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At first this seems like a stupid question: "Have you never used a magnifying glass on a sunny day?!"

But any lens will only ever make the focused image as intense as the target or weaker. Intensity here means the light per area of the generated image. This follows from laws of thermodynamics, and post #7 on this blog describes it. This is also why you can't make a lens-only pair of night vision goggles.

When you look at the sun through a telescope, the image becomes larger and brighter. The total amount of energy hitting my retina is higher, but the light per retina cell is typically lower (for most telescope magnifications). It can never be greater with passive optics.

So is the issue that my cornea (not my retina) will heat up too quickly and get damaged? I can't forsee how my retina would get damaged from this before my cornea would, but tons of reputable websites harp on my retina getting damaged. Or is that my aqueous humour can dissipate the heat quickly enough at low total energy even at high intensity, but not high total energy and medium intensity?

The answer here has implications for the relative danger of looking at a full eclipse through a telescope, but toward the end of totality you may accidentally view the sun partially for a split second as the moon moves out of the way. My understanding suggests this wouldn't be more damaging than just accidentally looking at the sun with the naked eye.

Edit: Useful links for understanding this question

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    $\begingroup$ "The total amount of energy hitting my retina is higher, but the light per retina cell is typically lower" Where are you getting this from? $\endgroup$
    – g s
    Commented Aug 14, 2023 at 4:52
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    $\begingroup$ "My understanding suggests this wouldn't be more damaging than just accidentally looking at the sun with the naked eye." — don't forget that total eclipse results in twilight-like overall brightness of the sky, which means it's likely that the pupil is dilated. This means the total brightness of the image of the appearing part of the Sun is higher. Additionally, while you can't keep looking at the full solar disk high in the sky, the total illumination from the just-appeared solar crescent after the eclipse is quite small that may not cause discomfort to force you blink or look away. $\endgroup$
    – Ruslan
    Commented Aug 14, 2023 at 5:19
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    $\begingroup$ As for the core question, note that larger image of the Sun at the comparable brightness to the case of naked-eye observation means more distance for heat to propagate away from the center of the image to unheated parts of the retina and sclera, which means increase of the temperature in the central part of the image — thus a more severe burn. $\endgroup$
    – Ruslan
    Commented Aug 14, 2023 at 5:25
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    $\begingroup$ "Also, it's not obvious the heat propagation problem is 2D." — well, the heat from the outer annuli of the image also needs to go somewhere. Heat dissipation will lead to heating of the surroundings, and thus reduce the temperature gradient that drives dissipation of heat from the center. Even with the naked eye you can damage the retina if you resist your reflexes and continue looking, and if you increase the area that you heat, you'll speed up the damage. $\endgroup$
    – Ruslan
    Commented Aug 14, 2023 at 8:31
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    $\begingroup$ "I can't forsee how my retina would get damaged from this before my cornea would". The cornea is transparent; the retina is not. Why would the cornea be damaged by sunlight? Furthermore, accounts from people who actually went blind from looking at the sun indicate that only the area of vision where the sun was focused on was actually was damaged, which wouldn't happen with corneal damage. Lastly, corneas can be replaced, not retinas. If looking at the sun caused corneal damage, it wouldn't be permanent. $\endgroup$
    – Rafael
    Commented Apr 12 at 14:18

8 Answers 8

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Summary

There are some traps only smart people can fall into. Radiative transfer is laden with traps. Here, I believe the key trap (which can be described in multiple ways) is between surface brightness being conflated with flux. You can think of surface brightness as flux per solid angle, which is very unintuitive.

Surface brightness is conserved through passive optics (obligatory caveat that there are situations which do not conserve surface brightness) and is energy per time per unit area per solid angle. Flux can increase, and is energy per time per unit area, and hence danger per unit area. Flux increasing is one of the reasons why we make bigger telescopes to see fainter objects.

The rest of this answer will go into more detail.

OP's conserved quantities, in a nutshell

From: https://en.wikipedia.org/wiki/Etendue

For example, a magnifying glass can increase the intensity of sunlight onto a small spot, but does so because, viewed from the spot that the light is concentrated onto, the apparent size of the sun is increased proportional to the concentration. "brightness" is defined as the optical power emitted per unit solid angle

Let's dissect this example. Etendue and surface brightness are conserved. The color of the Sun doesn't change through any passive optical device -- for blackbody radiation this conservation of color is the most intuitive way to imagine what we mean by "optical power emitted per unit solid angle." Obviously, by thermodynamics, we cannot use the Sun to passively create a spot hotter than the Sun.

