First a disclaimer: I KNOW that viewing the sun through a telescope will burn my eyes, so no need to convince me of that. I'm not going to try it. This question is about the optics and physics of why that is true.

A friend and I were discussing ways to safely view the upcoming solar eclipse, and I discovered a huge gap in my understanding of optics theory. Namely, these two truths:

  1. Telescopes cannot increase the surface brightness of an image, they can only increase total brightness by making the image bigger at the same surface brightness. No point in the bigger image will be brighter than it is with the naked eye, due to the conservation of étendue. See https://www.rocketmime.com/astronomy/Telescope/SurfaceBrightness.html.
  2. If you look directly at the sun through a magnifying telescope, you will quickly burn and permanently damage your eyes. However you can get away with looking at the sun with the naked eye for a second or two and not immediately go blind.

I accept both of these as true and am not challenging them, but seem to me to be in contradiction to each other, even from a purely optical and physical standpoint (the biology of the eye and human visual system being irrelevant). What am I missing?

To further clarify my question / confusion, consider this thought experiment. Let's construct a camera out of a closed cardboard box, with a small hole + lens on the front, and a simple sheet of paper on the back (inside) acting as the "film". Scenario 1: We point this camera at the sun, and it makes a tiny bright circle on the paper (a tiny image of the sun), but not bright enough to ignite and burn the paper because the aperture and lens are too small. Scenario 2: Next, we place a telescope in front of our camera, which magnifies the image and creates a larger circle on the paper.

The key question: Will any point of the paper in Scenario 2 become hotter than it was in Scenario 1? Is it possible cause the paper to ignite?

Truth 1 would indicate that the answer is no, while truth 2 would indicate that the answer is yes. What am I missing?

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    \$\begingroup\$ I am reasonably certain that I'd be unable to hold my gaze on the Sun for even one whole second when it was high in a clear sky. And, I don't know where you found the "two seconds...no permanent harm" info, but even if I could train myself not to blink, I wouldn't try it no matter who said it was safe. \$\endgroup\$ Commented Mar 19 at 19:34
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    \$\begingroup\$ This might be a physics stack exchange topic. My guess would be massively more total energy since you are gathering from something like a 4 inch disk instead of a 4 millimeter disk. So even if the brightness is invariant (which I'm not sure I agree with since apertures are a thing), there is more energized area on the retina and less ability to cool it. \$\endgroup\$
    – davolfman
    Commented Mar 19 at 20:51
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    \$\begingroup\$ Curiosity is good but ultimately this site is about solving photographic problems, and I personally don't really think that "Why can't I stare directly at an optically magnified view of the sun?" is a real photographic problem. I think this question needs a re-think at least and a re-write ideally. \$\endgroup\$
    – osullic
    Commented Mar 19 at 21:11
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    \$\begingroup\$ You obviously never burned holes in paper with a magnifying glass. \$\endgroup\$ Commented Mar 20 at 11:54
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    \$\begingroup\$ I apologize if this should have been in the physics stack exchange. I've seen other similar questions on here about how lens properties influence image brightness, and this seems related to those, since the main point of my confusion centers around the application of how magnification influences (or doesn't influence) image brightness. \$\endgroup\$ Commented Mar 21 at 3:03

5 Answers 5


Assuming your eye is part of the optic system with the telescope, and assuming the exit pupil of the telescope's eyepiece isn't larger than your eye's iris (that is, your eye doesn't contribute as a secondary aperture stop), then a telescope+eye system is no different than any other imaging system. \$\require{cancel}\$

As it turns out, with some back-of-the envelope calculations, you can show that for any (lossless) optical system that produces a photographic image of the sun (that is, we're not talking about solar concentrators for solar thermal power), the intensity of the sun's image on the sensor is approximately $${12\over N^2}\,{\mathrm{W}\over\mathrm{mm^2}}$$

That is, for any square-millimeter section of the image sensor (or your eye's retina) that is completely covered by part of the sun's image (i.e., inside the sun's disk), that area is receiving thermal energy of about 12 watts divided by the f-number of the lens/telescope squared. That's a lot of thermal energy for your retina, and for most imaging sensors. Some cameras sensors can handle that intensity for a short period of time because the actual image area is so small (such as with smaller phone cameras, or for very wide-angle lenses that produce such a small sun image).

The maths of the above value are fairly simple. Rectilinear lenses (any lens used to image a solar eclipse will be rectilinear, because you're interested in apparent magnification) obey the pinhole projection formula, which is simply that rays entering the lens at an angle \$\alpha\$ to the optical axis exit at the same angle. This implies that a subject of diameter \$d_\mathrm{s}\$ in focus at a distance \$u\$ from the camera will result in a image of diameter \$d_\mathrm{i}\$ (with a rear focus distance of \$v\$) according to similar triangles: $$ {d_\mathrm{s} \over u} = {d_\mathrm{i} \over v} $$ For subjects very far away (such as the astronomical objects), the rear focus distance is essentially identical to the lens's focal length \$f\$: \$v \approx f\$. Thus, astronomical objects of diameter \$d_\mathrm{s}\$ at a distance \$u\$ from the Earth will result in a focused image of diameter \$d_\mathrm{i} = f\cdot (d_\mathrm{s}/u)\$.

