Question

The average threshold of dark-adapted (scotopic) vision is 4.00 10-11 W/m2 at a central wavelength of 500 nm. If light with this intensity and wavelength enters the eye and the pupil is open to its maximum diameter of 7.20 mm, how many photons per second enter the eye

Answer 1

E = h f = h c / λ energy of single photon

n E = 4.00E-11 W/m^2 energy required for visibility

n h c / λ = 4.00E-11 number of photons required for visibility (per square meter)

n = 4.00E-11 * 5.00E-7 / (6.63E-34 * 3.00E8)

n = 20.0E-18 / 19.9 E-26 = 1.00E8

100 million photons would be required on 1 m^2 to create visibility

a / A = π * (3.6E-3)^2 / 1 = 4.07E-5 fraction of area available

4.07E-5 * 1.00E8 = 4.07E3 photons required