Originally Posted by **justme** Have you got a reference for that? |

Oooops... I made a mistake... it's the

*intensity* that varies with the inverse square law, not the amplitude. The amplitude will vary as the square root of intensity, so

$\displaystyle A(r) \propto \frac{1}{r}$ for a spherical wavefront and

$\displaystyle A(r) \propto \frac{1}{r^{1/2}}$ for a circular wavefront.

Note: it's a routine problem in 1st year undergraduate texts to calculate, for example, that the extra-terrestrial radiation at the Earth's surface is ~1400 W/m2. E.g.

Solar luminosity = $\displaystyle 3.828 \times 10^{26}$ W

Earth-Sun distance = $\displaystyle 1.5 \times 10^{11}$ m

Solar luminosity is spread out over a sphere with radius equal to the Earth-Sun distance, so

$\displaystyle L_{Earth} = \frac{L_{\odot}}{A} = \frac{L_{\odot}}{4\pi r^2} = \frac{3.828 \times 10^{26}}{4\pi \times (1.5 \times 10^{11})^2} = 1354 $W/m2

Most textbooks on waves explain the behaviour.