Mapping wavelength with the grating position

In each of the frames the peaks mean position was noted along with its micrometer setting for the grating.
These peaks were then matched to their actual wavelengths found from NIST.
A plot of the three parameters was the made and a function for a plane was fitted with the equation.

$\begin{equation} \lambda = f(m,\mu) = I + a_{0}m+b_{0}\mu+a_{1}m^2+b_{1}\mu^2 \end{equation}$

where is the setting of the micrometer for the grating, $\mu$ is the mean position of the spectral line and $I,a_{0},b_{0},a_{1},b_{1}$ are constants.

A plot of the wavelength against the grating position and the position on the CCD can be seen below:

HeliumMap.png:
This process was repeated for the Cadmium, Low and High pressure Mercury lamps
The values of the constants were found to be

Parameter/Lamp	Cadmium	Helium	Low Mercury	High Mercury
I	242±4	269±4	271±3	272±2
$a_{0}$	0.11±0.01	0.118±0.014	0.133±0.01	0.11±0.03
$b_{0}$	60.4±1.5	61.01±1.63	57.2±1.8	58.7±0.9
$a_{1}$	(-1.61±3.28)e-06	(-1.24±1.77)e-05	(-3.26±1.25)e-05	(-7.51±40.7)e-07
$b_{1}$	-0.51±0.16	-0.639±0.183	-0.015±0.28	-0.24±0.16
$\chi^2$	133	62643?	329	232

Scan of peak across camera

A peak was places at one side of the frame and then the grating position was changed so that the peak was scanned across the whole frame.
The width of the spectral line was measured at each of these points and then plotted against the position on the camera.
The plots below shows that the width of the line increases towards the edge of the frame.
These plots show a scan of a peak in the helium and cadmium spectrum.

HeliumScan1.png:

CadmiumScan1.png:

Convolution with Fourier Transform

To try to speed up the fitting of the convolution Fourier transfrms were used to convolute the functions instead.

$\begin{equation} f(x) \ast g(x) = \mathcal{F}(f(x))\mathcal{F}(g(x)) \end{equation}$

By numerically calculating the Fourier transforms, multiplying them and then calculating the inverse Fourier transform, this function was then fitted to the data?
For this the guess and fit parameters are

Parameters	Guess	Fit
Gauss Amplitude	23195	253 ±(4.0e05)
mean	40	39.64 ±0.001
Gauss Sigma	3	1.28 ±0.001
Semicirc Radius	4	3.96 ±0.001
Semicirc Amplitude	23195	601 ±(9.5e05)
background	400	455 ±0.14

Where the method of convoling with and without Fourier transforms gave the same fit parameters.
The fit gave $\chi^2 = 25383$
The times for each of the methods were

Method	Time/s
Convolute	0.1675
Fourier	0.1585
difference	0.009

Errors

Our errors are still coming out very small which is increasing our $\chi^2$ value massively.
Started by scaling the $N_{\rm{pe}} \pm \sqrt{N_{\rm{pe}}}$ to $N_{\rm{pe}}$ per pixel per second so that:

$\begin{equation} X_{\rm{pe}} = \frac{N_{\rm{pe}}}{l t}\end{equation}$

the error is then:

$\begin{equation} \sigma_{X_{\rm{pe}}} = \frac{\sigma_{N_{\rm{pe}}}}{l t}, \end{equation}$

where $\sigma_{N_{\rm{pe}}} = \sqrt{N_{\rm{pe}}}$ .

For example we had $N_{pe} = 5.5 \times 10^{6}$ , $\sigma_{N_{\rm{pe}}} = 2345$ , and
so $X_{\rm{pe}} = 21569$
and $\sigma_{X_{\rm{pe}}} = 9.19$
This seems a very small error?
when plotted they are not visible

The $\chi^2 = 25383$ , which is very large however even the best fit points don't go through the error bars, as shown below:

-- JosephBayley - 17 Nov 2015

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Topic revision: r6 - 03 Nov 2016 - JamesAngthopo

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