But we do know temperatures increase

However, we do know that a magnifying glass will raise the temperature of the dot, and a larger magnifying glass will raise the temperature of the dot more. What is happening, if surface brightness is conserved?

There are many ways to think about why temperatures increase:

  • The energy flux increased, and the temperature increases until outgoing blackbody radiation balances incoming flux.
  • From the point of view of a tiny observer in the dot, the half-degree angular diameter of the Sun has been replaced with a larger Sun. The biggest possible Sun by magnification would take up half the Sky... but let's go further to illustrate the point:if you imagined the Sun being all around you, it's like being inside the Sun, which means you would have the temperature of the Sun. This follows from thermodynamics. So, there is some continuum of temperatures you experience between the half-degree Sun at room temperature and the full-sky-around-you Sun at the temperature of the surface of the Sun. In short: making the Sun bigger from any perspective will increase the temperature of an observer at that perspective.
  • From the point of view of a tiny observer in the dot, the half-degree angular diameter of the Sun has been replaced with a larger Sun. deja vu. This is what the Sun would look like if you were on one of the closer planets. Say, Mercury or Venus. We know both are hotter, and now we can say why in a couple of ways. Again, this is to argue that the larger the Sun looks (which is the same as saying the Sun looks closer), the hotter things get.

The end result: of course surface brightness is conserved, but you can still increase flux and temperature up to some limit. Since we humans are fragile, and the Sun is hot, the limit is comfortably above what is needed to permanently damage your eyes.

Bringing it back to OP's statements

But any lens will only ever make the focused image as intense as the target or weaker.

Yes, refer to etendue or surface brightness etc.

Intensity here means the light per area of the generated image.

No. "Light per area" if taken to be "energy per time per area" would be flux. Intensity as you've called it in the prior sentence is energy per time per area per solid angle. Solid angle is the key distinction. This is confusing language, since the Wikipedia quote I myself quoted from uses Intensity to mean Flux. But if we are careful and precise and say things in terms of "energy per time per area per solid angle" vs "energy per time per area" we can avoid confusion and be consistent.

follows from laws of thermodynamics, and post #7 on this blog describes it. This is also why you can't make a lens-only pair of night vision goggles.

Re: surface brightness, yes!

When you look at the sun through a telescope, the image becomes larger and brighter.

Larger, but surface brightness is conserved, so not "brighter" per se. And larger = hotter!

The total amount of energy hitting my retina is higher, but the light per retina cell is typically lower (for most telescope

Both total energy per time and total energy per time per area and hence total energy per time per cell are higher.

So is the issue that my cornea (not my retina) will heat up too quickly and get damaged? I can't forsee how my retina would get damaged from this before my cornea would, but tons of reputable websites harp on my retina getting damaged. Or is that my aqueous humour can dissipate the heat quickly enough at low total energy even at high intensity, but not high total energy and medium intensity?

Closing up, this last point is just because the light passes through the cornea without interacting with it much, since it is transparent. For a given slice of cornea, there is energy flux into the slice, but an equal amount of energy flux out of the slice. Whereas your retina is made to absorb photons and collect energy, so your retina can burn. Though perhaps a biologist can better answer whether it's truly burning that makes you blind first, or whether it is some other subtle chemical reaction or physical reaction between photons and cells.

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  • $\begingroup$ You've explained that flux can increase compared to unfocused light, but the light at the retina is never unfocused. The flux at the focal plane of an imaging system depends on the speed of the system, not the aperture. Making a bigger telescope will only increase flux compared to a smaller one if the telescope is also faster. But the speed of the telescope + eye system is limited by the eye, so it can't be faster than the eye itself. $\endgroup$
    – Bart W
    Commented May 7 at 8:42
  • $\begingroup$ I don't think this flux argument is correct for several reasons: $\endgroup$ Commented May 7 at 17:07
  • $\begingroup$ Apparently I can't edit a response after 5 minutes, which is how long i took WHILE editing it. Disregard my above message. I think this flux post makes some minor errors I'll try to correct when I have more time, but is overall likely to be correct (the bigger magnifying glass achieving a burning temperature quickly analogy) As well as the fact that the temperature of the retina is what matters to burning it, NOT merely the light intensity. As another analogy: looking at the sun through a telescope very briefly should be fine. $\endgroup$ Commented May 7 at 17:19
  • $\begingroup$ And to answer my true original question: observing a tiny enough sliver of the sun even through a telescope should only be as dangerous as looking at the sun normally, depending on the exact ratios of course. So even if you make a small mistake while looking through a telescope and see part of the sun at the beginning or end of an eclipse, you should be fine. Thanks for your detailed response! $\endgroup$ Commented May 7 at 17:20
  • $\begingroup$ Cheers -- thank you for charitably reading my answer (charitable meaning that, even though I am imperfect, you took it seriously and got something out of it). $\endgroup$
    – Alwin
    Commented May 7 at 23:55
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Why specifically is looking through a telescope at the sun more dangerous than the naked eye?

