$^{*}$ shows you this is a decreasing function of $i$. $$\frac{\Delta n}{n}\simeq \Delta\theta\left(2\cot^2 i +\frac{n^2-1}{\sin^2 i}\right)^{1/2}$$ Of course, we don't know $n$ when we start the experiment, so all this shows is that the percentage error in $n$ should get smaller smoothly as $i$ increases.

$^{*}$ Replace $\sin^2 r$ with $\sin^2 i/n^2$ and $\cos^2 r$ with $1 - \sin^2 i/n^2$. $$ \frac{\Delta n}{n}\simeq \Delta\theta \left( \frac{1-\sin^2 i}{\sin^2 i} + \frac{1-\sin^2 i/n^2}{\sin^2 i/n^2}\right)^{1/2}$$ $$\frac{\Delta n}{n}\simeq \Delta\theta \left( \frac{1-\sin^2 i +n^2 -\sin^2 i}{\sin^2 i}\right)^{1/2}$$ $$\frac{\Delta n}{n}\simeq \Delta\theta \left( \frac{2 - 2\sin^2 i + (n^2 -1)}{\sin^2 i}\right)^{1/2} $$$$\frac{\Delta n}{n}\simeq \Delta\theta\left(2\cot^2 i +\frac{n^2-1}{\sin^2 i}\right)^{1/2}$$

$\endgroup$
  • $\begingroup$ Hi, may I know how do you get to the last expression? $\endgroup$ – WeiShan Ng Mar 31 '18 at 10:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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