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Unformatted text preview: AST 3722C  Spring 2008 Homework #10 – Extra Credit – solutions • 1. Suppose you are told that the nightsky brightness atop Mauna Kea in Hawaii is 21.5 magnitudes per square arcsecond in Vband. You want to find how many photons per second per square centimeter per square arcsecond this provides. Here are the steps involved in getting there. (a) Use handout #17 to convert the magnitude within 1 square arcsecond into a flux density. Assume that flux density is measured per wavelength interval. Your answer will effectively have units of erg/s/cm 2 / ˚ A/arcsec 2 . Let m sky = 21 . 5, and call this flux density F . F = 10 . 4 m sky f (0), so F = 9 . 14 × 10 18 erg/s/cm 2 / ˚ A/arcsec 2 . (b) Calculate the power per collecting area provided by that 1 square arcsecond of sky. Assume that when you look in Vband you’re looking through a filter that is a step function centered on λ = λ = 0 . 55 μ m and is Δ λ = 0 . 089 μ m wide. Also assume that the sky’s flux density is approximately constant at all wavelengths across the V bandpass. This means that to figure out power per collecting area within 1 square arcsecond, you simply need to multiply by the bandwidth Δ λ . Express your answer in erg/s/cm 2 /arcsec 2 . The bandwidth is 0.089 μ m which is 890 Angstroms. Note that we need to do this conversion since F has units of per Angstrom. Call this power per collecting area per square arcsecond F . So: F = F Δ λ = 8 . 14 × 10 15 erg/s/cm 2 /arcsec 2 . (c) Convert the power (i.e. photon energy per unit time) to a photon number per unit time by assuming that all photons in V band have approximately the same energy, i.e. the energy of a photon of wavelength λ . Express your answer in photons/s/cm 2 /arcsec 2 ....
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 Spring '09
 Fernandez
 Astronomy, Photon, Light, Chargecoupled device, flux density, square arcsecond

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