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:

$$I\left(t\right) \propto \left|E\left(t\right)\right|^2$$


How do I reconcile these two formulas?


1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.