With the telescope you have the same or slightly reduced irradiance in the image of the sun on your retina. The image of the sun is bigger. Therefore you will cause more damage. (Damage over a larger area)

You can also consider that the heat needs to conduct away from the entire image. It will take longer to do so from the center of the magnified image because the surrounding region of your retina is also hot. So it will stay hot longer causing more damage.

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  • $\begingroup$ (Damage over a larger area) This is the essence of my question. Looking at the sun for a split second does not seem to cause meaningful lasting damage to the part of my retina the light landed on. So the sun appearing, say on 10 different parts of my retina for an equal time should also not damage it more than looking at the sun 10 times over a long period. I alluded to the "dissipate" answer, except that even under a magnifying glass, the blink response time of 100ms seems more than sufficient to prevent substantial heating if I were to accidentally look at the sun through a telescope. $\endgroup$ Commented Aug 14, 2023 at 15:18
  • $\begingroup$ I now realize the real answer was your second paragraph about heat conduction, but I wasn't convinced at the time. I upvoted it at least. $\endgroup$ Commented May 7 at 17:27
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Working principle of refractive telescope, from Hyperphysics

Here you can see the working principle of a refractive telescope (there are many different kinds but I guess your question is mostly about this type of telescope).

Here what happens is that a far-away object irradiates onto an objective lens, which is set at a specific distance from an eyepiece lens, which in the end should recollimate the light so that it can be observed (if it is not collimated then you also have to account for how light would enter into your eye and that complicates the design by a few degrees of complication).

Here you can see a different interpretation of the same principle, but without the nice imaging into the human eye visual system:

Another model, from U. SD

Second thing to understand is the solar radiation irradiance as a function of wavelength:

Irradiance from the sun, from wikipedia

Here you can see that, roughly, solar irradiation is a function of many things: Wavelength (the color of the light), absorption depending on the height with respect to earth position, and in the wikipedia article where I got this picture from, you can also consider what time of the day it is, where in the world you are, etc.

For the sakes of simplification, I will pick just a value: 1 W/m² / nm. To make things even simpler, I will pic 1 W/m² across all wavelengths (meaning that all wavelengths reach the earth with the same power. This is not really physical as there are a number of processes happening in the world at all times, so bear with me that this is not realistic but within reason, approximate enough).

Next thing to consider is what is called optical power damage threshold (typically this is done for lasers so it can also be found as laser power damage threshold) which is a measure to determine how much power and for how long it has to be irradiating the eye to cause considerable damage.

https://www.ncbi.nlm.nih.gov/pmc/articles/PMC6484998/ In this paper you can find a control test with a 1319 nm continuous wave laser which was used to determine the laser power damage threshold on rabbits (:().

From the conclusions:

The rabbit retinal damage thresholds at 24-h post exposure for the ocular axial lengths of 15.97 and 17.25 mm were 1.06 and 1.79 J respectively. The obvious difference for the damage thresholds resulted from the dependences of pre-retinal absorption and retinal spot size on the ocular axial length. Detailed analysis indicated that a sufficient margin existed between the damage threshold and MPE for adult humans, but for the newborn eyes the safety factor may be less than 2.3. The obtained results could be used in the refinement of the safety standards for transitional NIR lasers.

So we know that laser damage will happen, roughly, at 1.06-1.7 J for 1319 nm. Considering our previous assumptions here, you can see that we have that the sun is reaching with 1 W/m² and damage happens at roughly 1.5J.

