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A beam of light of wavelength $\lambda$ and width $W$ needs to be focused at a distance $D$ to a spot not bigger than $w_S$, which stands for 'width of sail'.

Now, the diffraction limit says it clearly that a coherent beam will diverge at a rate no less than $\frac{\lambda}{W}$, and that the minimal target spot width will be $\sqrt{D \lambda}$. But my understanding from the comments in this question is that this applies in the far-field limit far away from the focal point, so my doubt is that if this hard bound on the spot size can be avoided by replacing the collimated beam by a focusing beam with an extremely long focal length.

So, my question boils down to the following: can the fundamental diffraction limit on spot size, be replaced with the extremely difficult, but possible in principle engineering problem of creating a focusing element with extremely high focal lengths, in such a way that for a specific length, the target spot size can be made smaller

A concrete scenario to make things more clear: the wavelength is $\lambda = 10^{-6}$ meters. The distance to the target is $D = 10^{12}$ meters ($10^{18}$ wavelengths). If the beam converges at focal length $D$, my hope is to keep the beam spot (which in this case coincides with the beam waist) at $10^2$ meters (which is 10 times smaller than the target beam spot if the laser would have been collimated, instead of focusing at $D$)

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3 Answers 3

Gaussian beam focussing with the large waist on the lens can be described the easiest with the relationship $$z_o \cdot z_o' = f^2,$$ where the $z_o$ are the Rayleigh lengths (http://en.wikipedia.org/wiki/Rayleigh_length) before and after the lens (assumption is that the bigger beam has the waist on the lens).

The beauty here is that geometric means are nice to remember. You have a large distance (the collimated side of the beam $z_o$), a very short distance (the focussed side of the beam $z_o'$) and a medium distance, the focal length f.

This can be derived trivially from the ABCD matrix rules for Gaussian beams and the ABCD matrices of a lens and a distance and is equivalent to the more common formula: $$ D' =\frac{ 4 \lambda f }{ \pi D},$$ where $D$ and $D'$ are the beam diameters before and after focus.

If your beam spot is at the waist D', then the mirror you are using to collimate it there, has to be much, much bigger. If your waist is indeed at the target spot, the focal length is $10^{12}$ m. With $D'= 10^2$ m, you get for the mirror size at $10^4$ m diameter.

So your problem is actually the other way around. Your only waist is on the mirror (a very large focal length is indistiguishable from infinity). Know what the largest telescope/mirror is that you can get (say $D=2 m$), then calculate the spot size in the far field.

$$ \omega(z) = \omega_0 \cdot \sqrt{1 + (\frac{z}{z_r})^2} .$$ I"m getting $ 3 \cdot 10^5 m = 300 km $ for the waist radius on target with your assumptions and an outgoing waist radius of 1 m.

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No. Diffraction limit says for a radiation field (radiating so that it is directly in and out, no coupling mechanism needed), its physical spot size is limited by its wavelength. Everything comes from the wavy nature of light in radiation field. No matter you focus it, transmit it or image it. You can think of this as you do a wave front calculation. You will find because of the phase term, delta x is never zero , instead it assumes a finite width for the incoming waves.

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picture the following scenario ($\lambda = 10^{-6}$ meters): at a distance of $10^{18}$ wavelengths ($10^{12}$ meters), a collimated source beam of a width of $10^{9}$ wavelengths will spread just by a factor of 2.4 or so. If i use a wider beam of say, $10^{10}$ the beam divergence will be ten times smaller, but the overall beam will still be larger, because we assumed it was a collimated beam. But what if the beam is slightly convergent instead of collimated? At a beam width of $10^{10}$ at the source, it could focus on a patch of only $10^{8}$ wavelengths at the target? is this wrong? –  lurscher Feb 7 '13 at 3:00
how did you get these $$10^8$$ and a factor of 2.4 thing?.. even though the diffraction says you can't focus your light up to some $$ 0.5\lambda/NA $$. In principle you can make NA very large by first, making the convergence angle very large -> 90degrees; and second, choosing a large refractive index after the lens. But it can't be more than 10 in reality. –  Bo Zeng Feb 7 '13 at 4:44
numerical aperture. –  Bo Zeng Feb 7 '13 at 4:49
the $10^8$ width figure is reached with the following argument: beam divergence is $\frac{\lambda}{W}$ (W the width of the source beam). if $\lambda = 10^{-6}$ meters and $D=10^{12}$ meters. If the beam width at the source is 10km (that is, $10^{-10}$ degrees of beam divergence) which implies a diffraction-limited spread of $10^{2}$ meters at the target. If the ray would have been collimated, this spread would add to the source beam width, which was bigger, at $10^4$ meters. But if the ray is slightly convergent, then surely the target spot should be smaller than that? –  lurscher Feb 7 '13 at 4:56

Unfortunately you're still constrained by quantum mechanics. Basically the beam diverges due to the hiesenberg uncertainty principle. This is a hard limit for coherent light. There has however been some work done on partially coherent light, such as that emitted by the sun. In some special cases it can be focused to beyond the diffration limit, but its not really useful for your application. By the way I'm very annoyed that someone else has thought of this now. OF course you still haven't got the details worked out. What you need to do is take a transparent FEP or ETFE plastic balloon up into a heliocentric orbit somewhere near the orbit of mars, fill the balloon with the boil off from a liquid nitrogen dewar. (Note no lasers, just sunlight.) And have two of them set so as to focus the beam half way to your target. This will give you the optimum spot size. By my calculations for a doulbe lens system setup as described with 550nm peak wavelength of sunlight, your spot would be about 2x the starting diameter after about 1.7AU for lenses 150m in diameter. If you want to take this out to alpha then you're talking about a lens ~110km to 500km in diameter. Enormous, but for something as monumental as interstellar travel, perhaps not exorbitant.

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i'm not following all your calculations but it seems you have thought about this before. Can you elaborate better how do you do your calculation? –  lurscher Apr 6 '13 at 7:04

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