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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.

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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...

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