The ratio of the Earth–Sun distance (\$1\,\mathrm{AU}\$) to the sun's diameter \$d_☉\$ is 107.5, or about \$108:1\$. So the diameter of the image of the sun on the camera's sensor is $$ d_\mathrm{i} = {f \over 108}\,. $$ Note that this is regardless of the lens's diameter, the lens's aperture size, or the size of the lens's entrance pupil.

For example, capturing the sun during an eclipse with a 500 mm lens will result in an image of \$500\,\mathrm{mm}/108 = 4.6\,\mathrm{mm}\$ on the sensor. Note that this is just the disc of the sun. This does not include the coronal halo, which is substantially larger, and usually of interest during an eclipse.

Roughly speaking, under direct sunlight on a clear day, the incoming solar radiation (insolation), the amount of power per area, is nominally \$\mathrm{I_s} = 1\,{\mathrm{kW/m^2}}\$. So aiming your camera directly at the sun, the lens's entrance pupil has a diameter \$D = f/N\$, and therefore experiences the power of \$\mathrm{I_s}\$ times the area of the entrance pupil: $$ \begin{align*} P &= \mathrm{I_s} \times \text{entrance pupil area} \\ &= \mathrm{I_s} \times \pi\;\left({D\over2}\right)^2 \\ &= \mathrm{I_s} \times {\pi\over4}\left({f\over N}\right)^2 \\ \end{align*} $$ All of that incoming power is focused onto the sun's image on the sensor. We already know the sun's image has a diameter of \$d_\mathrm{i}=f/108\$. So in terms of power per unit area of the sun's image on sensor, the sun-image sensor area experiences an irradiance of that power \$P\$ divided by the sun's image area (\$A_\mathrm{i} = \pi(\tfrac{d_\mathrm{i}}{2})^2\$): $$ \begin{align*} I &= {P \over \text{sun image area}} \\ &= P \times {1\over \pi} \left({2\over d_\mathrm{i}}\right)^2 \\ &= P \times {4\over\pi}\left({108\over f}\right)^2 \\ &= \mathrm{I_s}\times \bcancel{\pi\over 4}{\cancel{f^2}\over N^2}\bcancel{4\over\pi}{108^2\over \cancel{f^2}} \\ &= {1\,\mathrm{kW}\over \mathrm{m^2}} \cdot {108^2\over N^2} \\ &\approx {12\over N^2}\,{\mathrm{MW\over m^2}} = {12\over N^2}\,{\mathrm{W\over mm^2}} \end{align*} $$ This feels like a surprising fact, that regardless of the lens size, camera size, etc., the power intensity per unit area of the imaged sun on the sensor is only affected by the lens's aperture number. So a big DSLR with a fast \$f/2\$ lens results in \$3\,\mathrm{W/mm^2}\$, whereas an iPhone's \$f/1.78\$ lens results in 27% higher radiant power intensity, at \$3.8\,\mathrm{W/mm^2}\$. Of course, the difference is that the DSLR is producing that power intensity over a larger area on the sensor than the iPhone is. And assuming that the radiant heat distribution is approximately linear through the thin plane of the image sensor, the DSLR suffers from a larger Area/‌‌Circumference ratio of the sun's image on sensor. Thus, the thermal resistance to radiant heat dissipation issue scales, unfavorably, linearly with the lens's focal length.

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    \$\begingroup\$ Deleting my answer. This one is much more complete as it also has the formulas to the concepts. \$\endgroup\$ Commented Mar 22 at 9:35
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    \$\begingroup\$ Does the focussing process of the lens have a chance of making matters worse as it could potentially concentrate the beams tighter during the finding of focus, as it does not produce a sharp image, but a smaller area that the sunlight concentrates on and thus increasing the w/mm2 while decreasing the mm2? \$\endgroup\$ Commented Mar 22 at 9:41
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    \$\begingroup\$ @BenHershey You're going to have to unlearn "multiple lenses and apertures (like telescope + eye pupil)". Telescopes and binoculars aren't image forming systems like camera lenses are. A "pencil" of parallel rays entering a telescope or binoculars exit them also parallel; whereas parallel light entering a camera lens will converge to the lens's rear focus distance. \$\endgroup\$
    – scottbb
    Commented Mar 25 at 17:48
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    \$\begingroup\$ @BenHershey For telescopes, the amount of light collected by the front element describes the aperture size. Telescopy doesn't really care about f-number (N) like photography does, because amount of light entering the system is all that matters. Telescopy doesn't care about depth-of-field or other f-number issues. So "calculating overall N" in telescopy is kind of a nonsense concept. I only mentioned it as a throwaway in my answer, because my answer ignored real-world losses along elements in the optical path. \$\endgroup\$
    – scottbb
    Commented Mar 25 at 17:50
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    \$\begingroup\$ @MichaelC Okay... correction (to my previous statement): a telescope with an eyepiece is afocal, because the eye completes the optical system and makes the image. Removing the eyepiece from the telescope makes it converging focal, so mounting a camera onto the telescope works. \$\endgroup\$
    – scottbb
    Commented Mar 26 at 14:09

