1. Light and Radiation

Electromagnetic waves    (3/4)

So far we have examined the field quantities E and B. What is now

the intensity of light

seen with the eye of an observer, measured with a photodetector or taken as an image with a camera?

Light detection, visually or with an optical instrument, is not done with E and B directly. It is the energy of the electric and magnetic field which determines the intensity. The field energy is proportional to the squared fields, i.e. E2 and B2. Both the field quantities and their squares have very high frequencies, in the order of 500 terahertz with visible light. These frequencies are much too high to be seen by the photoreceptors of the eye.

Therefore, the eye and all electronic detectors for visible light perform a time-averaging over the squared field quantities. It is this average which is related to the intensity of light.

Frame rate of motion pictures ↓
This is shown in the following graph with the electric field E. For simplicity, we consider its time-dependence only:

E(t)= E o sin2π t T     (thin blue curve)

Its square corresponds to a sine function with half amplitude and half period T, shifted by 0.5 to positive values:

E 2 (t)= E o 2 sin 2 2π t T = 1 2 E o 2   ( 1−cos2π t T/2 )     (red curve)
Trigonometric relations ↓


Zoom Sign

Electric field E(t) plotted over time t (blue line), and its square E2(t) (red line). The amplitude is set to Eo=1, hence it is Eo2=1. The time-averaged square of E2(t) is ⟨ E 2 ⟩=0.5 (green line).

We use brackets ⟨ ... ⟩ to denote a signal average over time of E2(t). Since E2(t) is symmetric around E2(t)=0.5, it follows:

⟨ E 2 ⟩=0.5  E o 2     (green line)

It is this constant signal (as long as the wave amplitude remains constant) which is seen by the eye as the intensity of light.

Read more about the energy density and intensity of electromagnetic waves in supplement 1.4.