Eye on the Supernova

We are bounded in a nutshell of Infinite space: Blog Post #5, Worksheet # 2.1, Problem #4: Eye on the Supernova  

4. A supernova goes off and you can barely detect it with your eyes. Astronomers tell you that supernovae have a luminosity of $10^{42} \frac{erg}{s}$ ; what is the distance of the supernova? Assume the supernova emits most of its energy at the peak of the eye’s sensitivity and that it explodes isotropically.

For this problem, it is especially important we start setting up some constants and numbers that might make little sense at first, but as the problem progresses, their use will become apparent. These values are: $$L_\star = 10^{42} \frac{erg}{s} ,$$   $$\lambda_{(Wavelength of Supernova Light)}  = 500 nm = .5 \times 10^{-4} cm ,$$ by using the equation $c= \lambda \nu$, the frequency of the light the supernova emits is: $$\nu = 5.96 \times 10^{14}  s^{-1} ,$$ $$h_{(Plank’s Constant)} = 6.6 \times 10^{-27} erg \cdot s  ,$$ $$Radius_{Pupil}= .433 cm$$    $$Area_{eye} = \pi r_{eye}^2 = .589 cm^2 ,$$ $$n_{minimum photons the eye must receive}= 10 photons ,$$ and $$t_{eye exposure}= .1 s$$ (the amount of time necessary for the eye identify a single image, so if more than 1 image is presented in a .1 s interval, the eye cannot recognize individual images, like in a movie).

With these values, we can begin to plug into values that will first lead to the amount of photons emitted by the supernova per second and later the actual distance of the observer from the supernova. The equation that describes the energy of 1 photon is: $$E = h \nu  ,$$ and this equation can be manipulated with $E= \frac {L_\star}{n_{(Number of photons from Supernova)}}$ since $\frac {L_\star}{n}$ is another way of describing the individual energy of a photon. Plugging in values:

$$\frac {L_\star}{n_{(Number of photons from Supernova)}} = h \nu  ,$$

$$n_{(Number of photons from Supernova)}= \frac {L_\star}{ h \nu}   ,$$

$$n_{(Number of photons from Supernova)}= \frac {10^{42} \frac{erg}{s}}{6.6 \times 10^{-27} (erg \cdot s)  \cdot 5.96 \times 10^{14}  s^{-1}}  ,$$

$$n _{(Number of photons from Supernova)} = 2.5422 \times 10^{53} \frac{photons}{s} .$$  

With the number of photons emitted per second, this number can now be plugged into the photon flux equation: $$\frac{n}{4 \pi \delta^2} ,$$ which, as you can see, describes the amount of photons emitted divided by the surface area the shell of continuous photons occupies at the distance $\delta$ it has traveled.

This photon flux can also be calculated by understanding the amount of photons that are absorbed in the pupil in a given period, which is calculated with: $$\frac{n_ {minimum photons the eye must receive}}{A_{eye} t_{exposure}} .$$

And now these two equations are equalized since they are the same value: the photon flux, the photons in a specific area, $$\frac{n_{(Number of photons from Supernova)}}{4 \pi \delta^2} = \frac{n_ {minimum photons the eye must receive}}{A_{eye} t_{exposure}} ,$$ and solve for the distance $\delta$ from the observer to the supernova.

Now we solve for $\delta$:

$$\frac{n_{(Number of photons from Supernova)} \cdot A_{eye} \cdot t_{exposure}}{4 \pi \cdot {n_{minimum photons the eye must receive}}} = \delta^2 ,$$

$$\delta= \left( \frac{n_{(Number of photons from Supernova)} \cdot A_{eye} \cdot t_{exposure}}{4 \pi \cdot {n_{minimum photons the eye must receive}}}\right)^{\frac{1}{2}} ,$$

to now start plugging in the values:

$$\delta= \left( \frac{2.5422 \times 10^{53} \frac{photons}{s}  \cdot .589 cm^2 \cdot .1 s }{4 \pi \cdot {10 photons}}\right)^{\frac{1}{2}} ,$$

$$\delta= 1.1 \times 10^{25} cm$$  times the conversion factor of cm to pc

  $$\delta= 1.1 \times 10^{25} cm \cdot \frac{1}{3.094 \times 10^18 \frac{cm}{pc}} ,$$  equals

$$\delta= 3.55 \times 10^6 pc .$$ Which is how far away the Supernova is from your eye.