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Hello and thank you for your time: Let's assume I have a 100 squared meter circular flat mirror placed 1 AU from the sun, and let's assume my intention is to shine light on a circular area on another planet located at 3 AU's (it could be 2 AU's, this is totally arbitrary). Would this circular area still be a 100 squared meters patch on the ground of this hypothetical planet, or will the light rays diverge to cover a circular area of more than 100 meters and how much larger ? Please disregard orbits. The sun and the planet are flat, the mirror and everyhting is perfectly aligned, assume an ideal situation, also a perfect mirror; the answer I search for is if the light reflected form this flat mirror will behave like my laser pointer lighted surface does over distance (on earth inside the atmosphere) and grow in surface and lose intensity, in space with no atmosphere. Thank you very much for your answer if you choose to provide one.

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If the sun were a point source viewed from 1 AU, then it would be spatially coherent. (See Van Cittert-Zernike.) In that case, the light from the sun can be treated as a single plane wave illuminating the mirror. One can simply compute the Fourier transform of the circular aperture represented by the mirror to get the far field diffraction pattern that would illuminate the other planet, because far-field diffraction is given by Fraunhofer diffraction, which comes down to a Fourier transform. The result in this case is called an Airy pattern.

How big would this diffraction pattern be on the surface of the other planet? For this one can use a simple formula. The beam divergence angle produced by a circular aperture of radius $R$ is roughly proportional to $\theta=\lambda/R$, where $\lambda$ is the wavelength of the light. This gives you a tiny angle, but over distances of a few AU this can still imply a significant increase in size. The increase in size would be roughly given by $D=\lambda L/R$, where $L$ is the distance to the other planet.

Now, we know that the sun is not a point source at 1 AU, but a disc extending over about half a degree. This means we need to add the intensity patterns (because they are mutually incoherent) of all the plane waves whose propagation vectors fall within half a degree. The result is a convolution of the diffraction pattern's intensity with a disc. So it will smeer out the diffraction pattern.

Moreover, the radiation from the sun consists of a spectrum of wavelength. As a result the different wavelengths would produce diffration patterns of slightly different sizes. So we need to add the intensity patterns of all the different wavelengths in the spectrum. This also causes a smeering effect.

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Point sources obey the inverse square law:

point source

and at astronomical distances the energy dispersion of the incoherent light from a sun can be approximated by this formula.

So by your mirror, you are generating a secondary (reflecting) point source there. The energy it reflects will be at astronomical distances also a point source, depending on the energy content the area of the mirror caught, which depends on geometry and the inverse square law to be calculated.

Note the term incoherent. Laser light is coherent, i.e. the phase relationships are stable for a certain distance and do not obey the inverse square law. Even for lasers there is a distance , which when exceeded the beam starts diverging, the power loss again obeying the inverse square law. See the answer here .

$1.96×10^7$ Compared to the point source multiplier ($0.25$), it is easy to see why a laser spot is much more intense for the same amount of power. It is not because lasers don't follow the inverse square law. It is because the beam's directionality gives it a huge initial boost in intensity, after which it proceeds to fall off as the inverse of distance squared.

Laser light disperses much less, dependent on the laser and can be used for lunar communications.

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  • $\begingroup$ The coherence length does not determine when the laser beam starts to diverge. It is the Rayleigh length that does that. $\endgroup$ – flippiefanus Apr 5 '18 at 5:08
  • $\begingroup$ So, if we use the sun (forgetting about the mirror) using the sun's diameter would the following be true ? 0,00929826069 (diameter in Au's) at sun with an intensity of 1 - 0,03719304384 (diameter in Au's) at sun plus r with an intensity of 1/4th - 0,14877217104 (diameter in Au's) at sun plus 2r with an intensity of 1/9th - And does this mean that at a distance of 2 solar radiuses from the surface of the sun, the intensity of the sun is 1/9th of the original intensity and its coverage area is actually nine times that of the original source ? Sorry, I want to see if I understood correctly. $\endgroup$ – Ivo Jara Apr 5 '18 at 5:25
  • $\begingroup$ @intensity is per unit area, and I do not know what you mean by coverage area? The total area of the expanding sphere will be carrying the same energy, it is the intensity per unit area that gets less. $\endgroup$ – anna v Apr 5 '18 at 6:07
  • $\begingroup$ @flippiefanus A laser beam does not start to diverge, it always is diverging. What you mean is the Rayleigh range dictates when you can describe the beam in a different way. $\endgroup$ – MJC Apr 11 '18 at 14:23
  • $\begingroup$ @MattCliffe: No, a laser beam could also be converging. The phrase "starting to diverge" is a purely qualitative statement. The quantitative description of the beam (as a Gaussian beam) is valid in all regions. $\endgroup$ – flippiefanus Apr 12 '18 at 4:32
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It depends upon whether rays are parallel while falling upon the mirror.

If they are perfectly parallel, they will remain undeflected for long distances.

Even if sun is taken to be a flat plate of infinite area, it consists of several point sources. If you consider an area element dA it should emit light rays not only along the normal to its plane but also along lines making different angles with it. Thus, the light rays from such source cannot be perfectly parallel as far as my imagination proceeds.

Well, if you use some other optical instruments, in theory, you may get it happen.

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  • $\begingroup$ >If they are perfectly parallel, they will remain undeflected for long distances. They are not this is impossible. $\endgroup$ – MJC Apr 11 '18 at 14:21
  • $\begingroup$ @Matt Cliffe. Read correctly.. Light rays are not vectors and it always consists of rays in virtually all directions... $\endgroup$ – user9600383 Apr 26 '18 at 15:21
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    $\begingroup$ I did't say they were I just said what you said was wrong. $\endgroup$ – MJC Apr 30 '18 at 14:16
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Since the sun sends scattered light towards earth and not parallel light, each beam of light will reflect at slightly different angels and the beam of light on the other planet will not be the same size.

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