Question

If a 80 W lightbulb emits 3.5 % of the input energy as visible light (average wavelength 550 nm) uniformly in all directions, estimate how many photons per second of visible light will strike the pupil (4.0 mm diameter) of the eye of an observer 3.0 km away.

Answer 1

if 3.5% of the energy is in visible light, then the power ofvisible light is:
    P = 0.035 * 80W
       = 2.8Js^-1

if the average visible wavelength is 550nm, then the average energyper photon is:
    E = hc/λ = 6.626*10^-34Js x2.99792458*10⁸ms^-1 / 550*10^-9m
                = 3.61*10^-19J/photon

assuming the lightbulb emits isotropically, then at a distance of1000m away, the flux is just the power of visible light per unitarea is just P divided by the area of sphere of radius r =1000m

    f = 2.8Js^-1 / [4π(1000m)²]
       =2.228*10^-7Js^-1m^-2

therefore, the power entering the eye is just f times the area ofthe pupil (radius 2*10^-3m):
    P_eye = f *πr_eye²
           =2.228*10^-7Js^-1m^-2 *π(0.002m)²
           =2.799*10^-12Js^-1

therefore, the number of visible light photons striking the pupilis:
    N = P_eye / E
       = 2.799*10^-12Js^-1/ 3.61*10^-19J/photon
       = 7.75*10⁶photons persecond