Initially we wanted to find the errors in the spectra plot below. A discrete number of photons will hit the pixel in a certain amount of time, this number of photons follows a Poisson distribution. As the number of photoelectron also follows a Poisson the error is equal to the square root of the number of events, \sigma = \sqrt{N_{\rm{pe}}}. The spectra can then be plotted with the errors included.

- Spectra_error.png:

- Peak_fit.png:

amplitude = 2.45174694e+07

mean =32.1143568

sigma = 2.36989539

background = 1.44398361e+05

and the covariance matrix. The covariance matrix returns values for how much the fit parameters are correlated to each other. The diagonals of the covariance matrix return the variance of each of the fit parameters. For a Gaussian distribution the variance V[x], is equal to the square of the standard deviation, V[x} = (\sigma)^2. Therefore the error in the satndard deviation of the peak can be found from the squareroot of the correct element of the covariance matrix. The covariance matrix for this plot is

Amplitude | Mean | Sigma | Background | |

Amplitude | [1.53901604e+11 | -6.97141082e-01 | -1.63287377e+03 | -2.26365868e+08] |

Mean | [-6.97141082e-01 | 1.60878183e-03 | -2.72663680e-07 | 9.95929048e-03] |

Sigma | [-1.63287377e+03 | -2.72663680e-07 | 9.39810559e-04 | 2.33274473e+01] |

Background | [-2.26365868e+08 | 9.95929048e-03 | 2.33274473e+01 | 1.48494566e+07] |

.

.

.

The standard deviation of each of the fit parameters can then be found from the squareroot of the diagonal.

amplitude = (2.45174694 \pm 0.0392302949) e+07 pe

mean =32.1143568 \pm 0.0401096226

sigma = 2.36989539 \pm 0.0306563298

background = (1.44398361 \pm 0.0385349927) e+05 pe

**Plotting the standard deviation against the focus distance**

The standard deviation of the peak can then be plotted against the distance of the camera from the lens.

- Focus.png:

- Focus_fit.png:

**Plotting the standard deviation against camera angle**

The same process could then be repeated where the standard deviation of the peak can be plotted against the angle of the camera.

- Angle.png:

- Angle_fit.png:

**Mapping the pixel number to the wavelength of the light.**

To find the corresponding pixel number to wavelength in the above image the three lines were found to have values of 508.5822nm, 479.9912nm and 467.8149nm. By fittig a Gaussian to all three peaks as shown below the peak positions could be estimated.

- Spectra_position.png:

peak 1 = 168.80845947 \pm 2.65735394668

peak 2 = 282.11434879 \pm 2.37636473875

peak 3 = 553.879099215 \pm 2.8694730656

By then combining these peak values with the previously found peak values above it is possible to find the width of one frame in nm, this was found to be

frame width = 735 pixels = 80.9902944578 \pm 0.82257041263 nm

This frame can then be plotted with the correct wavelength values for each peak.

- Spectra_wavelength.png:

**Faint Peaks**

The plot below shows a strong peak on the left with two smaller peaks at pixel positions of approximately 100 and 400. These are the weaker lines on cadmium.

- Faint_Peaks.png:

- Faint_Peak_Fit.png:

This topic: Public > UserList > StewartBoogert > StewartBoogertAstronomy > StewartBoogertMSciProjects > StewartBoogertSpectroscopy2015 > 20151021_MSciSpectrocopyLab

Topic revision: r2 - 27 Oct 2015 - JosephBayley

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