#### Question

Estimating the following two numbers should be interesting. The first number will tell you why radio engineers do not need to worry much about photons! The second number tells you why our eye can never ‘count photons’, even in barely detectable light.

The number of photons entering the pupil of our eye per second corresponding to the minimum intensity of white light that we humans can perceive (∼10^{−10} W m^{−2}). Take the area of the pupil to be about 0.4 cm^{2}, and the average frequency of white light to be about 6 × 10^{14} Hz.

#### Solution

Intensity of light perceived by the human eye, *I* = 10^{−10} W m^{−2}

Area of a pupil, *A* = 0.4 cm^{2 }= 0.4 × 10^{−4} m^{2}

Frequency of white light, *ν*= 6 × 10^{14} Hz

The energy emitted by a photon is given as:

*E* = *h*ν

Where,

*h* = Planck’s constant = 6.6 × 10^{−34} Js

∴*E* = 6.6 × 10^{−34} × 6 × 10^{14}

= 3.96 × 10^{−19} J

Let *n* be the total number of photons falling per second, per unit area of the pupil.

The total energy per unit for *n* falling photons is given as:

*E* = *n* × 3.96 × 10^{−19} J s^{−1} m^{−2}

The energy per unit area per second is the intensity of light.

∴*E = I*

*n* × 3.96 × 10^{−19} = 10^{−10}

`n = 10^(-10)/(3.96 xx 10^(-19))`

= 2.52 × 10^{8} m^{2} s^{−1}

The total number of photons entering the pupil per second is given as:

*n*_{A }= *n × A*

= 2.52 × 10^{8 }× 0.4 × 10^{−4}

= 1.008 × 10^{4} s^{−1}

This number is not as large as the one found in problem **(a)**, but it is large enough for the human eye to never see the individual photons.