The average threshold of dark-adapted (scotopic) vision is [tex]4.00 \times 10^{-11} \, \text{W/m}^2[/tex] 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 8.00 mm, how many photons per second enter the eye?

The number of photons per second entering the eye can be calculated by finding the power incident on the eye and dividing it by the energy carried by each photon.

Calculating this value gives us approximately 3.8 × 1018 photons per second entering the eye.

Explanation:

To calculate the number of photons per second entering the eye, we need to calculate the power incident on the eye first. Using the given average threshold of dark-adapted vision and the maximum diameter of the pupil, we can calculate the power incident on the eye.

Power = Intensity × Area
Intensity = 4.00 × 10-11 W/m2
Area = π(0.008 m)2 = 6.28 × 10-5 m2

Plugging in these values, we get:
Power = (4.00 × 10-11 W/m2) × (6.28 × 10-5 m2) = 2.51 × 10-15 W

Since each photon carries an energy of E = hf, where h is Planck's constant (6.63 × 10-34 Js) and f is the frequency, we can calculate the number of photons per second using:
Number of photons per second = Power / E
Number of photons per second = (2.51 × 10-15 W) / (6.63 × 10-34 Js)

Calculating this value gives us approximately 3.8 × 1018 photons per second entering the eye.