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Intensity vs Pixel number CCD

  1. Sep 11, 2011 #1
    Hey Guys,

    i'm a first year astrophysics student and i got given this question to answer in a practical report:

    Examine the spectrum of the planetary nebula derived from the diffraction grating. You will produce a colour rendering of the spectrum together with a graphical representation of intensity versus pixel number.

    The pixel number scale can be replaced by a wavelength scale. Given that some of the bright emission lines are due to excited hydrogen and oxygen gas, determine a conversion between pixel number and wavelength. (Draw a graph showing pixel number on the x-axis and wavelength on the y-axis. The graph should be linear).

    now i have a pixel vs intensity graph which appears to have three sharp peaks. i'm not quite sure on what i'm supposed to go from here.

    thanks for your responses :)
  2. jcsd
  3. Sep 11, 2011 #2


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    Welcome to PF jaykob_hxc!

    The diffraction grating disperses the light in such a way that different wavelengths land in different places along a line on the CCD. So you're just calibrating your measurements into meaningful units by finding the scaling between "horizontal position along CCD" and "wavelength." To do this, you have to identify the spectral lines (intensity peaks) in your spectrum. These lines occur at known wavelengths (because they correspond to electron energy-level transitions in various gases that have been measured in labs here on Earth). Therefore, once you have identified the lines, you'll know what the wavelength difference is supposed to be between the line centres of any two adjacent lines. Compare that to the horizontal distance between them, and this gives you your scaling relation (or "calibration factor") between position on the detector and wavelength. Here's a hint for identifying the lines: planetary nebulae consist of gases whose atoms have been excited by energetic photons from a hot central source in the middle of the nebulae. There is a fairly standard set of "nebular emission lines" that results from this excitation. EDIT: and you've already been told that the lines present are probably hydrogen and oxygen emission lines.
    Last edited: Sep 11, 2011
  4. Sep 20, 2011 #3
    thanks for your help cepheid! :)

    my lecturer also put this up for anyone who reads this who was also stuck on a simillar question...

    If you do a plot of intensity vs pixel number (from the text file for
    the spectral profile that you produced in IRIS), you should see three main

    One of these (I think with the lowest pixel number) will be the zero-order
    image of the planetary nebula (which has the catalogue number NGC 6572), and you
    don't need to worry with this one in this context.

    The two other peaks will be due to hydrogen alpha, and to oxygen III. If the
    zero-order image has the lowest pixel number (i.e. x-position) of the three, the
    OIII line will be next highest, and the H-alpha line will have the highest pixel
    number or x-position.

    You will then have a difference in pixel number (difference in x-locations)
    corresponding to a difference in wavelength (given that OIII is 5007 Angstroms,
    and H-alpha is at 6563 Angstroms).

    The dispersion- which is what you're determining- is just (difference in
    wavelength)/(difference in pixel number), i.e. XX Angstroms/pixel. (Note that
    you need to estimate the pixel number that corresponds to the centre of each
    emission line, so probably the pixel number which has the highest intensity, in
    each of the OIII and H-alpha emission lines).

    For a spectrograph, the dispersion is one of the basic parameters that we need
    to know when analysing a spectrum, and using a source which has lines at known
    wavelengths (such as a planetary nebula) is one way to calibrate the
    spectrograph and its resulting images.

    also this link might help.

    thanks again!
  5. Sep 20, 2011 #4


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    Pixel saturation is a problem with all ccd cameras. Once a pixel is saturated, it bleeds into adjacent pixels, which confounds the data.
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