[Physics] Does a laser beam converge and diverge without lens

classical-electrodynamicselectromagnetismlaser

In the case of $TEM_{00}$ mode laser produces gaussian beam. I read in wikipedia that it converges till some point called 'beam waist'and then it diverges to infinity.

If there is a lens placed along the beam, then it converges to the focal point of lens and then it diverges. Here the beam can't converge to a single point but has a non-zero width.

  1. Is it because of the Hesienberg uncertainty principle that $\Delta x$ becomes zero which further implies $\Delta p$ tends to infinity?
    And the momentum can't be infinite, so it doesn't get focus to a single point. Am I right?

  2. I read that it happens even without a lens. How does focusing of beam happen without lens?

Best Answer

On point 1, you are correct. The uncertainty principle leads to the finite spot size when focusing a laser beam and the spread of the beam thereafter. In general, in order to produce a smaller spot, you need to focus the beam in at a high angle (short focal length lens), meaning a larger spread in transverse momentum of the photons. The same thing can be said for imaging optics. In order for a microscope to look at a small object, you need a short focal length microscope objective that can collect light at a high angular spread.

As for point 2, "focusing" of a laser beam without a lens can happen due to some non-linear effects if you are at very-high intensities in some type of medium (air or otherwise), but that is a different subject all together. However, spreading of a laser beam is an inherent property of the beam due to the uncertainty relationship as well. Because the laser has a finite spot size, it must also have a spread in the transverse momentum. This means that as the laser propagates, the beam will in general expand. The far-field angular divergence (full angle of the "cone" of the beam) is given by (for $TEM_{0,0}$):

$\theta = \frac{2\lambda_0}{\pi n w_0}$

where $\lambda_0$ is the wavelength, $n$ is the index of refraction and $w_0$ is the waist size of the beam. You can see here that the divergence angle is inversely proportional to the waist size of the beam, $w_0$. For larger beams, there is less divergence.