Question

Assuming that 10.0% of a 100 W light bulb's energy output is in the visible range (typical for incandescent bulbs) with an average wavelength of 580 nm, and that the photons spread out uniformly and are not absorbed by the atmosphere, how far away (in km) would you be if 414 photons per second enter the 3.30 mm diameter pupil of your eye? (This number easily stimulates the retina.)

Answer 1

Power light radiated, P = 0.1*100 W

= 10 W

given wavelength, lamda = 580 nm
N = 414

Energy of each photon = h*c/lamda

= 6.626*10^-34*3*10^8/(580*10^-9)

= 3.427*10^-19 J

Intensity of light at the location of eye, I = power transfered/Area

= N*energy of one photon/Area of the pupil

= 414*3.427*10^-19/(pi*0.0033^2/4)

= 1.659*10^-11 W/m^2

let r is the distance from bulb to eye.

now use, I = P/(4*pi*r^2)

r^2 = P/(4*pi*I)

r = sqrt(P/(4*pi*I) )

= sqrt(10/(4*pi*1.659*10^-11))

= 2.19*10^5 m

= 219 km <<<<<<<<<---------------------Answer