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When I look at the full Moon through my 10cm telescope, it is so bright that it hurts. Can large scientific telescopes observe the moon at all? Does that require special protection equipment? Or dedicated telescopes (or none at all)?

In particular, the E-ELT (European Extremely Large Telescope) will have a mirror with a 39m diameter. If it were pointed at the full moon, would that damage the science instruments? Would that generate a significant temperature at the focal point?

uhoh
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j6t
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3 Answers3

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No chance of damage (the giantness of the telescope makes little difference!), but these cameras can not shoot at a shutter speed of 1/1000 second, so the lit part of the Moon is out of reach due to overexposure.

Can large scientific telescopes observe the moon...

The bright part will probably be too bright to easily image with a deep field camera because it's designed for integration times of seconds to minutes.

The hardware won't be able to provide a 1/1000 second exposure, so only objects in shadows (or the unlit side of the Moon) have a chance of being exposed.

When I look at the full moon through my 10cm telescope, it is so bright that it hurts

Because of conservation of etendue (see below) the Moon has the same surface brightness when seen through any telescope or binocular. It's just that it's bigger and so is spread over a larger area of your retina.

It's just like looking at 100 full Moons in the sky, but each Moon is no brighter than the one we see now.

Put in less than precise but simple wording, magnification increases the size, but not the apparent brightness per unit area of extended objects.

...without being damaged?

There's no chance of damage.

This answer to Can a telescope ever increase the apparent luminance of an extended object? says No and explains that this is the result of conservation of etendue

In big telescopes, the focal planes are also pretty huge.

(units: mm)                aperture focal length   f/no.
Human eye                        6         17      2.8      
Vera C. Reuben telescope     8,360     10,310      1.23

So per square micron, the image of the moon will be $(2.8/1.23)^2 \approx 5$ times brighter on the worst case1 telescope's focal plane than on our retina (seen through a telescope or by eye), that's not going to hurt the silicon.

After all we often take outdoor photos with the Sun in the field of view and that doesn't even melt the polymer coatings and color filters on top of the CCD!

1lowest f/no. big telescope so brightest per unit area on the sensor.

LSST Focal Plane

Source

Suzanne Jacoby with the LSST focal plane array scale model. The array's diameter is 64 cm. This mosaic will provide over 3 gigapixels per image. The image of the moon (30 arcminutes) is present to show the scale of the field of view.

