D
Danielle D
Guest
Alpha Centauri A lies at a distance of 4.4 light-years and has an apparent brightness in our night sky of 2.7 x 10^-8 watts/m^2. Recall that 1 light-year = 9.5 x 10^12 km= 9.5 x 10^15 m.
PART A: Use the inverse square law for light to calculate the luminosity of Alpha Centauri A.
PART B: Suppose you have a light bulb that emits 75 watts of visible light. (Note: This is not the case for a standard 75-watt light bulb, in which most of the 75 watts goes to heat and only about 7-11 watts is emitted as visible light.) How far away would you have to put the light bulb for it to have the same apparent brightness as Alpha Centauri A in our sky? (Hint: Use 75 watts as L in the inverse square law for light, and use the apparent brightness given above for Alpha Centauri A. Then solve for the distance.)
You don't have to answer both parts if you don't know both
PART A: Use the inverse square law for light to calculate the luminosity of Alpha Centauri A.
PART B: Suppose you have a light bulb that emits 75 watts of visible light. (Note: This is not the case for a standard 75-watt light bulb, in which most of the 75 watts goes to heat and only about 7-11 watts is emitted as visible light.) How far away would you have to put the light bulb for it to have the same apparent brightness as Alpha Centauri A in our sky? (Hint: Use 75 watts as L in the inverse square law for light, and use the apparent brightness given above for Alpha Centauri A. Then solve for the distance.)
You don't have to answer both parts if you don't know both