The average threshold of dark-adapted (scotopic) vision is 4 × 10-11 W/m 2 at a central wavelength of 500 nm. If lightwith this intensity and wavelength enters the eye and the pupil is open to its maximum diameter of 8.50 mm, howmany photons per second enter the eye?

To solve this problem, we need to use the formula for the energy of a photon, which is E = hc/λ, where h is Planck's constant (6.63 x 10^-34 J*s), c is the speed of light (3 x 10^8 m/s), and λ is the wavelength of the light.

1. First, convert the wavelength from nm to m: 500 nm = 500 x 10^-9 m.

2. Next, calculate the energy of a single photon: E = (6.63 x 10^-34 J*s) * (3 x 10^8 m/s) / (500 x 10^-9 m) = 3.978 x 10^-19 J.

3. The power of the light entering the eye is given as 4 x 10^-11 W/m^2. A watt is a joule per second, so this is also the energy entering the eye per second per square meter.

4. The area of the pupil can be calculated using the formula for the area of a circle, A = πr^2, where r is the radius of the pupil. First, convert the diameter to radius: r = 8.50 mm / 2 = 4.25 mm = 4.25 x 10^-3 m. Then, calculate the area: A = π * (4.25 x 10^-3 m)^2 = 5.67 x 10^-5 m^2.

5. Multiply the power of the light by the area of the pupil to find the total energy entering the eye per second: (4 x 10^-11 J/s/m^2) * (5.67 x 10^-5 m^2) = 2.27 x 10^-15 J/s.

6. Finally, divide the total energy entering the eye per second by the energy of a single photon to find the number of photons entering the eye per second: (2.27 x 10^-15 J/s) / (3.978 x 10^-19 J) = 5.7 x 10^3 photons/s.

So, approximately 5700 photons per second enter the eye.