# Diffraction limited divergence angle

I was reading an article and came across the following paragraph:

"A thin lens of focal length f is used to collimate the light emerging from an optical fibre. The fibre has small core diameter and numerical aperture NA. It achieves a diffraction limited divergence angle θ for the collimated beam."

I am wondering how can there be a "diffraction limited divergence angle θ"? If the lens is placed at exactly focal length f away from the end of the optical fibre, shouldn't all the light be properly collimated with θ=0?

Divergence $\theta=0$ is not possible -- from a quantum mechanical perspective, it would violate the Heisenberg uncertainty principle for position and momentum of a photon. "Collimated" is not really a technical term; it is a qualitative description of a beam with a relatively small divergence angle.

The fact that the divergence is diffraction-limited means that the lens does not cause aberrations, so that the product of the beam width and beam divergence is as small as possible. This can be true regardless of how big the beam divergence is, and instead has to do with the beam quality. A diffraction-limited beam (e.g., a Gaussian) is one for which this product is equal to the smallest value allowed by the uncertainty principle.

• Oh I see... I've misinterpreted the term "diffraction-limited". By the way, where is the "diffraction" here, if the lens is assumed to be infinitely wide such that it covers all angles? – user148792 Apr 16 '18 at 19:52
• @delickcrow123 I think it just refers to the wave nature of the beam, even when it's not at an aperture or edge. The amplitude at any point still involves interference between different ray trajectories. – ostrichCamel Apr 16 '18 at 20:04
• @ostrichCamel Divergence θ=0 is not possible -- from a quantum mechanical perspective, it would violate the Heisenberg uncertainty principle for position and momentum of a photon. Maxwell's divergence equation also disallows a divergence of 0 – MJC Apr 19 '18 at 8:51
• @MattCliffe I agree that we could get the same result from the Maxwell equations, without invoking QM, but I don't think you can use only $\nabla\cdot E = \nabla\cdot B = 0$ to get there, if that's what you mean. Here, the term "divergence" refers to an angle, not the operator $\nabla\cdot$. – ostrichCamel Apr 19 '18 at 14:32
• @ostrichCamel you only need $\nabla \cdot \bar{E} =\frac{\rho_{free}}{\epsilon_0}$ to show that beams cannot have zero angular divergence. – MJC Apr 19 '18 at 14:53

In classical electromagnetism the uncertainty principle does not apply as it represents the limit of $h \rightarrow 0$.

For cases where the paraxial approximation applies the minimum beam waist is $2 w_0=\frac{4 \lambda f}{\pi d}$ for focal length $f$ and collimating lens diameter $d$. http://www.ophiropt.com/laser-measurement-instruments/beam-profilers/knowledge-center/tutorial/what-is-m2

You can actually show that no collimated beam exists using ray optics and some radiometry.

Perfect collimation would require that the chief ray angle be zero and the marginal ray height be nonzero angle be zero. If your chief ray angle is zero, that means your beam came from a source of zero size.

From radiometry we know that would require a source of infinite radiant exitance. Outside that exception, there would be no energy in the perfectly collimated beam.

As others have mentioned, even if that source existed, you would then be defeated by wave optics which would disallow a perfectly collimated beam anyway.

Practically speaking, however, you can generally pick beam diameters sufficiently large to produce "zero" divergence as far as your application is concerned.

Also, a diffraction limited divergence is in contrast to a worse performing beam. If you have a non-Gaussian beam (i.e. M^2 > 1) the beam will diverge more.

EDIT: It's actually more simple than that. I originally used the wrong definition of what a perfectly collimated beam would be. If you wanted a completely collimated beam (all rays traveling parallel to the optical axis), that would mean your marginal and chief ray angles would be zero, which would make the value of your Lagrange invariant zero, which means you'd have zero light.

• "Collimation means the chief ray angle is zero and the marginal ray height is nonzero" – MJC Apr 19 '18 at 14:56
• Please explain what the issue is with that sentence. – Daniel Apr 19 '18 at 16:11
• Sorry I started to do so and then got bored. Explaining things, or explaining why something cannot happen in terms of ray optics is bad. Also it just shows how useless the term collimation is. It is colloquial and ambiguous. – MJC Apr 19 '18 at 16:15
• I think there's a strong case for using the simplest model possible to explain the behavior you see. I'd use rays over QM if it works. However, I did remember one issue with my answer that I'll remedy (I used the incorrect criterion for collimation). – Daniel Apr 19 '18 at 18:13
• I agree with the simplest model, however I do think the model should suit the problem and in this case I don't think it does (I am more than happy to concede I am wrong :) ). My main gripe is over the use of the word. – MJC Apr 20 '18 at 8:03