Since we want oranges Vs. oranges for our exercise, you need to convert one until somehow. A watt is a unit of energy per time unit, so 1 W/m² is 1 J/s/m² or J/sm². Since we have J in our damage threshold, we need to convert the solar irradiation to a measureable power.

1 W/m² is 1e-6 W/mm² = 1e-6 J/s mm².

https://www.amazon.de/-/en/Monocular-Telescope-Waterproof-Outdoor-Hunting/dp/B07N1M47CG

A typical 40x commercially available cheap telescope will give you a 60mm objective. That would make, roughly, an area of $pi r^2$ = 11300 mm². The impinging power onto the objective lens would be 0.011 W (1e-6 * 11300 mm²), or 0.011 J/s.

With a magnification of 40x, we can roughly state that the telescope will roughly give 0.44J/s.

You can clearly see from the quick calculation exercise where the problem comes from: While the solar irradiation is relatively low (or low enough not to cause severe damage in case you look at the sun directly for less than a second), if you put it into an optical system that increases the concentration of light to a smaller area. Here you can see a lot of parallels with laser safety regulations where there are some types of lasers where the natural reaction of the eyelids to close is sufficient to avoid permanent damage, but the longer that you keep your eyesight onto the eyepiece, and the higher the magnification, the higher the power that reaches your eye and can therefore damage it irreversibly.

As a final note, here it is important to note that a) some calculations are just desk calculations and may differ from the actual situation. The calculations are just to give a rough idea of the physical processes but in terms of irradiance, magnification, etc. There may be more accurate calcuations in the literature.

There is for example a review on the effects of UV in eyesight: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3872277/.

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  • $\begingroup$ Thanks for the detailed response. However, a laser is substantially different than looking at the sun through a telescope, because while the laser appears as a point source, and therefor will land mostly as a point source on the retina, the sun will be spread out on the retina, which is a core detail of the question. $\endgroup$ Commented Aug 14, 2023 at 15:28
  • $\begingroup$ If you look at the second image, you can see that there is a "bundle of rays" instead of a single point. In optics you typically consider the object as a set of parallel rays which are roughly the size of the object. For far-away objects, you can approximate the light it emits is roughly a point source (a bit of oversimplification, you can see a related answer: computergraphics.stackexchange.com/questions/2424/…). For far away "point sources", light will be collected by the telescope and follow the rough estimation I provided. $\endgroup$
    – ondas
    Commented Aug 15, 2023 at 12:33
  • $\begingroup$ Approximating the sun as a point source is not appropriate here. It isn't a point source, and if it was, you'd always do permanent damage to your retina by looking at the sun even briefly with the naked eye. So approximating it immediately gives the wrong answer even in the base case. $\endgroup$ Commented Aug 16, 2023 at 16:29
  • $\begingroup$ For a far-away object, approximating the light source as a point is valid for the Inverse-square law: en.wikipedia.org/wiki/Inverse-square_law, this is a good first approximation for celestial bodies and valid across solutions (Here point source does not mean that the point-like illumination reaches the earth/retina, rather it means that light is generated by a point and will radiate until it reaches the measurement point). $\endgroup$
    – ondas
    Commented Aug 21, 2023 at 7:45
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During the transit of Venus a few years ago, I damaged the eyepiece of my very amateur telescope by directing it at the sun to observe the transit. The plastic around the inside of the lens of the eyepiece became pitted where the sun's image had rested during adjustment and had melted the plastic. The same thing would've occurred if I had placed my eye there!

The energy of the suns rays falling on the $100 \,\text{mm}$ diameter aperture being concentrated at a point only $2 \,\text{mm}$ diameter was enough to melt the plastic. It's just the geometry of the area of a circular disk: the sun's energy over $100 \,\text{mm}$ disk of the aperture was being concentrated into a $2 \,\text{mm}$ disk at the eyepiece. That's 2500 times concentration: $$\text{area} = \pi r^2 \\ \text{ratio of areas} = \frac{a_2} {a_1} = \frac{\left(\frac{100}{2}\right)^2\pi}{\left(\frac{2}{2}\right)^2\pi} = 2500$$

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I'm not sure what there is to answer ...

The total amount of energy hitting my retina is higher, but the light per retina cell is typically lower (for most telescope magnifications).

