All the light you can barely see

We are bounded in a nutshell of Infinite space: Blog Post #22, Worksheet # 7.1, Problem #6: All the light you can barely see


6. If your telescope can detect optical magnitudes $m_V \leq 21$ how far away, in parsecs, can you detect a Type Ia supernova with your telescope? (HINT: The Sun’s absolute magnitude is $M_V =4.83$ .)


Here, we’ll be going back to some of our earlier problems involving distance moduli and Luminosity and magnitude comparisons. From previous problems and known astronomical facts, we have:


$$L_{Ia} = 3.1308 \times 10^{43} \frac{ergs}{s} ,$$ the luminosity of the sun is: $$L_\odot = 3.846 \times 10^{33} \frac{ergs}{s} ,$$ and the equation for comparing absolute magnitudes is $$M_\star - M_\odot = -10^{0.4} \log_10 \left(\frac{L_\star}{L_\odot}\right) .$$ Having these values, as well as the Absolute Magnitude of the sun, given by the problem, $M_V =4.83$, then we can simply solve for the absolute magnitude of the Supernova. $$M_{Ia} - M_\odot = -10^{0.4} \log_{10} \left(\frac{L_{Ia}}{L_\odot}\right) ,$$ $$M_{Ia}= -10^{0.4} \log_{10} \left(\frac{L_{Ia}}{L_\odot}\right) + M_\odot,$$ $$M_{Ia}= -10^{0.4} \log_{10} \left(\frac{3.1308 \times 10^{43} \frac{ergs}{s}}{3.846 \times 10^{33} \frac{ergs}{s}}\right) + 4.83,$$ $$M_{Ia}= -24.8944 + 4.83,$$ $$M_{Ia} = -20.061.$$


Now knowing the absolute magnitude of the supernova, we can now solve for the maximum distance when the max apparent magnitude that can be recorded is 21. Using $$m – M = 5 \log_{10} (d) -5 ,$$ $$d \leq 10^{\frac{m- M  +5}{5}} ,$$ $$d\leq 10^{\frac{21- (-20.06)  +5}{5}} ,$$ $$d\leq 10^{9.212} ,$$ $$d \leq 1.629 \times 10^ 9 pc = 1.629 \times 10^6 kpc ,$$ which is the max distance our telescopes can see with present technology, for now.