This website shows a telescope projecting the sun onto a blackboard: https://astronomyconnect.com/forums/articles/2-three-ways-to-safely-observe-the-sun.21/

Why isn't the board catching fire? You can easily start a fire on a sunny day by targeting the focal point of a magnifying glass onto something flammable. Why isn't the telescope in this picture doing the same thing?

Telescope projecting sun on a blackboard.
Photo by Luis Fernández García

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    $\begingroup$ That should be Tangentially related xkcd what if Can you use a magnifying glass and moonlight to light a fire? $\endgroup$
    – MSalters
    Nov 10 '20 at 16:49
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    $\begingroup$ As a matter of fact, that's exactly how my neighbor's house caught fire when he forgot to bring in his telescope (thankfully, they caught it quickly). So it does happen. $\endgroup$ Nov 11 '20 at 0:57
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    $\begingroup$ I had a similar set up with binoculars, unfortunately I forgot to take the lens cap off the eyepiece I wasn't using. It now has a small hole melted into it. $\endgroup$ Nov 11 '20 at 12:04
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    $\begingroup$ @DarrelHoffman that's not a blackboard - it's white where it's not in shadow. Compare to the bottom right edge of the paper box that's also in full sun Besides a lot of blackboards are black-painted wood $\endgroup$
    – Chris H
    Nov 11 '20 at 14:23
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    $\begingroup$ The answers below are good at explaining why it doesn’t cause a fire, but please note that if the telescope isn’t aligned properly for the image to be projected by it (i.e. the “projection” stays inside the tube), it may cause heat accumulation on parts of the telescope. This happened to me once, as the focus was on a plastic ring in the eyepiece rather than in the middle of said ring, so the ring started melting and smoke came out. I have also seen photos of a wooden-tube telescope that completely burnt down because of a similar event. $\endgroup$ Nov 20 '20 at 23:46

For a magnifying lens or mirror to be able to ignite something with light from the Sun, its surface area must be large relative to the square of the focal length. Solar energy will be spread throughout the projected image, and the size of that image will be essentially proportional to the focal length, making its area proportional to the square of focal length. A typical hand magnifier will have a relatively short focal length, making the projected image quite small. Telescopes, however, are designed to emulate lenses with much longer focal lengths so as to produce larger images. The amount of heating from a telescope will be maximized when it's properly focused, but if the light is spread through a 64mm-diameter image it will be less than 1/1000 as powerful as it would be if it were focused with a shorter lens to produce an image which is only 2mm in diameter.

Incidentally, a factor which makes the "Archimedes death ray" improbable as a means of focusing solar energy to directly ignite ships is that the size of the projected image of the Sun would increase with the distance to the enemy ships. On the other hand, the amount of focused solar energy needed to temporarily or permanently blind people is far below the amount required to ignite things. If the crew of a ship had flaming projectiles they wanted to launch at a town, but sunlight focused by the townspeople's shields were to blind some crew members at an inopportune time, it's not hard to imagine that the ship's crew might accidentally set fire to their own ship or other nearby ships. I think it entirely plausible that people witnessing the battle from shore might have observed that solar energy was being focused on ships, and that the ships subsequently ignited; it's not hard to imagine that such people would conclude that the solar energy ignited the ships whether it actually did or not.


It could start a fire if the screen is at the focal point of the optical system. That is how you light fires with a magnifying glass.

Here, the blackboard is likely away from the focal point, so you can see the shape of the eclipse (and you get a bigger image) without setting things on fire.

Although this is fairly safe, there are a few things to pay attention to:

  • If you do this, make sure nobody can walk between the telescope and the screen, because if they go near the focal point, they could get very hot.
  • Doing this will cause your telescope to heat up. If there are any plastic parts, they can melt.
  • The telescope in the picture seems to have a small opening. Don't do this with a big telescope. You don't need to collect a lot of light.

Not an answer to the question, but an important note: Observing the Sun is the most hazardous thing you can do in astronomy. Make sure you know what you are doing before you try.

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    $\begingroup$ +1.It should be noted, that the projected image is larger than the telescope aperture. Thus the intensity per unit area on the projection plane is less than direct illumination $\endgroup$ Nov 10 '20 at 11:50
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    $\begingroup$ Also, am image is *never *in the focal point as by the image equation 1/f = 1/b + 1/g $\endgroup$ Nov 10 '20 at 11:53
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    $\begingroup$ @planetmaker That is mathematically true but irrelevant, since when observing the sun with the setup shown the picture, b is about 100,000,000,000 times bigger than f. $\endgroup$
    – alephzero
    Nov 10 '20 at 17:11
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    $\begingroup$ I have another rule I follow: never, ever, leave an open telescope (solar or not) unsupervised while the sun is up, especially around the public. If you must leave it (e.g. bathroom break), put on all the covers and remove any eyepiece. People are too curious for their own good. Even without the public around, I heard of one person leaving their telescope “pointing away”, walking off, and the sun moving into alignment. Thankfully nobody got hurt but plastic. And don’t forget the finder! Just take it out. $\endgroup$ Nov 11 '20 at 1:14
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    $\begingroup$ @EuroMicelli as such we always leave the telescope and the dome's opening such that the sun will never 'walk' into it, but leave them further Eastern than the Sun - even when we are present and not immediately observing during day. $\endgroup$ Nov 11 '20 at 12:04

The key quantity, as others have noted, is the ratio of the objective lens area to the area of the Sun image. Suppose you use a magnifying glass of 70 mm diameter and 180 mm focal length. The Sun's apparent angular diameter is 32 arcmin or 9.3 mrad; the focused Sun image diameter is $$ 180~\text{mm} \times 0.0093 = 1.7~\text{mm} $$ As the lens area is 1750 times as large, the image gets hot quickly, with predictable results.

