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--- wiki:astronomy:observational_astronomy:observational_astronomy:extinction [2023/11/14 13:02] – Roy Prouty
+++ wiki:astronomy:observational_astronomy:observational_astronomy:extinction [2023/11/14 13:11] (current) – Roy Prouty
@@ Line 34: / Line 34: @@
 So blazing forward with this equivalence, we can re-write the Pogson Equation in terms of this atmospheric extinction relationship we've just derived.
-$$m_b-m_t = -2.5 \log_{10}\frac{I_\lambda^b}{I_\lambda^t} = -2.5 \log_{10}\exp{\Biggl(- \sec{z} \int_b^t \alpha_\lambda(x) dx\Biggr)}$$
+$$m_b-m_t = -2.5 \log_{10}\frac{I_\lambda^b}{I_\lambda^t} = -2.5 \log_{10}\Biggl[\exp{\Biggl(- \sec{z} \int_b^t \alpha_\lambda(x) dx\Biggr)}\Biggr]$$
+Let's carefully take the log-base-10 of the exponential...
+$$m_b-m_t = -2.5 \log_{10}(e)\biggl(- \sec{z} \int_b^t \alpha_\lambda(x) dx\biggr)$$
+Finally, recalling that we want to calibrate to the top-of-the-atmosphere, we find
+$$m_t = m_b + 2.5 \log_{10}(e)\biggl(- \sec{z} \int_b^t \alpha_\lambda(x) dx\biggr)$$
+So we find that the top-of-the-atmosphere magnitude is our bottom-of-the-atmosphere = top-of-telescope magnitude with some additional magnitude that was "extincted" away from incidence as the radiance traversed the atmosphere.\\ \\
+Note that the $F$ or $I$ used in the above equations are //not// counts. They must be some unit of power. So radiance, flux, energy per second (power), or even counts-per-second.
+----
+Sources
+  - Lifted with minor modification from: https://slittlefair.staff.shef.ac.uk/teaching/phy241/lectures/l07/