The lens focuses the light energy to a small area. It does this by moving light energy from elsewhere (i.e. where the light would have gone if the lens were not there). Accordingly, you could say that the light per retina cell is typically lower, but some retina cells would be receiving significantly more energy - and so they're still going to be damaged.

Compare: if we put your head in a $-250 ^\circ C$ container and the rest of your body in a $25 ^\circ C$ room, we could argue that the heat energy per body cell is typically sufficient for comfort, but that would be no solace, because you'd still be dead.

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    $\begingroup$ some retina cells would be receiving significantly more energy No. The question starts out by explaining that this is not true. $\endgroup$ Commented Aug 14, 2023 at 15:10
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When people tell you not to look at the sun through a telescope, that's not because of the telescope. It's because of the sun.

You're 100% correct that a telescope (or any passive optical system) can't increase the surface brightness of an object. But the sun is bright. It is in fact so stupidly bright that eclipse glasses need to reflect back 99.999% of the incoming light in order to reduce the brightness to a comfortably safe level for our eyes. Film-based solar filters for telescopes use exactly the same material as eclipse glasses, except they need more of it because telescopes are bigger.

A telescope's reduction of (surface) brightness is due to magnification and transmission losses. Those effects are orders of magnitude away from achieving the same kind of brightness reduction. Additionally, transmission losses will be at least partially due to absorption instead of reflection, meaning that parts of your optical system are absorbing gigantic amounts of heat.

So, from the perspective of optics, looking at the sun with an unfiltered telescope is not more dangerous than without. Both can be done exactly twice: once with your left eye and once with your right. After that, most people will have run out of eyes.

That being said, there are some additional things I can think of that can make using the telescope more dangerous than not using it:

  1. The danger is not only in your eyes. Pointing an unfiltered telescope at the sun turns it into a fire hazard. A telescope is a device for collecting and focusing light. All light rays that enter the telescope sufficiently parallel to the optical axis will get focused in what is approximately a plane. When looking at the sun unfiltered, the energy density in and around this focal plane can get high enough to melt or burn nearby optical elements. This is not an exaggeration. Multiple light buckets have been turned into water buckets by necessity of dousing the flames resulting from the sun accidentally wandering into the field of view.
  2. As noted in other answers, the larger area of the magnified image means a bigger part of the retina is in danger of being burned.
  3. We have an instinctive reaction to avert or close our eyes while looking at bright things. Personally, I can't comfortably look anywhere close to where the sun is in the sky. And yet, people are able to look at the sun through an unfiltered telescope until they jump back in pain because their retina is burning. Either these people foolhardily ignore this instinct or the telescopic view, having an apparent field of view that doesn't nearly fill the entire retina, bypasses it somehow. While this one is more speculation on my part, it's very possible that this plays a role in endangering your eyesight more than not using the telescope would. I am surely not going to test it, and I strongly discourage anyone else from doing so.

I will end this by noting that the damage dealt, as is very usually the case, depends on the duration of exposure. I don't know where the cutoff is, and I don't think anyone is very keen on trying to find out. It is reasonable to think that an accidental exposure of only a fraction of a second is short enough to not be dangerous. However, there is absolutely zero reason to expect the damage to start building up only when the pain starts. There are plenty of things that we don't experience as painful that can cause damage. And one can never be too careful when looking at the sun. You only have two eyes, and you can't regrow them.

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You asked:

So is the issue that my cornea (not my retina) will heat up too quickly and get damaged?

and

Or is that my aqueous humor can dissipate the heat quickly enough at low total energy even at high intensity, but not high total energy and medium intensity?

The reason the cornea is not damaged has to do with the transparency of different materials to different wavelengths.

Glass is opaque to UV radiation, so the sun's rays on your cornea will have been filtered by the optical glass, and will only contain visible light and infrared. Your cornea and aqueous humor are transparent to those frequencies, so the rays pass through without loss of energy (heating). The damage occurs to your retina due to the IR radiation being absorbed by the retina.

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When you look at the sun through a telescope, the image becomes larger and brighter. The total amount of energy hitting my retina is higher, but the light per retina cell is typically lower (for most telescope magnifications). It can never be greater with passive optics.

Indeed, the amount of energy that arrives to the retina is determined by the energy flux allowed through the telescope opening aperture. All this energy is then focused on the retina. If a small loupe on a sunny day is sufficient to set paper on fire, one should not be surprized that similar effect is achieved by a telescope with a larger aperture.

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