How far away are those supernovae? Blog 22, Worksheet 7.1, Problem 6

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

What we know is $M_{ V \odot}  = 4.83$ and $L_{WD} = 10^{10} L_{\odot}$.
To solve, we can revisit week 5 to solve this problem. We know that absolute magnitude is related to luminosity: $$M_{WD} - M_{\odot} = -2.5log \left( \frac{L_{WD}}{L_{\odot}} \right)$$ So $$M_{WD} =  -2.5log \left( \frac{10^{10} L_{\odot}}{L_{\odot}} \right) + M_{\odot}$$   $$M_{WD} =  -2.5log \left( 10^{10} \right) + 8.83 = -25 + 4.83 = -20.17$$ Now that we have a (rather large) absolute magnitude and a limit on our ability to detect apparent magnitudes, we can find distance. $$m - M = 5log(d) - 5$$ $$d = 10^{\frac{m-M +5}{5}}$$ $$d = 10^{\frac{21 + 20.17+5}{5}} = 1.7 \times 10^9 \: pc$$ By comparison, the center of the galaxy is about 8 kilo-parsecs away so we would get a decent range within our galaxy.

Source

Supernovae can be amazingly bright.