"Yep, I see that."

In this problem, we were asked to find the power output of a light bulb in a large, dark cave.  We were told that the bulb is 1 kilometer away and that the human eye has to receive 10 photons to register a source of light.  We were also given the equation for finding the energy of a proton, E = hv, where h is Plank's constant in erg*s and v is the frequency of the light.

Knowing the relationship between frequency, wavelength, and the speed of light to be c = vλ, I rearranged the Energy equation to read E = hc/λ.  I then solved to find the energy in ergs using the wavelength λ=580 nm, the average wavelength in the visual spectrum.  I multiplied this energy by 10, the number of photons reaching the eye if the light is barely visible.

Power is a quantity in units of energy per time.  Flux is in units of power per unit of area.  To find the power output, I first found the flux of the light by dividing the energy of 10 photons by the area of of the light's "surface" from 1 kilometer away, which is 1 x 10^10 square centimeters.  This left me with a flux of 3.3 x 10^-22 erg/(s*square cm).  To convert flux to luminosity, the astronomer's measure of power, I used the Inverse Square Law: L = F4πd^2.  With this equation, I found the luminosity to be 4 x 10^-11 ergs/s.