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Data Reduction III: Atmospheric Extinction
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Instrument Magnitude
Now invent an instrument-filter magnitude.
Recall the definition of a magnitude
$$m_1 - m_2 = -2.5\log_{10}\frac{F_1}{F_2}$$
Observe that the collection of source #2 terms amounts to some constant
By properties of logarithms: $$m_1 - m_2 = -2.5\log_{10}F_1 - 2.5\log_{10}F_2$$
Grouping all source #2 terms: $$m_1 = -2.5\log_{10}F_1 + (m_2 - 2.5\log_{10}F_2)$$
Identify $m_2 - 2.5\log_{10}F_2 = C$ as an arbitrary constant. $$m = -2.5\log_{10}F + C$$
Atmospheric Extinction
Imagine breaking the vertical extent of the atmosphere up into infinitely many thin, parallel slabs of varying composition and character. As a ray of light enters the atmosphere and traverses some path through these layers, the ray interacts with each layer.
The Beer-Lambert Law gives us the relationship between the intensity of light rays entering a medium and the intensity of the resultant outgoing light rays. If the medium can be characterized with some extinction coefficient $\kappa$ in units of $\frac{\%}{m}$, then the change in intensity for every unit length of the medium traversed is $\Delta{I} = - \kappa I(x=0)\cdot \Delta{x}$, where $I(x=0)$ can be taken as the intensity of the light ray as it enters the medium.
In differential form: $$\frac{dI}{dx} = -\kappa(x)I(x)$$
In the diagram above, you can see that any angle relative to zenith ($z$) actually forces the light ray to traverse a longer path through the atmosphere by a factor of $\sec{(z)}$. So the Beer-Lambert Law tells us that the differential amount of intensity lost to the extinctive character of the layers of the atmosphere is:
$$dI = -\kappa(x)I(x)\sec{(z)}dx$$
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