faq7 - FREQUENTLY ASKED QUESTIONS February 10, 2012 Content...

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FREQUENTLY ASKED QUESTIONS February 10, 2012 Content Questions I’m unclear on the variance: is the width of the distribution σ 2 or σ ? The “variance” is σ 2 , and the “standard deviation” is σ , the square root of the variance. The integral of a Gaussian going from μ - σ to μ + σ corresponds to 68% of the total area. There are various definitions of “width” that are different depending on the specific distribution– but in casual speech “width” would correspond more to σ than to σ 2 . How did you get those expressions for ˙ n s and ˙ n b ? These come from converting the specific flux (energy per area per time per wavelength) to number of photons per seen in your detector unit time. For the number of detected signal photons: The first factor is λF λ hc : this is F λ , which is energy per area per time per wavelength, converted to number of photons per area per time per wavelength: since E = = hc/λ is energy per photon, there are λ/hc photons per unit energy. Δ λ is the “bandwidth”: it’s the wavelength interval you’re looking at (this assumes a small enough interval that the other factors in the expression can be considered roughly constant as a function of wavelength– see next question). π
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This note was uploaded on 02/26/2012 for the course PHYSICS 105 taught by Professor Kruse,m during the Spring '08 term at Duke.

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faq7 - FREQUENTLY ASKED QUESTIONS February 10, 2012 Content...

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