_{T}watts uniformly in all directions. For a ray in any given direction we need to construct a flux tube with a solid angle ΔΩ about the ray in order to define the quantities. The flux traveling through the tube will be ΔΦ = Φ

_{T}· ΔΩ/4π since the solid angle for all directions is 4π and ΔΩ/4π is the fraction of Φ

_{T}that enters the tube. Calculations are simplified by defining the Intensity, I = Φ

_{T}/4π, which is the same in all directions and at all points along a ray from the source. The flux through the tube is ΔΦ = I·ΔΩ and this flux will pass through all cross-sections of the tube.

The normal cross-sectional area for a flux tube is ΔS = R

^{2}ΔΩ where R is the distance of the surface from the point source. The Exitance of a finite source is defined as the amount of flux traveling through a unit of surface area or,

M = ΔΦ/ΔS = I·ΔΩ/R

^{2}·ΔΩ = I/R

^{2}.

The same is true for any cross-section of the flux tube and its Irradiance is,

E = I/R

^{2}.

The quantity of flux passing through an element of surface, ΔS, normal to the ray at a distance R from it will be,

ΔΦ = E ΔS = I·ΔS/R

^{2}

which is the inverse square law.

The sides of a flux tube are determined by rays which are normal to a cross-sectional surface and so the inverse square law applies only to spherical sources. If one had a large plane which was the source of the light then the rays coming from it would not diverge and the cross-sectional areas would all be the same size. The irradiance would have approximately the same value as one moved away from the plane. The inverse square law will work for plane surfaces unless the distance from the source is very large compared to its dimensions.

For multiple sources one needs to consider the flux tubes from all of them and the angles that they make with the irradiated surface.

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