1
$\begingroup$

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

$\endgroup$
10
  • 2
    $\begingroup$ "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
    Aug 14 at 4:52
  • $\begingroup$ @gs some details appear in the linked forum post. The basic gist of the argument is that a telescope lens (having a larger aperture than the eyeball) will collect more light (energy/s) than the eyeball alone. But magnification acts to spread the energy over a larger area on the retina and de-magnification would result in clipping on some aperture in the eyeball which is not the retina. The upshot seems to be that, because the eye has a finite entrance pupil, it is impossible to increase the intensity of the image of the sun on the retina with passive optics. $\endgroup$
    – Jagerber48
    Aug 14 at 5:15
  • $\begingroup$ This seems highly related to en.wikipedia.org/wiki/Etendue#Conservation_of_etendue. I have to admit that the argument in the link seems surprising yet somewhat convincing to me. I need to ponder it some more. $\endgroup$
    – Jagerber48
    Aug 14 at 5:15
  • 1
    $\begingroup$ 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
    Aug 14 at 5:25
  • 1
    $\begingroup$ "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
    Aug 14 at 8:31

3 Answers 3

2
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ some retina cells would be receiving significantly more energy No. The question starts out by explaining that this is not true. $\endgroup$ Aug 14 at 15:10
2
$\begingroup$

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/.

$\endgroup$
4
  • $\begingroup$ 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. $\endgroup$ Aug 14 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
    Aug 15 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. $\endgroup$ Aug 16 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
    Aug 21 at 7:45
1
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ (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. $\endgroup$ Aug 14 at 15:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.