Glorfindel
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uhoh
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    "that doesn't even melt the polymer coatings and color filters on top of the CCD" ... perhaps because of very short shutter speeds, which you have established are not available to scientific instruments. See also https://photo.stackexchange.com/questions/4016/can-the-sun-damage-the-camera-sensor-under-what-conditions – Eric Towers Aug 30 '21 at 12:18
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    A link on the photo.stackexchange page Eric Towers referenced, showing the damage done to rental lenses by people taking photos of a solar eclipse: https://www.lensrentals.com/blog/2017/09/rental-camera-gear-destroyed-by-the-solar-eclipse-of-2017/ – Peter Erwin Aug 30 '21 at 12:27
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    And this video, showing how easy it is to damage a DSLR by attaching it to a large telephoto lens pointing it at the Sun (even with the shutter closed, you can totally destroy the sensor): https://www.youtube.com/watch?v=2TO_yZDxryQ&t=109s – Peter Erwin Aug 30 '21 at 12:30
  • Can't increase apparent brightness? Did you never start fires with a magnifying glass when you were a kid? – jamesqf Aug 30 '21 at 15:06
  • @jamesqf the reason our eyes do not start on fire are several; 1) our retina is substantially transparent; the light decays over a distance longer than it does in brown grass, 2) our eyes are aqueous which has a higher heat capacity than grass, 3) there is intervening material before the retina that while it is sort-of transparent, actually absorbs a significant amount of light before the retina (esp. IR and UV) 4) we don't (try not to) stare at the sun for an extended period of time. There are probably others. – uhoh Aug 30 '21 at 16:43
  • @jamesqf rather an an oblique comment, why not express your hesitation in the form of a Stack Exchange question, perhaps here or in Physics SE and see where it goes! – uhoh Aug 30 '21 at 16:44
  • @EricTowers point-and-shoot cameras don't have physical shutters, neither do cell phones, neither do many mirrorless DLSRs. None of these cameras is damaged when the sun gets in the frame for seconds or longer, otherwise nobody would buy them! Imagine trying to sell a camera that you couldn't take outdoors on a sunny day! – uhoh Aug 30 '21 at 16:47
  • @PeterErwin why not post all that stuff as an answer to my Photography SE question? That way photographers with experience in these matters can vote on it and vet it? – uhoh Aug 30 '21 at 16:48
  • @jamesqf h̶e̶a̶t̶ ̶c̶a̶p̶a̶c̶i̶t̶y̶ thermal conductivity and specific heat. It's why you need dry grass to start your fires and you can't use a live tree (unless it has some pretty dry, thin bark). – uhoh Aug 30 '21 at 16:54
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    @uhoh: No, the reason our eyes don't start on fire (besides being made of soggy wet stuff) is that they don't have a large enough area to collect sufficient solar energy and concentrate it. Increase the area substantially, and you can set almost anything on fire (or melt/vaporize it): https://en.wikipedia.org/wiki/Odeillo_solar_furnace And I don't express this as a question, because I don't HAVE a question. I was just trying to get you to think about why your answer just might be wrong. – jamesqf Aug 30 '21 at 21:54
  • @jamesqf disagree so I've just asked What's the difference between a solar furnace and my eyeball if I glance at the Sun? Besides water at the focal plane, is it basically just f/no.? Let's see what happens. – uhoh Aug 30 '21 at 23:35
  • @uhoh The links I posted came from one of the answers (3rd highest ranked) on the Photography SE page Eric Towers linked to. So they've already been vetted. – Peter Erwin Aug 30 '21 at 23:42
  • @PeterErwin okay we can smoke a very thin sheet of black anodized stainless steel but that's not what an image sensor is; it's mostly transparent material on top of silicon. The attenuation length of visible light is several microns to tens of microns in near IR compared to probably nanometers in the anodized steel, and the thermal conductivity of silicon is better, and much of the absorbed energy is converted first to carriers which move and are collected before they are thermalized. It's an apples and wax apples comparison. – uhoh Aug 30 '21 at 23:55
  • @uhoh From the linked blog post: "The most common problem we’ve encountered with damage done by the eclipse was sensors being destroyed by the heat. We warned everyone in a blog post to buy a solar filter for your lens, and also sent out mass emails and fliers explaining what you need to adequately protect the equipment. But not everyone follows the rules, and as a result, we have quite a few destroyed sensors." – Peter Erwin Aug 31 '21 at 00:10
  • @PeterErwin okay I will think about how to ask a new question somewhere and cite that blogpost, something along the lines of "how come my f/2 iPhone 6 point-and-shoot camera has never been damaged by shooting into the Sun, but other people's cameras get damaged? The problem with asking that in Photography SE is that sometimes Physics doesn't trump firmly-held beliefs, so I might ask in Physics SE. Let's see how things go with my solar furnace eyeball question first. – uhoh Aug 31 '21 at 00:14
  • @PeterErwin As mentioned there, it may have to do with scale; the intensity of an f/2 image at the focal plane should be the same for a tiny and huge lens, but for the part of the intensity that's thermalized the circumference to area ratio has scale; heat can dissipate from the edges of a 40 micron spot much more easily than from a 2000 micron spot. I think that may be the key to the difference. Security cameras and sky cameras can look at the Sun all day every day for decades... – uhoh Aug 31 '21 at 00:19
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    Wow! A GIGApixel camera is awesomely big :D – The_Sympathizer Sep 01 '21 at 00:23
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It could damage the instruments, but it can't generate very significant temperature. You can't use optics (mirrors, lenses) to heat something more than the temperature of the thing itself! If you could you would be transferring heat from something cold to something hot, without doing any work, and that breaks the second law of thermodynamics.

The surface of the moon is at about 100*C, so thermodynamically it is impossible to use moonlight to heat something above that temperature. In practice, even with a large mirror, you don't get lots of heat. At least not enough to melt the CCD.

However, the instruments are not designed to deal with that much light for a long time. This could be handled by simply using very short exposures.

