The pupil area of the human eye is ~ 1 × 10^-5 m². The minimum light intensity that the eye is sensitive to is ~ 1 × 10^-10 Wm^-2. How many photons per second of wavelength 550 nm does this correspond to?

Question

Quizplus · Accepted Answer

The energy of a single photon is given by E = hc/λ, where h is Planck's constant, c is the speed of light, and λ is the wavelength. The number of photons per second is then given by P = I/E, where I is the light intensity. Plugging in the given values, we get:P = (1 × 10^-10 Wm^-2) / ((6.63 × 10^-34 Js) / (550 × 10^-9 m)) = 3000 s^-1Therefore, the correct answer is C.

The Pupil Area of the Human Eye Is ~ 1