By lurscher

2013-02-04 16:56:41 8 Comments

There are many applications for orbital space mirrors in astronomy (better telescopes) and space propulsion (solar power for deep space probes), but this is limited by the minimum beam divergence achievable with current technology

So, i'm trying to understand what physical and technological limits exists in our capacity to build mirrors that can keep as small beam divergence as possible. For instance, a sail probe to Saturn would require that the beam doesn't significantly diverge above 300m-600m (the biggest sail we can conceive of building in the inmediate future) at distances as big as 5-6 AU ($10^{11} - 10^{13}$ meters)

What is the best beam focusing divergence we can achieve for solar light with mirrors right now, and what limits the improvement of this? technological limits? fundamental physical limits?

Edit let's assume the concrete case of a wavelength of $10^{-6}$ meters, and a distance of $10^{12}$ meters (Neptune orbit). Can't i, for instance, build a focusing element with a focal length of $10^{12}$ meters that would push the far field beam divergence at farther distances from the focal point? Is this a manufacturing limitation of focusing element engineering (not enough precision to build lens made of atoms with the required focal length), or something more instrinsic, say, a focal point cannot be farther than some finite distance that depends on the wavelength?


@Kent Nickerson 2020-05-16 17:19:25

As comments point out, a far field focus is subject to a "diffraction limit" which subtends an angle (in radians) of roughly $w/D$, where $w$ is wavelength of radiation and $D$ is diameter of the mirror, whether the focal length is finite or not. Let's call this angle $A$. A beam reflected from the mirror initially has diameter $D$ but at a far distance $S$, the beam width will be $AS$ and dominated by diffraction. This is somewhat counter intuitive, because a smaller far beam requires a wider initial beam. You can get an approximate optimal value of $D$ for smallest spot by calculating the point where $D=AS$, which yields $D=\sqrt{wS}$ - a result mentioned in comments.

$D=\sqrt{wS}$ also applies to the optimal diameter of the pinhole in a pinhole camera for maximum resolution, where $S$ is distance to film/sensor.

Related Questions

Sponsored Content

4 Answered Questions

1 Answered Questions

[SOLVED] Making a bright beam of artificial white light

3 Answered Questions

2 Answered Questions

[SOLVED] limits on a gauss box of light

1 Answered Questions

[SOLVED] Polarization and mirrors

Sponsored Content