Here's one way you could do it. If you know the luminosity (i.e. wattage) of the lamp, you could calculate the intensity like this:
[tex]I = \dfrac{L}{4\pi r^2}[/tex]
This of course assumes that the radiation is distributed isotropically, which means you're not enclosing the lamp in a mirror or anything. Next, you can multiply the intensity by the surface area of a human pupil to find the total power of the light entering the eye:
[tex]P = IA[/tex]
[tex]P = \dfrac{LA}{4\pi r^2}[/tex]
Now, you can impose light quantization ([itex]E = nhc/\lambda[/itex]), and write:
[tex]P = \dfrac{hc}{\lambda}\dfrac{dn}{dt}[/tex]
[tex]\dfrac{LA}{4\pi r^2} = \dfrac{hc}{\lambda}\dfrac{dn}{dt}[/tex]
[tex]\dfrac{dn}{dt} = \dfrac{\lambda}{hc}\dfrac{LA}{4\pi r^2}[/tex]
Now we talk to the physiologists, who tell us that the human eye can see as little as ten photons per second (if the source is flashing). So you go and measure the area of the human pupil, plug in the 500 meters and the luminosity of the bulb as well as the wavelength (i.e. color) of the light, and see if you've got at least ten photons per second.
Of course my calculation didn't really take into account certain wave effects, like diffraction through the pupil, so this is really only a first approximation.