Question

The threshold of dark-adapted (scotopic) vision is 3.3 ? 10^?11 W/m² at a central wavelength of 500 nm. If light with this intensity and wavelength enters the eye when the pupil is open to its maximum diameter of 8.3 mm, how many photons per second enter the eye?

photons/s

Answer 1

maximum area of pupil through which light can enter = pi *r² = 3.14*(8.3*10^-3 )² /4 = 54.078 *10^-6m²

energy per unit time(power) which can enter = intensity * area = 3.3*10^-11 *54.078 *10^-6   = 178.459 * 10^-17 W

[ intensity is nothing but energy per unit time and per unit area i.e. power per unit area ]

energy of one photon with wavelength 500 nm = h*c / wavelength = 1240 ev/500 = 2.48 ev

numbers of photons per second = power incident / energy of one photon = 178.459 * 10^-17 / 2.48 ev

= 44.97 *10² photons per second