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A Lambert source has the from of an infi...

A Lambert source has the from of an infinite plane. Its luminance is equal to `L`. Find the illuminance of an area element oriented parallel to the given source.

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Consider an element of area `dS` at point `P`. It emits light of flux
`d phi = LdS Omega cos theta`
`= LdS (dA)/(h^(2) sec^(2) theta).cos^(2) theta`
`= (LdS dA)/(h^(2)) cos^(4) theta`
in the direction of the surface element `dA` at `O`. The total illuminance at `O` is then
`E = int (LdS)/(h^(2)) cos^(4) theta`
But `dS = 2pi rdr = 2pi h tan theta d(h tan theta)`
`= 2pi h^(2) sec^(2) theta tan theta d theta`
Subsitution gives `E = 2pi L int _(0)^(pi//2)sin theta cos theta d theta = pi L`
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