<p>If you insist on observing the exploding Betelgeuse at peak brightness, you could potentially damage your eye. The complete answer enters the realm of physiology. Here I'll discuss the astronomical parts:</p>

<p>Betelgeuse will explode as a type II supernova, the typical brightness of which is around <span class="math-container">$M \sim -17$</span>. With a distance of <span class="math-container">$d\simeq200\,\mathrm{pc}$</span>, its distance modulus is
<span class="math-container">$$
\mu = 5\log(d/\mathrm{pc}) - 5 \simeq 6.5,
$$</span>
so its apparent magnitude will be
<span class="math-container">$$
m = M + \mu \simeq -10.5.
$$</span></p>

<p>For these calculations I assume that the Sun is the threshold for damaging your eye (a brief look at the Sun is okay, a longer look will cause permanent damage. But… physiology…). The Sun has an apparent magnitude of <span class="math-container">$m_\odot = -26.7$</span>, i.e. it is <span class="math-container">$\Delta m = 16.2$</span> magnitudes brighter. In other words, Betelgeuse will be
<span class="math-container">$$
f = 10^{\Delta m/2.5} \simeq 3\times10^6
$$</span>
times dimmer than the Sun.</p>

<p>However, the Sun is an extended source, spanning an angle of roughly <span class="math-container">$\theta_\mathrm{Sun} = 32$</span> arcminutes across. In contrast, Betelgeuse is a point source, which when transferred through the atmosphere and the telescope, is spread out over <span class="math-container">$\theta_\mathrm{Bet} \sim$</span> a few square arcseconds. Thus its light will be more concentrated; i.e. it will be much brighter, but it will hit a much smaller area of your retina. However, your eye will also move around, smearing out the light. Not being a physiologist, for the sake of this calculation I assume that the light is smeared out over a disk 1 arcminute across (about the size of a planet seen from Earth).</p>

<p>Thus, the factor <span class="math-container">$f$</span> will itself be a factor <span class="math-container">$(\theta_\mathrm{Sun} / \theta_\mathrm{Bet})^2 \simeq 1000$</span> times larger — that is, Betelgeuse is only <span class="math-container">$\sim 3\,000$</span> times dimmer than the Sun.</p>

<p>Hence, for our assumptions your eye will be damaged if you observe exploding Betelgeuse through a telescope with an area <span class="math-container">$\sim 3\,000$</span> larger — or roughly 55 times wider —  than your pupil. In bright light, the pupil contracts to roughly 3 mm in diameter, so if observing through a telescope of 16 cm or larger, you could damage your eye. </p>

<p>Based on evolutionary models of Betelgeuse, <a href="https://arxiv.org/abs/1406.3143v2" rel="noreferrer">Dolan et al. (2016)</a> estimate an apparent magnitude of <span class="math-container">$m=-12.4$</span>, i.e. roughly 6 times brighter than our estimate. This would mean that you only need a 7 cm telescope to damage your eye.</p>

<p>However, 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 you have plenty of time to look away.</p>

CC BY-SA 4.0
2018-12-10T08:07:36.633Z
I'm not sure this is the right calculation. Betelgeuse subtends only a tiny angle, meaning that when it's in focus all its light is focused on effectively a single point on the retina, whereas the sun's light is spread over an area. I'd be surprised if that didn't make it dangerous when using a much smaller telescope.
N. Virgo
CC BY-SA 4.0
2018-12-10T13:12:58.003Z
Additionally, this answer assumes that the brightness of the sun is the minimum required to go blind. For a complete answer, we should probably consider how much brightness the human eye can withstand without being blinded. It could be the case that a much smaller telescope than the one you propose is sufficient.
JBentley
CC BY-SA 4.0
2018-12-10T14:08:34.763Z
@Nathaniel Yes, you're right. I'm not sure what the completely correct approach would be. A point source — which when seen through the atmosphere would span a few square arcsec — would be much brighter, but damage only a tiny fraction of your eye.
pela
CC BY-SA 4.0
2018-12-10T14:13:03.200Z
@JBentley That's true, that's an assumption. But considering that you _can_ actually look at the Sun for a brief period of time without permanently damaging your eye, I think makes this threshold okay, at least for an order-of-magnitude estimate.
pela
CC BY-SA 4.0
2018-12-10T15:17:26.003Z
I edited to take into account these very relevant comments.
pela
CC BY-SA 4.0
2018-12-10T18:17:28.913Z
Yes but *could you look away?* It would be amazing.
S. Imp
CC BY-SA 4.0
2018-12-08T23:57:24.797Z
2018-12-11T09:55:26.417Z
2018-12-11T09:55:26.417Z
pela
<p>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.</p>

