Why do telescopes use mirrors that simply reflect photons, when they instead could be covered with large sensors to register them? Reflection is all good and well, all thanks to silver and beryllium for that. But wouldn't it be better to electronically register the photons directly instead of after having them bounce around between stupid mirrors? Would any data be lost in a pure CCD-telescope without any mirrors or lenses?

Couldn't a large wired CCD light sensor send the detected signals further in a smarter way than a stupid physically reflecting surface can do? It's the same photons and the telescope itself doesn't generate any new information about the distant galaxies it reflects. Why physically bend mirrors for adaptive optics, instead of bending the raw binary data with an algorithm for the same effect?

  • $\begingroup$ "the telescope itself doesn't generate any new information about the distant galaxies it reflects" - bear in mind a CCD on its own is just a solar panel - it measures light intensity, but not where the light is coming from. It needs focused light, this is really basic optics. Read the answer from DJohnM again as I think he hit the nail on the head. $\endgroup$ – Andy Mar 30 '16 at 10:01
  • $\begingroup$ By the way there is a CCD that detects direction - a plenoptic camera (plenoptic.info/index.html) but note it uses microlenses on the CCD itself so lenses are still involved. (Plus a whole load of other stuff.) $\endgroup$ – Andy Mar 30 '16 at 10:06

The CCD has no way of recording the direction, the point in the sky, from which a photon is coming.

Say you point your mirror-less telescope at the Moon. Every point on the moon's surface would be reflecting photons onto every part of the CCD at the same time.

You've just created an expensive, sensitive, ambient light meter. There would be no image information whatsoever.

  • $\begingroup$ I still don't get it why stupid silicon in glass as a middleman is better for the photons than the smart silicon in electronics in which it finally ends up in before we can interpret it. Let the CCD have the same shape as a primary mirror would've had, if that helps, and register the photons electronically instead of simply bouncing them. $\endgroup$ – LocalFluff Mar 30 '16 at 5:50
  • 1
    $\begingroup$ It doesn't help.... And the "stupid silicon" is smart enough to receive myriad photons from all over the object and send each one to exactly the right spot on the film or CCD chip to form an image. $\endgroup$ – DJohnM Mar 30 '16 at 6:38
  • $\begingroup$ The photons are sent from the celestial object. Why re-send them? Is the data loss from reflection maybe too small to be a priority problem? $\endgroup$ – LocalFluff Mar 30 '16 at 14:29
  • $\begingroup$ The photons from various objects across the field of view are effectively random until the wavefronts are collected and sorted by a "computer" which realigns all the waves and puts them to a flat plane - optics. (and reflection losses are quite small these days.) $\endgroup$ – Andy Mar 30 '16 at 15:30

To answer your question, we need to first show the job each mirror is doing.

First up, the Newtonian (lovingly called the "Newt", and invented by Sir Ike Newton):


Two mirrors in this design, not surprisingly labeled as primary and secondary.

The job of the primary mirror is NOT merely to reflect light, but to concentrate the diffuse photons onto a much smaller point. This makes really dim objects brighter, and is the first step in magnification. (Further magnification is done by the eyepiece, which is similar to a small refracting telescope.)

In the case of the Newt, the secondary mirror reflects the now concentrated photons to a more convenient point for viewing. Without the secondary mirror your head would get in the way of the view. A secondary mirror is not necessary, and in fact many telescopes will place instruments, such as CCD's, at this "prime focus" point.

In the case of the Hubble Space Telescope, the secondary mirror reflects the concentrated photons to the scope's instruments, where they can work their magic.

In all reflecting telescope designs the primary mirror uses the laws of physics to give the end user, whether it's the human eye, or research gear, as many concentrated photons as possible, maximizing what we can see/detect. The bigger the primary mirror, the more concentrated the photons, and the more we have to work with.

When it comes to seeing what we call "the dim fuzzies", bigger IS better!

