1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

CCD quantum efficiency

  1. Oct 24, 2005 #1
    Hey, this is just a quick question on the quantum efficiency of CCDs. First of all, you are given the quantum efficiency of the CCD and the throughput value of the filter. So do you just multiply the throughput by the original quantum efficiency to obtain the quantum efficiency of the filter/CCD combination??

    For example, if you had a quantum efficiency of 80% and a throughput of 50% for a particular wavelength, then the overall quantum efficiency (at that wavelength) would be:

    QE = (0.8*0.5)*100 = 40%

    Is that right?

    Last edited: Oct 25, 2005
  2. jcsd
  3. Oct 25, 2005 #2
    haha no takers?

    Well I just have one more question whether anyone can help or not : )

    Given that you know the total count rate from a star and you know the amount of photons detected based on the visible/red filter you're using how do you calculate V-R?

    By the way the above question is just an estimation and you give your assumptions.

    I saw in the lecture notes that you could relate apparent magnitude to the ratio of the detected count rate to the total count rate.

    V-R = Mv - Mr
    Given that the star's distance isn't changing, the subtraction of the absolute magnitudes is equivalent to the subtraction of the apparent magnitudes.

    ie V-R = mv-mr

    Where mv = -2.5*log(Sv/So) and mr = -2.5log(Sr/So)

    In the above expressions Sv is the count rate detected through the peak wavelength of the visible filter and Sr is the similar, but with a red filter. "So" is just the total count rate from the star.

    I guess the only assumption that I can gather from this is that there is no absorption through the interstellar medium.

    Sorry if this is a muddled explanation.

    Thanks for any feedback
  4. Oct 25, 2005 #3


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    It looks right, but I'm not sure I would use the term "quantum efficiency" to describe that combination. It will give you the fraction of incoming photons that are detected.

    Actually, the apparent magnitude scale needs to be normalized to some standard. In other words, we want to be able to compare the apparent magnitudes from one star to the next. This would be impossible to do if every apparent magnitude were normalized to the total brightness of the star in question. Rather, the "So" in your equation would have to be the normalization of the magnitude system. This is often taken to be the count rate from Vega in the band in question, but it depends on the magnitude system being used.

    For this problem, however, it doesn't matter. If you're trying to solve for the color, you just get:

    [tex]m_V - m_R = -2.5 log(\frac{S_V}{S_0}) + 2.5 log(\frac{S_R}{S_0}) = -2.5 log(\frac{S_V}{S_R})[/tex]

    Notice that the normalization cancels out. This is because you're now comparing the star's flux from one band to another (not from one star to another).
  5. Oct 25, 2005 #4
    Haha sort of had it... : )

    Thanks for pointing that out man.

    Just out of interest though, is the Vega magnitude system the most widely used? Or do different magnitude systems only suite different situations??
  6. Oct 25, 2005 #5


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    The system normalized to Vega is the Johnson magnitude system. It used to be the most widely used, but the Sloan Digital Sky Survey magnitude system (u, g, r, i, and z) has been heavily used of late. It does, as you suggest, depend on the application.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook