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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 youre looking through a filter that is a step function centered on = = 0 . 55 m and is = 0 . 089 m wide. Also assume that the skys 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 . Express your answer in erg/s/cm 2 /arcsec 2 . The bandwidth is 0.089 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 = 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 . Express your answer in photons/s/cm 2 /arcsec 2 ....
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This note was uploaded on 11/09/2009 for the course AST 4700 taught by Professor Fernandez during the Spring '09 term at University of Central Florida.
- Spring '09