  • $\begingroup$ Bigger is better, but couldn't semi-intelligent electronics de-diffuse the photons better than a stupid geometrically reflective surface can? How is it better that a photon from a distant galaxy hits a silicon atom in a mirror before it hits, and is electronically recorded, on a silicon atom on a CCD or similar light sensitive device? Just like all the points on a mirror can produce a focus, I suppose that with some math (not my job), a CCD array could do that too. $\endgroup$ – LocalFluff Mar 29 '16 at 20:09
  • 2
    $\begingroup$ There's a lot more to astronomy observations than just recording photons. For example, look at prisms, grisms, gratings, and filters. A CCD just records photons, without noting the wavelength. $\endgroup$ – Donald.McLean Mar 29 '16 at 20:13
  • $\begingroup$ @Donald.McLean Why not capture all the photons directly, why play ping pong with them? $\endgroup$ – LocalFluff Mar 29 '16 at 20:29
  • $\begingroup$ For much the same reason airplanes take off with a head wind; to give the technology a free boost from Mother Nature. Further, our level of CCD tech is more like a Piper Cub, which can use all the help it can get, than like a jet fighter. Often what we are working on is already on the edge of what we can detect; CCD's by their lonesome are incapable of doing what you propose without the optical assist. $\endgroup$ – SkyGuide Mar 29 '16 at 21:52
  • 2
    $\begingroup$ The math and physics of how and why mirrors and lenses work is actually pretty advanced stuff - MUCH harder to explain than to use. On the flip side, in the radio bands, astronomers have to do all the math that the lenses and mirrors do pretty much for free. The universe is a weird place. $\endgroup$ – Donald.McLean Mar 30 '16 at 12:24

I actually found a concept of 2D easy-to-scale telescope some time ago (here's the link). I guess we will slowly abandon refractor telescopes, for, as I understand, we are pushing them to their limits right now and it is getting really hard to make a bigger one (because of how hard it is to make a sufficiently sized mirror of the quality needed). BUT it worth noting, that I am not an expert in any way, so I suggest someone, who knows more about the topic, edit this answer.

EDIT: There's a really good point in the DJohnM's answer, so I thought I'll add that the thing I linked here (SPIDER) isn't just a big 2D array of CCDs; it actually does have a tiny lense over each of the detectors and each of those measures light in a bunch of different wavelength, so it can preserve information about direction and wavelength of light. So the answer to initial question is no, we can't just build a CCD array instead of full-sized telescope, but the idea to make telescopes scalable in two dimensions instead of three seems to be a good one and threre are people working on it.


Since neither the word "phase" nor "interference" is mentioned in any other answer here, I'll approach it from that direction.

In this answer I said

In an imaging optical telescope (or any imaging system including eyes) every pixel is illuminated simultaneously and directly by all areas of the aperture. From a given point in the distance a telescope will (try to) preserve the phase of all paths reaching the pixel so that the resulting intensity corresponds to the incoming power. This allows the system to obtain the best resolution.

What that means is that the curved mirrors of a reflecting telescope are designed so that all the paths from a distant object in a given direction reach a pixel in phase. The paths from any other point in the sky reach the pixel completely out of phase and cancel to zero. That's why each pixel corresponds to a given direction.

Without those curved mirrors you can't make an image because the CCD's pixels convert the wave's information to intensity only, and loose all phase information. Without any information on phase, there is no way to combine the signals in each pixel to reconstruct the incident wave.

Radio telescope arrays can be though of like your pixels, but those signals are digitized to a bit stream that maintains phase information. The correlator computer takes all those phases and reconstructs the image. If each dish in the array was equipped with a bolometer instead of an RF amplifier and baseband converter, phase information would be lost and no matter how large your baseline you wouldn't have interference.


I you just set out a CCD in a room, each pixel will record photons from every direction. With this, you will be able to record the amount of ambient light, but you won't get an image of the room.

Now if you want to have an image, for each pixel, all the photons have to be coming from the same direction. And for each direction, all the photons coming from that direction have to fall on the same pixel. For that to happen, you can use a camera obscura.

But if you only use the photons coming from one direction for each pixel, you won't be collecting much light, so your image will be rather dark. This is ok if you are taking a picture of a sunny landscape, but if you want to take a picture of stars, you need to collect all the light you can get.

This is where the telescope comes in ! A telescope will collect all of the photons from all of the directions, and reflect them in such a way that all the photons coming from a certain direction will end up on the same pixel. This way, you can have an image that is neither blurry nor dark.

  • $\begingroup$ +1 This demonstration is 99.44% correct (pardon the oblique reference). There are the whole camera that can see around corners, and coded mask technologies that don't require imaging optics, but it's probably beyond the scope of the current question. $\endgroup$ – uhoh Feb 21 '20 at 3:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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