What is the formula for the intensity of light, and how are amplitude, frequency and number of photons considered? One formula for light intensity is$$
I = \frac{nfh}{At}
\,,$$where:

*

*$n$ is the number of photons;


*$h$ is Planck's constant;


*$f$ is the frequency;


*$A$ is the incident area;


*$t$ is time.
Another formula describes intensity as a function of the magnitude of electric field squared:
$$I\left(t\right) \propto \left|E\left(t\right)\right|^2$$
$$I=\left|S\right|=\frac{\left|E\right|^2}{Z_0}$$
How do I reconcile these two formulas?
 A: Classical/Wave Model
An electromagnetic wave is composed of an oscillating electric and magnetic fields, which are orthogonal. Our field equations might be described by
$$\mathbf{E}(x,t) = {E_0}\sin\left(kx-\omega t\right)\mathbf{\hat x}$$
and
$$\mathbf{B}(x,t) = {B_0}\sin\left(kx-\omega t\right)\mathbf{\hat y}.$$
Here the frequency is given by $f = \frac{\omega}{2\pi}$ and the wavelength by $\lambda = \frac{2\pi}{k}$. The amplitues are given by $E_0$ and $B_0$. These equations form a plane wave which has a total intensity, at any point in time, as given by the Poynting vector
$$ \mathbf{S} = \frac{1}{\mu_0}\left(\mathbf{E} \times \mathbf{B}\right). $$
The time-average of the Poynting vector turns out to be
$$ I(t) = \left< \mathbf{S}(t) \right> = \frac{1}{2c\mu_0} E_0^2.$$
This is the equation you mention. There are no photons to be counted in this paradigm, for photons are waves and not particles by classical electrodynamics theory.
Particle/Quantum Model
In the high-energy limit, photons act more like particles than waves.
The intensity is defined as power per unit area, and power is defined as energy per unit time. Thus:
$$I = \frac{P}{A} = \frac{E}{\Delta t} \frac{1}{A}.$$
The energy of a photon is $E = hf$, so the total intensity for $n$ photons is
$$I = n \cdot \frac{hf}{A\Delta t}. $$
In this model, photons are only counted, and not seen as waves. Thus there is no amplitude to be considered.
