# Does a laser beam converge and diverge without lens?

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?

• for your point 2: I think you can have a focusing gaussian beam in the output of a cavity formed by 2 curved mirrors with the same sign of curvature : something like ((> . Somehow equivalent to say that the "lens" is within the cavity. Many lasers have the waist at the output face (with a plano-curve cavity). – scrx2 Nov 21 '15 at 21:13

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.