James K
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  • Yes, already linked to it. – James K Aug 29 '21 at 23:05
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    The thermodynamic argument only applies to the temperature of the sun, not the moon. I realize that Munroe says otherwise, but as far as I can tell he's simply wrong. The temperature of the moon rocks depends on the light they absorb, while moonlight is the light they don't absorb, and there's no particular connection between the two. – benrg Aug 29 '21 at 23:24
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    Exactly, as a thought experiment imagine a giant mirror in space instead of the Moon, it would melt stuff regardless of whether it's cold or not. Moon is just a not-very-good mirror. – JohnEye Aug 30 '21 at 07:47
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    @JohnEye but the not-very-good part is important. – user253751 Aug 30 '21 at 08:46
  • @user253751 Why? If only 1% of the surface of any given rock reflects sunlight, it's still sunlight and it's still hot. You just need a bigger telescope. What you cannot do with a big telescope is focus the black-body radiation of an object with some temperature to a higher temperature than what the object has. People should stop taking Munroe's stuff as gospel. He's an uber-nerd, but he's still human. – JohnEye Aug 30 '21 at 09:22
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    @JohnEye The second half of the what-if article addresses your exact argument. What about it is incorrect? – Spacedog Aug 30 '21 at 16:03
  • I think Munroe is correct when he discusses "conservation of etendue"; I suspect he's wrong in the second-to-last paragraph, as benrg and JohnEye suggest. – Peter Erwin Aug 30 '21 at 23:56
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    @Spacedog At least with a proper mirror the argument would be clearly wrong (and I do note that Munroe didn't make it): Looking into a mirror from close up is like looking into the object proper; the mirror is optically equivalent to a glass pane and the argument would be "you cannot make anything hotter than the glass pane" which is obviously wrong as can be demonstrated with a lens behind a window. Now the question is whether that line of thought also applies to scattered reflection. – Peter - Reinstate Monica Aug 31 '21 at 17:16
  • @Spacedog This answer claims that while Munroe says "but, you say, the Moon is not a blackbody!" he continues to treat it as such in the rest of his argument. – Peter - Reinstate Monica Aug 31 '21 at 17:32
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    @Peter - Reinstate Monica: Suppose we have a large number of mirrors at our disposal. Say each mirror is 1 m^2, and so intercepts 1 KW of solar radiation. (We're doing this thought experiment in space.) The energy is reflected to a 1 m^2 spot, so the spot recieves 1 KW/m^2. Add a second mirror, it gets 2 KW/m^2. Increase the number of mirrors to a million, and that spot is getting 1 GW/m^2, no? And (ignoring problems of stable orbits) we could increase the number until we're focusing a large fraction of the sun's output on that 1 square meter. What happens? – jamesqf Aug 31 '21 at 18:46
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    Is the moon not just reflecting light from something that is ... a whole lot hotter than 100°C? – rackandboneman Sep 01 '21 at 01:11
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It probably depends on the details of the equipped sensor(s).

Let's assume the sensor size in the focal area is around 10cm x 10cm (a typical size, if you look at VLT instrumentation).

The sun is about 400.000 times brighter than the moon at 1370W/m².

So the irradiation at the focal area hence is:

$$ P = \frac{1370W/m²}{400.000} \cdot \frac{\left(\frac{39m}{2}\right)^2\cdot\pi}{0.01m²} = 410W/m² $$

Thus we end up with an irradiation intensity about 1/3 of that of the naked sun - that's something typical CCDs can still cope with - though I'd caution to do so for extended exposures and not attach an objective for visual observation - it would damage your eyes. Depending on the detailed thermal properties of the sensor, its cooling and fixation might sustain damage when some elements become overheated due to prolonged exposure. This is an especially likely scenario if filters are involved as often employed - they are not designed to absorb much energy, they are designed for low light situations and wavelength precision.

planetmaker
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  • "an irradiation intensity about 1/3 of that of the naked sun" When I stand outside in the Sun, my skin is slightly warmed by sunlight, but I don't see how 1/3 of that is going to sustain damage to a CCD. You might check how much power is dissipated by the on-chip readout circuitry per square meter and see if it's not actually larger. Light itself is certainly not going to hurt it. – uhoh Aug 29 '21 at 23:20