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 $m$ 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 2424 2694 2713 2722
$a_{0}$ 0.110.01 0.1180.014 0.1330.01 0.110.03
$b_{0}$ 60.41.5 61.011.63 57.21.8 58.70.9
$a_{1}$ (-1.613.28)e-06 (-1.241.77)e-05 (-3.261.25)e-05 (-7.5140.7)e-07
$b_{1}$ -0.510.16 -0.6390.183 -0.0150.28 -0.240.16
$\chi^2$ 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.

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


  • 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 $N_{pe} = 5.5 \times 10^{6}$, $\sigma_{N_{\rm{pe}}} = 2345$, $l = 510$ and $t = 0.5s$
  • 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:


Errorbar3.png -- JosephBayley - 17 Nov 2015

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Topic revision: r6 - 03 Nov 2016 - JamesAngthopo

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