1
$\begingroup$

In the context of a THz Gaussian beam $(1)$, it is stated:

Terahertz output radiation at higher frequency produced a narrower collimated beam owing to diffraction effects (...)

What kind of diffraction effects cause higher radiation to produce narrower collimated beams?

I'm more interested in an answer based on the diffraction effect, but mathematically this still makes no sense to me, because according to the formulas for the Rayleigh range and for the spot size parameter (respectively shown below),

$$z_R = \frac{\pi \omega_0^2 n}{\lambda} \tag{1}$$

$$w (z) = w_0 \sqrt{1+ \left( \frac{z}{z_R}\right)^2} \tag{2}$$

an increasing frequency should lead to a greater spot size.

Where is this reasoning wrong?


  1. Broadband high-resolution terahertz single-pixel imaging. A.Vallés, J.He, S. Ohno, T.Omatsu, and K. Miyamoto. Opt. Express 28, 28868-28881,(2020), Opt. Express eprint.
$\endgroup$

2 Answers 2

1
$\begingroup$

Look also at the expression for divergence:

\begin{equation} \theta=\frac{\lambda}{\pi w_0 } \end{equation}

for 2 beams with the same waist $w_0$, the one with the lower wavelength (higher frequency) will also have a lower divergence.

With the equations you posted you can also see that 2 beams with the same waist, the Rayleigh range will be longer for that with the lowest wavelength (higher frequency).

Hope this helps.

$\endgroup$
0
$\begingroup$

First, the math. The Rayleigh range $z_R = \pi w_0^2n/\lambda$ gets larger when the wavelength $\lambda$ gets smaller. So, increasing frequency makes for a larger $z_R$. Now, when we look at the waist formula, $$w=w_0\sqrt{1 + \left(\frac{z}{z_R}\right)^2}$$ we can see that a larger Rayleigh range will cause the beam to expand less for the same $z$.

As for the physical reason this happens, let me be a little hand-wavey. Think back to the double-slit experiment. Why is the central peak of long-wavelength light wider than short-wavelength light? It's because it takes larger path differences for two rays to destructively interfere due to the longer wavelength. It's the same with a coherent beam. It takes larger divergences from the beam axis for different parts of the long-wavelength wavefront to destructively interfere. This allows long wavelengths to diverge faster than short wavelengths.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.