# How can an axicon plus a convex lens produce an optical vortex?

The open access Nature paper Generation of “perfect” vortex of variable size and its effect in angular spectrum of the down-converted photons describes the use of an optical axicon (cone-shaped lens) plus a normal convex lens to produce an optical vortex beam.

The introduction states that an axicon alone can be used to convert a 0th order Laguerre-Gaussian beam (what I'd call an ordinary gaussian beam) into a 0th order Bessel-Gaussian beam, and that the addition of a simple convex lens can then turn the Bessel-Gaussian beam into a vortex beam.

Among different techniques, Fourier transformation of the Bessel-Gauss (BG) beam of different orders is the simplest technique to generate perfect vortices. The axicon, an optical element with a conical surface, converts Laguerre-Gaussian (LG) beams into BG beams. The order of the BG is the same as that of the input LG beams. The Gaussian (zero order LG beam) input beam results in 0th order BG beam. Fourier transformation of such BG beams using a lens of focal length, f, results perfect vortex at the back focal plane with field amplitude distribution given as:

$$E(\rho,\theta)=i^{l-1}\frac{w_g}{w_0}exp(il\theta)exp\left(-\frac{\rho^2+\rho_r^2}{w_0^2}\right)I_l\left(\frac{2\rho_r\rho}{w_0^2}\right)$$

Since both components are axially symmetric, where does the "vorticity" or $exp(i\theta)$ come from?

How does this work? I'm flummoxed.

• It is not the lens that produces the vortex. The vortex is already present in the LG mode. The axicon then converts it into a Bessel beam with the same order vortex and the lens then Fourier transforms it to producte the perfect vortex, still having the same order vortex. – flippiefanus Jul 28 '17 at 16:46
• Yes, you are right. There is a $exp(il\theta)$ term in the definition that I've overlooked, and it seems I've only seen intensity plots where it's not visible. It is here though: i.stack.imgur.com/cGDLA.png Thank you for de-flummoxing me. – uhoh Jul 28 '17 at 16:54
• @flippiefanus were you going to write that up as an answer? If not, I can do it. It's just good to have something to accept rather than leave this dangling. Thanks! – uhoh Jul 29 '17 at 6:13