# Best collimation at focal point of lens?

From geometrical optics it is clear that we get the best collimation when the object lies at the focal point of the lens. I know that this carries over into cases where diffraction happens, but have no idea how to proof it. So how can we proof that the divergence angle is smallest at the focal point when we are considering both diffraction (at the lens) and the results from geometric optics?

EDIT-Clarification of what is been asked

Let us consider a more specific case:

Consider a source that creates a cone of light apex angle $\theta$, travelling towards an infinite lens. After passing through the lens the maximum angle between one of the diffracted (and refracted) 'rays' is $\theta_d$. I am trying to show that the position of the source that minimises the angel $\theta_d$ is when it is at the focal point. How can I show this?

• Collimation of light -- are we using the term in this sense? I don't think of the beam waist as being collimated, though it is normal to the surface. Please clarify. – Peter Diehr Mar 16 '16 at 11:30
• @PeterDiehr No I think we are using it in the same sense. By best collimation I mean that the radius of the beam undergoes the least amount of variation. – Quantum spaghettification Mar 16 '16 at 11:33
• If you want perfect collimation with diffraction, you need Gaussian beams, so it's not a simple focal point/angle problem. – CuriousOne Mar 16 '16 at 11:48
• And it's not at the focal point of the lens; one uses an infinite conjugate microscope lens for best results, or you can build a simple telescope. The goal is to minimize divergence. – Peter Diehr Mar 16 '16 at 11:51
• But beam divergence isn't defined at the waist; OTOH there you could look at the phase front. I think "collimation" is the wrong word to use in this context. – Peter Diehr Mar 16 '16 at 12:11

Geometric optics gives the direction of each ray in isolation. When you consider diffraction, the result depends on the intensity distribution in the cross section of the beam.

The comments have talked about Gaussian beams because that is the most common distribution for laser beams. So we we will consider an ideal case - Gaussian beams, optics with no aberrations, and optics with large enough diameter that the aperture does not significantly truncate the beam. In that case, If a Gaussian beam goes in, a Gaussian beam comes out the other side of the lens. The picture is from this.

All Gaussian beams have this relationship between the diameter of the beam waist and the divergence angle of the beam at a large distance from the waist. θ=1.22λ/D. To make θ small, make D big.

In the picture, the diameter D is smaller than the beam diameter at the lens. This is because the waist on the right is not at distance f from the lens. Geometric optics shows that rays will converge. Diffraction of a Gaussian profile produces the beam on the left.

If the beam waist was a distance f from the lens, the beam waist would be right at the lens. It would be a little bigger, so the divergence angle would be a little smaller.

If you are not limited to Gaussian beams, you can get the divergance angle to 0. This is done by using a beam with a Bessel cross section. See this post for more information.

Note I am the OP

The geometric angle of deviation is given by: $$\theta_{dg}=\theta \frac{u}{|v|}$$ $$=\theta\left| 1-\frac{u}{f}\right|$$ If we let $u=f+\Delta$ then: $$\theta_{dg}=\theta\frac{ |\Delta|}{f}$$ Now taking $\Delta$ to be small we can assume the angular deviation due to diffraction is given by: $$\theta_{dd}=\frac{1.22 \lambda}{D}$$ $$=\frac{1.22 \lambda}{\theta u}$$ $$=\frac{1.22\lambda}{\theta f}\left( 1-\frac{\Delta }{f}\right)$$ So the total angular deviation is given by: $$\theta_d=\frac{1.22 \lambda}{\theta f} +\theta \frac{|\Delta|}{f}-\frac{1.22\lambda\Delta}{\theta f^2}$$

Looking at this formula you may be a bit worried that we can 'beat the diffraction limit' if $\Delta\gt 0$. This is infact not the case. If we get to the stage where this becomes a problem the rays will cross over the axis soon enough that the angle that once' seemed to be negative is now positive i.e. in such a case our angle along distance away from the lens would be given by: $$\theta_d=\frac{1.22 \lambda}{\theta f} +\theta \frac{|\Delta|}{f}+\frac{1.22\lambda|\Delta|}{\theta f^2}$$