Now suppose the pictured telescope has a focal length of 500 mm and an aperture of 50 mm (f/10). If you remove the eyepiece and put a card at the prime focus instead, you get a 4.6 mm diameter image illuminated 115 times as strongly as direct sunlight. With a little patience you could still start a fire that way.

If you put the eyepiece back in, you can do eyepiece projection as shown in the picture. Suppose the eyepiece has a nominal focal length of 20 mm and the image is projected 500 mm away. Then you can calculate an effective focal length of 12 m and an effective focal ratio of f/240. The 112 mm diameter Sun image is illuminated only 1/5 as strongly as the direct sunlight entering the 50 mm objective. Not only does the projected image not get hot, but it needs a shade to keep direct sunlight from degrading the contrast. However, the prime-focus image inside the eyepiece is just as concentrated as before, so the time pointed at the Sun should be limited to avoid damaging the eyepiece.

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    $\begingroup$ I would expect that if an eyepiece has a reticle or other material object on the focal plane, it would be very quickly destroyed by being pointed directly at the Sun. If there's nothing but air at the focal plane, however, is there any reason the eyepiece should particularly care about how much energy is focused on the air there? $\endgroup$
    – supercat
    Nov 10 '20 at 23:28
  • $\begingroup$ @supercat there's probably an interesting new question there! Solar observatories can have much larger apertures than this and will certainly have intermediate focal planes where intensity might be far higher than this, or might not be (I think it scales as f/no squared) I wonder if they use a dry clean air source or something similar for the environment around high intensity intermediate focal planes? Especially the ones that eventually project on to large areas for viewing/sketching (older observatories) $\endgroup$
    – uhoh
    Nov 11 '20 at 0:07
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    $\begingroup$ @supercat The out-of-focus f/10 cross-section 20 mm behind prime focus is 6.6 mm, still 57x as intense as direct sunlight. $\endgroup$
    – Mike G
    Nov 11 '20 at 1:04
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    $\begingroup$ @supercat, keep in mind that air is not vacuum. Air will heat significantly around the focal point; then heat will transfer to the surroundings (convection then conduction). Prolonged exposure will damage components near the focal point. And an eyepiece (reticle or not) would be quickly damaged if installed without solar filters at the aperture (never use so called “eyepiece solar filter” they are very dangerous). $\endgroup$ Nov 11 '20 at 1:34
  • $\begingroup$ @EuroMicelli: Certainly if one concentrates sunlight enough in air it will get hot, but I would have expected that concentrations far greater than 50x would be needed before that became enough of a problem to cause damage except in devices that would be exceptionally sensitive to differential thermal expansion. $\endgroup$
    – supercat
    Nov 11 '20 at 15:54

It all depends on how concentrated the energy is. Sun light travels through a lens like this: (left to right)

Lenses and focal points

The closer to the focal point the surface of the board is, the more the light that hits the surface of the lens will be concentrated into a smaller area on the board. More light means more heat, which will start a fire. If the board is further away from the focal point, even though the light hits a larger area, each individual molecule receives less of it, thus no point on the board becomes hot enough to catch on fire.

In the case of telescopes, the lens is at the far end of the telescope, and the focal point is at the end you look through (or fairly close). Thus it's easy to tell how focused the light is going to be based on how many telescope widths away the board is.

(A lot of other answers are getting pretty technical in their explanation and used the same pronouns when referring to the lens and the board. I thought these overcomplicated things and that a simple answer might be better.)

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    $\begingroup$ +1 for a simple, accurate explanation with a diagram $\endgroup$ Nov 11 '20 at 12:07
  • $\begingroup$ From your answer, is the image projected on the screen upside down? $\endgroup$
    – Bookaholic
    Nov 11 '20 at 21:52
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    $\begingroup$ -1, because I'm pretty sure this answer is wrong. If the board was not (approximately) at the focal point, the image would be blurred, which it clearly isn't. Instead, as supercat and Mike G correctly note below, what matters is the size of the image relative to the size of the objective lens that collects the light. In this case, the image is projected relatively far from the telescope, and is thus relatively large. If the telescope was refocused to project a smaller image at a closer distance, the light intensity would increase. $\endgroup$ Nov 12 '20 at 0:52
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    $\begingroup$ "From your answer, is the image projected on the screen upside down?" That depends if this is a homemade telescope with only a single lens then yes. However, telescopes usually have at least one other lens you put your eye up to, that probably puts things right side up, and makes sure the light diverges after leaving the tube. So I'm going to say no. That being said I am no expert on telescopes and only looked through them a handful of times in my life. $\endgroup$
    – Caston
    Nov 12 '20 at 5:18
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    $\begingroup$ @ llmari Karonen While what you say is correct, it's really hard to tell if the sun is in focus since it's such a simple shape (a crescent), and the photo isn't an amazing resolution. Furthermore, given that the light is coming in virtually straight at the lens (since the light source is the sun), can the focused image be larger than the surface area of the lens? Since the image does seem pretty sharp though, do we know if there actually is a lens in said telescope? Usually, this setup only requires a hole, doesn't it? Perhaps it's nothing but a hollowed-out tube. $\endgroup$
    – Caston
    Nov 12 '20 at 5:37

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