There is no significant difference between looking at the sun with the naked eye and taking a picture of it with a digital camera. There are only two variables that control the exposure, the duration of time which light is allowed to pass, and the diaphragm restriction limiting the amount of light that passes.

Magnification is not a factor... i.e. the same source recorded larger at the same exposure is more light; but it is not more light per area. The magnification itself does not change the amount of light transmitted/recorded. Increased magnification actually spreads the light out over a larger area and reduces the exposure (light/area); which is why it must be compensated for with a larger diaphragm restriction (same f#), or longer exposure (same diaphragm restriction/smaller f#).

The answer to the key question is "No." That scenario is essentially the same thing as adding a teleconverter (magnifier) to a lens... it increases (magnifies) the recorded subject/image area and reduces the light density, resulting in a slower (smaller) effective f#.

I believe your error is in the understanding of statement #2. If you look directly at the sun with a telescope or your eyes you will quickly burn and permanently damage your eyes (retinas). And looking at the sun in either manner will not immediately make you go blind.

Going blind is unlikely, what is likely is permanent damage of the retina at the size/area the sun was focused on it... i.e. a permanent (second) blind spot. Temporary/more widespread blindness is also likely due to the bleaching of the photoreceptors (cones/rods), but the eye will recover from that.

It is also possible to cause permanent damage to the eye (solar retinopathy) with exposures of a few seconds or more. But this is a chemical reaction (phototoxicity), and not caused by thermal energy. That is why it is also commonly referred to as "eclipse retinopathy"... because it is possible to look at a near total eclipse (or with insufficient protection) for a longer period without discomfort/thermal damage; while also unknowingly damaging your eyes. The symptoms of this kind of damage takes time to develop, it is not "immediate." And this is the major risk of knockoff/uncertified solar glasses and DIY solutions.

  • \$\begingroup\$ Thanks. That seems to make a lot of sense. Can you provide any relevant terminology, equations, or source links to back up the claim about magnification not being a factor for exposure? I see a lot of conflicting info on the internet around this topic which is why I asked it here. (The claim that magnification makes things brighter is all over the place.) If you can provide some sources, I'll very likely accept yours as the answer. :) \$\endgroup\$ Commented Mar 21 at 3:12
  • \$\begingroup\$ You provided a very good reference in your post. There is a difference between using a magnifying glass to focus the sun on a surface (to burn something) and viewing an image that was magnified. In the first case it is certainly possible to burn something; but the magnification only determines the size of the burn; not the rate/severity. However, when you view an image created by a lens you are not viewing focused light; it is the eye that focuses the light on the retina, and it cannot already be in focus when entering the eye. \$\endgroup\$ Commented Mar 21 at 12:33
  • \$\begingroup\$ I.e. the image/light source must be ≥ 25cm away... For instance, the optical distance (light path and defocus) of a camera's viewfinder is generally set to be approximately 1 meter. This greater distance allows most people without perfect eyesight (mild presbyopia/farsightedness) to still focus the image/light. And when you view a TV, the image on the screen is in focus, but it is radiating the light outwards and that light is not in focus entering the eye. \$\endgroup\$ Commented Mar 21 at 12:38
  • \$\begingroup\$ I also edited my answer to add a little bit more about magnification in imaging. Increased magnification has the opposite effect of concentrating the light... i.e. it is not possible to concentrate light and increase it's intensity with imaging (the fire from moonlight post and your etendue link). It is possible with a reflector, but the result is not an image of the source(s). \$\endgroup\$ Commented Mar 21 at 13:46
  • \$\begingroup\$ That makes a lot of sense. So, just to clarify the claim that magnification is not a factor for exposure: does that only hold for images that are in focus? Is it possible to intensify exposure in a smaller area (concentrate the light) with lenses if the image is out of focus? For a point source of light like the sun and a single lens, the smallest and most concentrated light occurs at focus (at least, seems to be from my tinkering). But I'm not sure if that's still true when multiple lenses are at play. \$\endgroup\$ Commented Mar 23 at 3:57

I find xkcd‘s take on this very telling, it starts by asking a different question: How hot can you make something using optics?

