# Do waves of light overlap?

Hopefully I can phrase this question so that it can be understood and not appear to be entirely uneducated.

From what I understand when light hits an object it reflects that light toward you. Let's say we have 1,000 white golf balls on the moon clumped together, each one reflecting it's own light, and we zoom in as far as our technology will allow. We now have an image of a bright patch on the moon. Is it possible to further resolve the image by zooming in to an already zoomed image in real-time and see each golf ball? Or do light waves reach a point where they become saturated, overlap and can't be distinguished from one another? I ask this because I am conducting an experiment in which I have created a massive distilled water lens with incredible focussing power. However I still get the same image only larger with minimal increase in resolution. Any insight into this matter would be greatly appreciated.

• Do you have a website where you talk about this water lens? It sounds interesting, like using spinning mercury to make a large telescope mirror. – Mark H Jul 24 '19 at 9:55
• I am working on a liquid lens concept that allows me to change the focal distance by just turning a dial. Unlike a spinning liquid mirror I can point my telescope in any direction. – ComeauConcepts Jul 24 '19 at 10:48
• What do you mean by, "zooming in to an already zoomed image?" If the "already zoomed image" is an image that you have previously captured (e.g., with a camera), then the resolution that you captured is the resolution that you captured. You won't be able to get any more. – Solomon Slow Jul 24 '19 at 12:06
• If you are asking how to construct a telescope that will reveal the individual golf balls, then I suggest that you google "resolution limit of telescope." I don't remember all of the details, but I believe that in order to resolve golf balls on the moon, you will have to build a very large telescope. – Solomon Slow Jul 24 '19 at 12:08

Light from any source is always resolvable given an adequate optical system. There is no perfect optical system even in principle. Even if you have perfectly shaped mirrors and lenses with no aberration and put your telescope in space so there is no atmospheric interference, the finite size of the telescope blurs the image. Light diffracts around sharp edges, and the edge of the aperture of your telescope is such a sharp edge.

Below is a simulated image of a point source of white light created by an optical system with a finite aperture. A point on an object is never focused to a point in the image plane. The resolving power of a telescope is dependent on the size of the disc created in the image plane. Once the telescope is focused and forms an image, the image cannot get any better by magnifying the image plane. The aperture as already blurred the image before any of the optics can have their effect.

Image taken from here.

So, since the image is blurred by the edge of the telescopes aperture, a larger aperture will have a greater resolving power since the edge will interact with less of the light. This resolution can be calculated with the following equation:

$$\theta = 1.220\frac{\lambda}{D}$$ where $$\theta$$ is the minimum angle between two objects you are looking at, $$\lambda$$ is the wavelength of light, and $$D$$ is the diameter of the circular aperture of the telescope (the size of your lens). If you want to double your image resolution (image things half the currently resolvable size), you need an aperture twice as wide. So, in order to image a golf ball on the moon would require an aperture of $$D = 1.220\frac{\lambda}{\theta} = 1.220\frac{550\,\textrm{nm}}{10^{-10}\,\textrm{rad}} = 6710\,\textrm{m}.$$ Your lens would have to be 7 kilometers or 4 miles wide.

• Your insight is greatly appreciated! This answer tells me that I am on the right track. However it leads me to one last question. Conventional astronomy uses aperatures and structures maintain the positioning of the optic system components. Does just the lens itself act as an aperature? If you had a completely suspended lens system would the resolution improve? – ComeauConcepts Jul 24 '19 at 10:05
• @ComeauConcepts The aperture is the maximum width of light light enters the system. So, the diameter of a free-floating lens would count as $D$ in the equation. – Mark H Jul 24 '19 at 10:08

In principle it would be possible to zoom in and resolve individual golf balls, but you would need a pretty good telescope even to see a group of 1,000 that far away. I doubt if Palomar or Mount Wilson could resolve a single golf ball on the moon, but there are fantastic, state-of-the-art telescopes now being built in Hawaii, Chile and elsewhere which probably could. Whether you could persuade the astronomers operating them to use their precious instrument to search for golf balls on the moon is another thing.

• So what you're saying is that it is the lack of image resolving technology that prevents me from being able to see individual objects at great distance? Assuming that I am already using the best tech there is. – ComeauConcepts Jul 24 '19 at 8:38
• You're not using the best technology, it costs billions of dollars! I doubt if even Trump could afford it. – Michael Walsby Jul 24 '19 at 9:17

A golfball on the moon extends an angle of 4 cm over 400.000 km, that is $$10^{-7}$$ radians. Best seeing conditions on earth are ~0.4 arcseconds, so $$2 \cdot 10^{-6}$$ radians, so you cannot optically resolve a golf ball on the moon even if your telescope has a sizeable NA. However, that is probably not what you are trying to do.

What is your focal length, NA and what are your lens aberrations? If you know this you can find the psf of your system. If your object is small enough it will image as the psf. I assume you did not construct a lens without figuring this out in the design phase.

• Technically what I am trying to achieve is placing a hypothetical microscope over the image that a telescope brings to focus, effectively zooming into the image that is already resolved. Resolving in stages to eliminate the necessity for a ridiculously large telescope. – ComeauConcepts Jul 24 '19 at 9:13
• Telescopes these days use CCDs to create the images. For any decent telescope I would guess that you can assume that the CCD already captures the light at the best resolution that the telescope can manage, otherwise you would be wasting money on a big mirror (the expensive bit) that captures light that you can't image and they probably thought of that. – rghome Jul 24 '19 at 9:39
• @rghome, on the other hand, the resolution of the CCD could be better than the resolution of the telescope. Either because (a) "oversampling" simplifies some image processing algorithm that they want to perform or (b) the price was right. – Solomon Slow Jul 24 '19 at 12:11
• @ComeauConcepts, I just noticed that you said, "...a hypothetical microscope." There's a name for that: It's called the "eyepiece" of a telescope. They eyepiece is a microscope that is focused on the real image formed by the telescope's primary. Unfortunately, the diffraction limit, which is explained in other answers, is a limit to how much detail is there in that real-image for you to see. – Solomon Slow Jul 25 '19 at 14:21