<p>Given the great distance, the decrease in brightness due to distance, and the countless amounts of dust &amp; 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.</p>

<p>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!</p>

CC BY-SA 4.0
2018-12-10T05:25:03.150Z
Does it matter that the brightness comparable to the moon will be concentrated in a star-sized point?
Barmar
CC BY-SA 4.0
2018-12-10T18:15:17.213Z
I am not qualified to say. I will speculate that its intensity will be diffused quite a bit by all the dust & gas between us and Betelgeuse -- it's 640 light-years away. On the other hand, the moon looks *extremely* bright in my own telescope. Bright enough to dazzle the eyes quite a bit.
S. Imp
CC BY-SA 4.0
2018-12-10T19:15:48.140Z
I assumed you'd already included the diffusion in your estimate of the apparent magnitude.
Barmar
CC BY-SA 4.0
2018-12-10T20:09:03.157Z
My calculation was a simple one. I simply assumed that Betelgeuse might increase in brightness by a million, and converted that increase to apparent magnitude. Check the [third paragraph here](https://www.astronomynotes.com/starprop/s4.htm). This sort of sidesteps complicated diffusion calculations by just assuming diffusion will not change if the thing gets brighter.
S. Imp
CC BY-SA 4.0
2018-12-10T20:13:03.620Z
That seems like a reasonable assumption for broad estimates like this. In reality, diffusion is probably dependent on light wavelength, which I assume changes during supernova. But if we're just dealing with orders of magnitude, it's probably not significant.
Barmar
CC BY-SA 4.0
2018-12-08T22:52:22.877Z
2018-12-08T22:52:22.877Z
S. Imp
<p>From Wikipedia: The visual light curves of the different supernova types all depend at late times on radioactive heating, but they vary in shape and amplitude because of the underlying mechanisms, the way that visible radiation is produced, the epoch of its observation, and the transparency of the ejected material. The light curves can be significantly different at other wavelengths. <strong>For example, at ultraviolet wavelengths there is an early extremely luminous peak lasting only a few hours corresponding to the breakout of the shock launched by the initial event, but that breakout is hardly detectable optically.</strong></p>

<p>I'm not sure I'd want to be looking at Betelgeuse through a large telescope when this happens. Visible light is not what damages your eye. UV is. </p>

CC BY-SA 4.0
2020-01-11T19:43:15.153Z
Well, that certainly depends on the brightness. Most UV is absorbed in the atmosphere, so if you stare at the Sun, optical light _is_ what damages the eye.
pela
CC BY-SA 4.0
2020-01-11T19:13:52.020Z
2020-01-11T19:13:52.020Z
Wilfred Gaines
<p>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 light curves taken from Wikipedia:
<a href="https://i.sstatic.net/GH7QI.png" rel="noreferrer"><img src="https://i.sstatic.net/GH7QI.png" alt="Luminous output of different types of Supernovae"></a></p>

<p>The rise is fast on an astronomical scale – several orders of magnitude over a period of roughly ten days – but <em>very</em> 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.</p>

<p>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 is from the radioactive decay of the <em>huge</em> mass of ejected nickel-56 (half life 6 days).</p>

<p>The <a href="https://en.wikipedia.org/wiki/Supernova" rel="noreferrer">Wikipedia article on supernovae</a> is quite good and covers this all in more detail.</p>

CC BY-SA 4.0
2018-12-10T13:10:30.303Z
This answer explains that the brightness increases gradually, but doesn't seem to answer the question as to whether or not a person watching Betelgeuse would go blind (unless we count "no, it would not be a problem" as the answer).
JBentley
CC BY-SA 4.0
2018-12-10T13:44:59.897Z
Note that the OP asked if there'd be an "unpleasant surprise" -- the answer is, "No, there would be no surprise." You would not be blinded unless you really wanted to be. Obviously if you stare into a bright enough light with a big enough telescope for a long enough time you can blind yourself in one eye -- you'd have to do it all over again with the other, of course. I don't feel that constitutes a "surprise."
Mark Olson
CC BY-SA 4.0
2018-12-10T19:03:10.930Z
@JBentley It's not fast enough and it's not bright enough to blind you.
Florin Andrei
CC BY-SA 4.0
2018-12-27T10:46:24.537Z
@Mark Olson : Yep. OP is almost surely imagining it being like a camera flashbulb (or perhaps, a nuke?) "suddenly" going off and you are looking into said flashbulb with your telescope. But it's not.
The_Sympathizer
CC BY-SA 4.0
2018-12-08T23:22:56.307Z
2018-12-10T01:51:56.673Z
2018-12-10T01:51:56.673Z
Mark Olson
<p>Would that person go blind?   </p>

