Bdtyktl

Would that person go blind?

Neutrino detectors and the abundance of Neutrinos would detect the upcoming visible show about 3 hours before any visible signs, so there would be time to point certain telescopes that could handle the brightness towards it.

I'm curious if an individual with a telescope pointed in that direction would have an unpleasant surprise. Would the scientific community be wise to not announce the massive stellar explosion until after it's visible to avoid potential negative effects from over-eager amateur astronomers.

I realize this is a kind of silly question and it might depend too much on the telescope, but I'm curious.

No, it would not be a problem. Supernovae are not at all like flashbulbs -- they brighten over a period of many days and dim again even more slowly. Here are a number of different lightcurves taken from Wikipedia:

The rise is fast on an astronomical scale -- several orders of magnitude over a period of roughly ten days -- but very slow on a human scale. An amateur looking at it would not notice any significant change in brightness, but if the same person came back a few hours later or the next night, the change would be very evident.

As far as we can tell, the reason is that the light at peak brightness is caused by emissions from material blown off by the explosion. For example, in Type 1a SNe, most of the light if from the radioactive decay of the //huge// mass of ejected nickel-56 (half life 6 days).

The Wikipedia article on supernovae is quite good and covers this all in more detail.

Unless you observe it through a 5–10 meter class telescope during peak brightness, you'll be fine.

Betelgeuse will explode as a type II supernova, the typical brightness of which is around $M sim -17$. With a distance of $dsimeq200,mathrm{pc}$, its distance modulus is
$$
mu = 5log(d/mathrm{pc}) - 5 simeq 6.5,
$$
so its apparent magnitude will be
$$
m = M + mu simeq -10.5.
$$
The Sun has an apparent magnitude of $m_odot = -26.7$, i.e. it is $Delta m = 16.2$ magnitudes brighter. In other words, Betelgeuse will be
$$
f = 10^{Delta m/2.5} simeq 3times10^6
$$
times dimmer than the Sun. That means that if you observe Betelgeuse through a telescope with an area three million times larger than your pupil, it will look as bright as the Sun. Taking 6 mm for the diameter of the pupil, that means a telescope of roughly 10.5 meters. Such large telescopes exist, but these are not made for looking through with your naked eye.

Based on evolutionary models of Betelgeuse, Dolan et al. (2016) estimate an apparent magnitude of $m=-12.4$. This would mean that you'd need a 4.5 meter telescope, which still wouldn't be one meant for your naked eye.

Moreover, as Mark writes in his answer, supernovae don't increase to their peak brightness in matters of seconds, but rather in matters days (roughly half a mag per day), so there is no real risk.

Brightness varies inversely with the square of the distance. Betelgeuse is about 642.5 light years away and has an apparent magnitude of 0.42. My grasp of apparent magnitude concepts is a bit wobbly, but I believe if it grew a million times as bright, it might have an apparent magnitude of -14.5 or so, which is a lot more like the brightness of the moon than the sun.

Given the great distance, the decrease in brightness due to distance, and the countless amounts of dust & gas between earth and Betelgeuse, I think you'd probably be fine. You might be dazzled by its brightness -- a bit like looking at a light bulb, I imagine -- but I doubt it would cause any physical harm.

EDIT: I hope a real astronomer sounds off here. I'm not sure what kind of supernova we might expect from Betelgeuse, but apparently supernovas (supernovae?) can achieve a theoretical brightness equal to 5 trillion suns!