Find 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 $\left(-{10}^{-10}W{m}^{-2}\right)$. Take the area of the pupil to be about $0.4c{m}^2$ and the average frequency of white light to be about $6\times {10}^{14}Hz$.

Minimum intensity,

$I={10}^{-10}W{m}^{-2}$

Area of pupil,

$A=0.4c{m}^2=0.4\times {10}^{-4}m$

Average frequency,

$v=6\times {10}^{-14}Hz$

Energy of one photon

$E=hv=6.63\times {10}^{-34}\times 6\times {10}^{14}J=4\times {10}^{-19}J$

Let $n$ be the number of photons crossing per square metre area per second.

Now Intensity = Energy incident per square metre area per second

$\Rightarrow I=\text{ Total Energy of }n\text{ photons }$

$\Rightarrow I=n\times \text{ Energy of one photon }$

$\Rightarrow n=\frac{\text{ Intensity }}{\text{ Energy of one photon }}$

$\Rightarrow I=\frac{{10}^{-10}W{m}^{-2}}{4\times {10}^{-19}J}=2.5\times {10}^8{m}^{-2}{s}^{-1}$

So, the number of photons entering the pupil of our eye per second is

$N=n\left(\text{ Area of the pupil }\right)$

$\Rightarrow N=2.5\times {10}^8\times 0.4\times {10}^{-4}{s}^{-1}={10}^4{s}^{-1}$