<p>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.    </p>

<p>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.</p>

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

CC BY-SA 4.0
2018-12-08T23:35:04.450Z
"**When** it goes ..." - It's 642.5 light years away, so it would need to have **already** gone supernova over 550 years ago ... But we know what you meant, and Mark's answer is OK, as is the other.
Rob
CC BY-SA 4.0
2018-12-09T13:09:37.333Z
@Rob Depends on your reference frame ;) ( https://en.wikipedia.org/wiki/Relativity_of_simultaneity )
jpa
<telescope><amateur-observing><supernova>
What would happen if someone had a telescope and watched Betelgeuse when it goes supernova?
userLTK
# What would happen if someone had a telescope and watched Betelgeuse when it goes supernova?

(Asked by: username_0 on 2018-12-08)

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.

**Tags:** telescope, amateur-observing, supernova

> **Comments:**
> **username_1**: "**When** it goes ..." - It's 642.5 light years away, so it would need to have **already** gone supernova over 550 years ago ... But we know what you meant, and Mark's answer is OK, as is the other.  
> **username_2**: @username_1 Depends on your reference frame ;) ( https://en.wikipedia.org/wiki/Relativity_of_simultaneity )  

---

## 4 answers

### Answer by username_9 on 2018-12-08. Score: 39 | Accepted answer

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 light curves taken from Wikipedia:
[![Luminous output of different types of Supernovae](https://i.sstatic.net/GH7QI.png)](https://i.sstatic.net/GH7QI.png)

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 is from the radioactive decay of the *huge* mass of ejected nickel-56 (half life 6 days).

The [Wikipedia article on supernovae](https://en.wikipedia.org/wiki/Supernova) is quite good and covers this all in more detail.

> **Comments:**
> **username_5**: This answer explains that the brightness increases gradually, but doesn't seem to answer the question as to whether or not a person watching Betelgeuse would go blind (unless we count "no, it would not be a problem" as the answer).  
> **username_9**: Note that the OP asked if there'd be an "unpleasant surprise" -- the answer is, "No, there would be no surprise." You would not be blinded unless you really wanted to be. Obviously if you stare into a bright enough light with a big enough telescope for a long enough time you can blind yourself in one eye -- you'd have to do it all over again with the other, of course. I don't feel that constitutes a "surprise."  
> **username_10**: @username_5 It's not fast enough and it's not bright enough to blind you.  
> **username_11**: @username_9 : Yep. OP is almost surely imagining it being like a camera flashbulb (or perhaps, a nuke?) "suddenly" going off and you are looking into said flashbulb with your telescope. But it's not.  

---

### Answer by username_3 on 2018-12-08. Score: 25

If you insist on observing the exploding Betelgeuse at peak brightness, you could potentially damage your eye. The complete answer enters the realm of physiology. Here I'll discuss the astronomical parts:

Betelgeuse will explode as a type II supernova, the typical brightness of which is around $M \sim -17$. With a distance of $d\simeq200\,\mathrm{pc}$, its distance modulus is
$$
\mu = 5\log(d/\mathrm{pc}) - 5 \simeq 6.5,
$$
so its apparent magnitude will be
$$
m = M + \mu \simeq -10.5.
$$

For these calculations I assume that the Sun is the threshold for damaging your eye (a brief look at the Sun is okay, a longer look will cause permanent damage. But… physiology…). 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
$$
f = 10^{\Delta m/2.5} \simeq 3\times10^6
$$
times dimmer than the Sun.

