Physics Asked by Jesús Portillo on December 17, 2020
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:
$$Ileft(tright) propto left|Eleft(tright)right|^2$$
$$I=left|Sright|=frac{left|Eright|^2}{Z_0}$$
How do I reconcile these two formulas?
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 $$mathbf{E}(x,t) = {E_0}sinleft(kx-omega tright)mathbf{hat x}$$ and $$mathbf{B}(x,t) = {B_0}sinleft(kx-omega tright)mathbf{hat y}.$$ Here the frequency is given by $f = frac{2pi}{omega}$ and the wavelength by $lambda = frac{2pi}{k}$. 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 $$ I(t) = left< mathbf{S}(t) right> = frac{1}{2cmu_0} E_0^2.$$
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 $$I = n cdot frac{hf}{ADelta t}. $$ In this model, photons are only counted, and not seen as waves. Thus there is no amplitude to be considered.
Correct answer by zhutchens1 on December 17, 2020
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