# 2D and 3D EM wave intensity and amplitude of $\mathbf E$ and $\mathbf B$ fields

For a spherical wave such as light from the sun or a light bulb, we can use the following formula to calculate its intensity: $$I ({Watts}/{m^2}) = \frac{P}{4\pi{r^2}}$$ Given the Sun's power = $$4\times10^{26}Watts$$ and distance to earth = $$1.5\times10^{11}m$$, the sunlight intensity is approximately 1.4 $${kW}$$/$$m^2$$ above the earth's atmosphere. To find out the amplitude of electric field $$E_0$$ and magnetic field $$B_0$$, we get the time average of the energy flux or Poynting vector: \begin{align} I ({Watts}/{m^2}) &= <\mathbf S(r,t)>=<\frac{1}{\mu}\vec{E} \times \vec{B}> =\frac{c\varepsilon{E_0}^2}{2} = \frac{c{B_0}^2}{2\mu} =\frac{{E_0}{B_0}}{2\mu} =c\varepsilon{E_{rms}}^2 \\\\ E(r,t) &= E_0cos(\mathbf k \cdot \mathbf r-\omega t) \\\\ B(r,t) &= B_0cos(\mathbf k \cdot \mathbf r-\omega t) \end{align} So the sunlight's $$E_0 = 1.027 kV/m$$ and $$B_0 = E_0/c = 3.4 \times 10^{-6} T$$. By comparison, a 65Watts light bulb at sphere radius of 3m has $$I=0.57W/m^2$$ with $$E_0=20.7V/m$$ and $$B_0=6.9 \times 10^{-8} T$$.

My question is since the intensity falls off with $$4 \pi r^2$$ as the radiation spreads out uniformly in 3-D, shouldn't the $$E_0$$ and $$B_0$$ decrease with $$1/r$$ according to the above equations, i.e., $$E(r,t) = E_0(1/r)cos(\mathbf k \cdot \mathbf r-\omega t)$$. By the same token, for a surface wave in 2-D, the intensity falls off with $$2 \pi r$$, so the $$E_0$$ and $$B_0$$ should decrease with $$1/{\sqrt r}$$. On the other hand, a focused laser beam are plane waves in 1-D, the intensity remains constant, so do $$E_0$$ and $$B_0$$. If this is true, it is hard to picture a 2-D or 3-D wave would have a decreasing amplitude of $$E_0$$ and $$B_0$$ as it propagates on.

An ideal plane wave does not satisfy the propagation law $$\frac{1}{r}$$ for its $$E$$ and $$B$$ fields, this is because the source of an ideal plane wave is an infinite plane with infinite energy emitting a wave of infinite energy. Obviously that is non-physical. What is physical is a provable result that asymptotically any EM source of a finite size emits waves in any direction that are sufficiently far away from the source behave locally like a plane wave in the so-called far-field. This is the basis of all antenna design and all radar systems; see the Rayleigh criterion for the far-field of an antenna where the Fraunhoffer diffraction dominates. Because of its large aperture diameter to wavelength ratio a laser's far-field, typically, is very very very far away...