However, the Sun is an extended source, spanning an angle of roughly $\theta\_\mathrm{Sun} = 32$ arcminutes across. In contrast, Betelgeuse is a point source, which when transferred through the atmosphere and the telescope, is spread out over $\theta\_\mathrm{Bet} \sim$ a few square arcseconds. Thus its light will be more concentrated; i.e. it will be much brighter, but it will hit a much smaller area of your retina. However, your eye will also move around, smearing out the light. Not being a physiologist, for the sake of this calculation I assume that the light is smeared out over a disk 1 arcminute across (about the size of a planet seen from Earth).

Thus, the factor $f$ will itself be a factor $(\theta\_\mathrm{Sun} / \theta\_\mathrm{Bet})^2 \simeq 1000$ times larger — that is, Betelgeuse is only $\sim 3\,000$ times dimmer than the Sun.

Hence, for our assumptions your eye will be damaged if you observe exploding Betelgeuse through a telescope with an area $\sim 3\,000$ larger — or roughly 55 times wider — than your pupil. In bright light, the pupil contracts to roughly 3 mm in diameter, so if observing through a telescope of 16 cm or larger, you could damage your eye.

Based on evolutionary models of Betelgeuse, [Dolan et al. (2016)](https://arxiv.org/abs/1406.3143v2) estimate an apparent magnitude of $m=-12.4$, i.e. roughly 6 times brighter than our estimate. This would mean that you only need a 7 cm telescope to damage your eye.

However, 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 you have plenty of time to look away.

> **Comments:**
> **username_4**: I'm not sure this is the right calculation. Betelgeuse subtends only a tiny angle, meaning that when it's in focus all its light is focused on effectively a single point on the retina, whereas the sun's light is spread over an area. I'd be surprised if that didn't make it dangerous when using a much smaller telescope.  
> **username_5**: Additionally, this answer assumes that the brightness of the sun is the minimum required to go blind. For a complete answer, we should probably consider how much brightness the human eye can withstand without being blinded. It could be the case that a much smaller telescope than the one you propose is sufficient.  
> **username_3**: @Nathaniel Yes, you're right. I'm not sure what the completely correct approach would be. A point source — which when seen through the atmosphere would span a few square arcsec — would be much brighter, but damage only a tiny fraction of your eye.  
> **username_3**: @username_5 That's true, that's an assumption. But considering that you _can_ actually look at the Sun for a brief period of time without permanently damaging your eye, I think makes this threshold okay, at least for an order-of-magnitude estimate.  
> **username_3**: I edited to take into account these very relevant comments.  
> **username_6**: Yes but *could you look away?* It would be amazing.  

---

### Answer by username_6 on 2018-12-08. Score: 4

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!

> **Comments:**
> **username_7**: Does it matter that the brightness comparable to the moon will be concentrated in a star-sized point?  
> **username_6**: I am not qualified to say. I will speculate that its intensity will be diffused quite a bit by all the dust & gas between us and Betelgeuse -- it's 640 light-years away. On the other hand, the moon looks *extremely* bright in my own telescope. Bright enough to dazzle the eyes quite a bit.  
> **username_7**: I assumed you'd already included the diffusion in your estimate of the apparent magnitude.  
> **username_6**: My calculation was a simple one. I simply assumed that Betelgeuse might increase in brightness by a million, and converted that increase to apparent magnitude. Check the [third paragraph here](https://www.astronomynotes.com/starprop/s4.htm). This sort of sidesteps complicated diffusion calculations by just assuming diffusion will not change if the thing gets brighter.  
> **username_7**: That seems like a reasonable assumption for broad estimates like this. In reality, diffusion is probably dependent on light wavelength, which I assume changes during supernova. But if we're just dealing with orders of magnitude, it's probably not significant.  

---

### Answer by username_8 on 2020-01-11. Score: 1

From Wikipedia: The visual light curves of the different supernova types all depend at late times on radioactive heating, but they vary in shape and amplitude because of the underlying mechanisms, the way that visible radiation is produced, the epoch of its observation, and the transparency of the ejected material. The light curves can be significantly different at other wavelengths. **For example, at ultraviolet wavelengths there is an early extremely luminous peak lasting only a few hours corresponding to the breakout of the shock launched by the initial event, but that breakout is hardly detectable optically.**

I'm not sure I'd want to be looking at Betelgeuse through a large telescope when this happens. Visible light is not what damages your eye. UV is.

> **Comments:**
> **username_3**: Well, that certainly depends on the brightness. Most UV is absorbed in the atmosphere, so if you stare at the Sun, optical light _is_ what damages the eye.  

---