The answer is: You can‘t make something hotter using optics than the temperature of the surface of the source. This means that you can‘t use optics to make fire with moonlight.

But why? The thermodynamic argument is that optics work „for free“ (they don‘t need energy). If it was hotter, you would somehow gain energy.

The crux is, that the image in optics is not a point, but has some extent:

Basically optics is reversible, so you can‘t use optics to concentrate light from two different points A and B (on the sun) onto a single point E on earth, because if light goes back through the optical element, it must come from either A or B, but cannot go back to both.

Same applies to „bundling“ light, see XKCD.

But what optics can do, is to make you almost as hot as the surface of the source. And that means, as hot as the surface of the sun. And that temperature is - well - not good for your eyes (Proteins start to denaturate at about 60°C).

  • \$\begingroup\$ Thanks. I'm a big fan of xkcd and am familiar with the fire from moonlight article. It's not obvious to me though whether that answers the "key question" in my post. Obviously if your eye retina reaches the temperature of the sun, that's bad. But my question is whether adding magnification in front of your eye can get your retina hotter than simply looking at the sun with your eye. Or more generally, does magnification actually increase image brightness or is this a common myth? \$\endgroup\$ Commented Mar 21 at 3:08

Light from a distant object like a star arrives at the telescope as a bundle of parallel rays. The front lens (objective lens) captures these rays. As they traverse the objective lens they are refracted (Latin to bend inward). The revised path of these rays sketches out a cone of light.

The focal length of this objective lens is taken as a measurement lens-to-apex of this cone when the object being viewed is a far distance; like a star said to be at distance infinity.

The image of a star being viewed forms at this apex. We can view it by allowing the cone of image forming rays to play on a white screen of ground glass (sheet of glass with roughened surface). We can also place a short focal length lens downstream of this apex and view a magnified image of the subject.

The brightness of this image is primarily due to the fact that the objective lens has a surface area that gathers light rays much like a funnel. In other words, the larger the diameter of the objective lens the more light gathered; thus the brighter the object's image. Astronomers, wishing to see dim objects design telescopes with gigantic diameter.

Binoculars and terrestrial telescopes have small manageable diameter objective lenses. We are interested in how well they function in dim light. A relative brightness math formula to the rescue.

Say you are peering at night; sailors want binoculars with superior image brightness. The chosen magic numbers are 7X with a 50mm diameter objective. The 50mm (2 inch) front lens gathers the dim light. The eyepiece magnifies and objects appear 7 times closer; thus 7X magnification. The light that reaches our eye come through a aperture that is 50 ÷ 7 = 7.1mm in diameter. Now the average human eye, at night has a iris opening of about 7mm (depending on age). Thus, the 7X50 binocular, or any such combination with a 7mm exit pupil, passes this test.

Another formula used is Relative Brightness. What is the relative brightness of 7X50 binoculars? (50 ÷ 7)^2 = 51. How does this compare to 10X35 binoculars? Math is (35 ÷10)^2 = 12.25. Thus, the smaller binocular is 12.25 ÷ 51 = 0.24 = about 25% of brightness of a pair of 7X50 binoculars.

When light hits objects it can traverse or reflect away or it can be absorbed -- any combination of these. When light is absorbed, its energy is converted to heat. If you are peering at the sun with telescope or binoculars, if precautions are not taken, the brightness of the image will be converted to heat and you likely will suffer eye damage.

Use approved filters to be safe, not sorry.


There are different kinds of telescopes and different elements to control brightness, mainly filters and diaphragms.

  1. Natural filters

The first filter is the atmosphere. You can see a sunset with the naked eye, but at noon there is less atmosphere to filter the light.

The physics behind that is the light gets absorbed by dust particles and the obliquity of the rays needs to go through more and more atmosphere.

  1. Artificial filters, like neutral density or specialized filters block some wavelengths and diminish the power of the wavelengths that actually pass.

  2. Diaphragms. The wider the aperture the more light passes. The narrower the aperture the fewer rays pass. There is a solar observatory that basically is a pinhole, that projects the sun at about 60cm (I do not remember the name of the observatory) but is limited in capabilities because the distance from the pinhole to the plane is quite big so it is (I recall) a big hole on the ground, so it only can see the sun at noon, something like that.

  3. But by optics you are probably referring to lenses.

A starting point would be the area of the entrance of the lens and the area of the projected image.

A magnifying glass of 6cm that projects the sun as a dot of 3mm is focusing the rays 20x20 times or 400 times.

Any lens that projects the sun on an area less than the area of the lens is magnifying light. If the area is the same as the lens, the light will be theoretically 1:1, on a DSLR this would be around a 50mm lens, and if it is bigger then the power per area unit would start to diminish.

And from there you can make some combinations reducing the diaphragm of your lens.


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