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homework10solutions

# homework10solutions - AST 3722C Spring 2008 Homework#10 –...

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Unformatted text preview: AST 3722C - Spring 2008 Homework #10 – Extra Credit – solutions • 1. Suppose you are told that the night-sky brightness atop Mauna Kea in Hawaii is 21.5 magnitudes per square arcsecond in V-band. 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 V-band 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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homework10solutions - AST 3722C Spring 2008 Homework#10 –...

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