Is it possible to reduce the divergence of a red laser pointer beam from 40 degrees to 1 degree?

Question

I would like to find out if it is possible to reduce the divergence of a red laser pointer beam from 40 degrees to 1 degree.

@Ruslan wrote on October 3 2013 that because the active zone of a red diode laser has diameter of order of several micrometers the divergence of a red laser pointer beam is currently 40 degrees.

Due to Heisenberg's uncertainty principle $\Delta\times\Delta p \gtrsim \hslash/2$, one can't really make a quantum have zero momentum in any direction. So you can't say that photons go in the same direction - this is just a simplified description of laser operation. In reality, the thinner the beam, the higher the divergence.

Compare e.g. a DPSS laser (e.g. green laser pointer) with a diode laser (e.g. a red laser pointer).

In a DPSS laser the active material will have diameter of order of hundreds of micrometer, and the exiting beam will start from even smaller diameter for various reasons. The divergence is quite small: if you remove the collimating lens, your light image from a green laser pointer will be several centimeters after the light goes several meters. Divergence angle would be $\lambda/d = 532\ \mathrm{nm}/100\ \mu\textrm{m} \approx 0.3°$.

If you try doing the same with a red laser pointer, you'll see that its light diverges quite a lot: after going several centimeters in direction of propagation, it'll already give image of several centimeters. The reason for this is that active zone of diode laser has diameter of order of several micrometers. This makes output beam quite thin, making Δx small and thus Δp high, and this is what leads to high divergence. Divergence angle would be $\lambda/d = 640\ \mathrm{nm}/1\ \mu\textrm{m} \approx 40°$. Actual angle would depend on which transverse direction you select, because active zone is ∼10× longer in one direction than in another.

In general, the thicker your starting laser beam, the more collimated it is, so if you manage to make a (visible wavelength) laser with beam starting at 1cm thickness, you'll have almost perfectly collimated laser beam.

I would like to know if there is a method to enlarge the active zone of a red diode laser from a diameter of several micrometers to 30 micrometers. Also, could a lens with short focal length be used to reduce the divergence of a red laser pointer beam from 40 degrees to 1 degree?

Answer 1

Of course, a lens with a short focal length can be used. The easiest way is to take the standard lens that comes with most laser pointers and move it a bit closer to the laser crystal.

If you want the beam waist of the output beam to be exactly at the lens aperture, then you'll have to do a bit more work and find a suitable microlens. For a beam with 1 deg ($\theta=0.017$ rad) divergence, you'd need the beam to have a diameter $d\approx \lambda/\theta=37~\mathrm{\mu m}$ at the lens. The lens would be at a distance of 52 microns from the laser crystal. So that would be rather difficult to achieve.

Update: clarification on beam waist and divergence. Here is what a Gaussian laser beam looks like this (image by DrBob on Wikipedia):

With a typical laser, the output coupler is at the narrowest part, the beam waist. The divergence angle $\Theta$ is a fixed parameter of the beam, no matter how far you propagate it. If you put a lens in the beam waist or within the Rayleigh length $z_R$, you can't decrease the far-field divergence angle $\Theta$, although you can focus it with a short focal length and achieve an even higher divergence after the focus. If you're not interested in the beam diameter near the beam waist, only in the spot size at 100 m away, then I wouldn't bother with microlenses and just use a regular lens at a few millimeters from the laser aperture.

I can't comment on modifying the diameter of the laser crystal itself; if it is possible at all, it will require advanced semiconductor processing techniques.

By the way: using Heisenberg to describe beam divergence as a function of beam diameter and wavelength doesn't make much sense, because Planck's constant cancels out once you write out $\Delta p$. The rule of thumb $\theta=\lambda/d$ works just as well for acoustic waves, which are not at all quantum mechanical.