Physics of Focusing a Laser

There is a limit to how small you can focus an ideal single-mode laser beam. The product of the divergence half-angle $\Theta$ and the radius $w_0$ of the beam at its waist (narrowest point) is constant for any given beam. (This quantity is called the beam parameter product, and is related to the $M^2$ beam quality measure you may have heard of.) For an ideal Gaussian ("diffraction-limited") beam, it is:

$$\Theta w_0 = \lambda/\pi$$

So, to answer what I interpret as your main question:

Let's say that I have a laser beam of some given power that starts with some diameter $D_0$ at the point of emission and increases to $D_f$ at some distance $r$ away. Would this be sufficient information to imply a limit to the power per unit area (W/m^2) that could be obtained through focusing and what would that be?

The answer is no.

The parameters you have given are sufficient for calculating $\Theta$, but only if $r$ is large enough so that the points at which you measure the diameter are in each other's far field.

You would also need to know the beam radius at the waist, so you could calculate the beam parameter product. Then, to get the minimum spot size, you would need to refocus the beam so that it is maximally convergent. The absolute limit is the fictitious divergence half-angle of $\pi/2$, or 90 degrees, although in practice the theory breaks down for half-angles of more than 30 degrees (this number is from Wikipedia) since the paraxial approximation stops being valid. For an ideal beam at this impossible opening half-angle, this gives you a minimum waist radius of $2\lambda/\pi^2$. So yes, it does depend on the wavelength.

What lens characteristics and approaches would someone look for in order to do this with a laser pointer?

You need a lens with a very short focal length. This gives you the largest convergence. Note that the more convergent the beam, and the smaller the waist size, the smaller the Rayleigh range is. That is, the beam radius will get very small, but it won't stay very small, it'll get bigger very quickly as you move away from the focus. (The Rayleigh range is the distance over which the beam radius increases by $\sqrt{2}$.

In addition, thinking of a Gaussian beam as being "straight" is not quite correct. There is always a waist, always a Rayleigh range less than infinity, and always a nonzero divergence angle.

EDIT

Also, it is important to realize that there is no difference between an unfocused and a focused Gaussian beam. Refocusing a Gaussian beam with a lens just moves and resizes the waist.

The aperture size of the laser is not the same as the waist size. If the beam is more or less collimated, then the aperture will still be larger, because the waist radius is usually defined in terms of the radius at which the intensity drops to $1/e^2$ of its peak value. If the beam is cut off by an aperture at that radius, then even if it were close to diffraction-limited, it certainly wouldn't be anymore. So, apertures are always larger.

The waist is the thinnest point of the beam. Usually this point is inside the laser cavity, or outside the laser if there are focusing optics involved, which there often are. So still, the answer to your question is no. You are not missing the definition of $\lambda$; rather, you are comparing your minimum waist radius to the value of $2\lambda/\pi^2$ that I said was "impossible". I called it impossible, because to make a beam converging that strongly, you would need a lens with a focal length of zero!

Let's try a more realistic example with some numbers. Take your red laser pointer with $\lambda$ = 671 nm. Laser pointer beams are often crappy, but not so crappy as you might think, if they are single-mode. Let's assume that this particular laser pointer has an $M^2$ ("beam quality parameter", which is the beam parameter product divided by the ideal beam parameter product of $\lambda/\pi$) of 1.5. A quick Google search didn't give me typical $M^2$s of red laser pointers, but this doesn't seem to me to be too much off the mark.

Note that if you know the $M^2$ and measure the divergence of a beam, then you can calculate the waist radius. We are going to do that now. Suppose the laser pointer beam is nearly collimated: you measure a divergence of 0.3 milliradians, about 0.017 degrees. Then the waist size is

$$ w_0 = \frac{M^2 \lambda} {\pi\Theta} = \frac{1.5 \times 671 \times 10^{-9}} {\pi \times 3 \times 10^{-4}} \approx 1\,\text{mm}. $$

In this case, they probably designed the laser pointer with an aperture radius of 2 or 3 mm.

Now suppose you focus your collimated beam with a 1 cm focal length positive lens, which is quite a strong lens. The beam's new waist will be at the lens's focal length. That means you can calculate the divergence half-angle: it is the smaller acute angle of a right triangle with legs 1 mm and 10 mm. So,

$$\tan\Theta = 1/10,$$

or $\Theta\approx$ 6 degrees. Applying the formula once more to calculate the waist yields a waist radius of 3.2 microns, which is quite small indeed.

A "safe" laser pointer might have a power of 1 mW. The peak intensity is equal to $2P/\pi w_0^2$, so before the lens the peak intensity is about 600 W/m^2. After the lens it is about 100000 times larger.

So, to summarize:

• yes, there is a fundamental limit to the intensity, and it does depend on the wavelength, but you cannot even come close with a real-world cheap laser pointer.
• you need to know two of any of these quantities: divergence half-angle, waist radius, Rayleigh range, beam parameter product.
• really, the minimum size and maximum intensity depend quite heavily on what optics you use and how good they are.

I'm going to start with a specific example. Here is a laser pointer you can buy online.

• Cost is $12.50
• 1.0 mRad = 0.05 degrees
• wavelength: 645 nm
• output 5 mW
• beam diameter: 1.1 mm

Now, I am having great difficulty with this problem, and I'm still in disagreement with what ptomato has written as well as many other things I find online. My question, however, is still an objective and definable one, which is more or less "can the above laser cut steel with the appropriate lens?" Anonymous Coward gave a good reference, which is a guide to Gaussian optics. This document uses 1983 Sidney A. Self, Focusing of spherical Gaussian beams as its primary reference. The equations are the same throughout and are congruent to what you find in the prior Wikipedia links and such.

The problem is that they seem to keep giving equations for a diffraction limited case, saying nothing to clearly identify that's what they're talking about. In a diffraction limited case of course you need the wavelength, but my anticipation is that a $12.50 laser will not be diffraction limited.

Nonetheless, I'll still get down some of the equations here.

For starters, the waist radius is denoted $\omega_0$, which is the radius at the $1/e^2$ intensity as pointed out before, and the divergence half-angle is denoted $\Theta$. For this particular device those values are known. See page 5 of the guide, it notes the location of the beam waist is typically designed to be close to the output surface of the laser. I will be assuming that this is a real Gaussian beam and I thus won't be working with aperture. Again, Gaussian beam, not diffraction limited, and highly imperfect. I'll still be forced to take the reported "beam diameter" to be the representative of the $1/e^2$ radius, as well as being at the waist. This could be fairly wrong, but only by the same factor that my answer will be.

$$\omega_0 = 0.6 mm$$ $$\Theta = 0.001 rad$$

The literature goes into definition of several other values, as well as an expression for $\Theta$ itself. See page 3 of the guide to get the following equation without the $M^2$.

$$\Theta = \frac{M^2 \lambda}{\pi \omega_0}$$ $$M^2 = \frac{\Theta \pi \omega_0}{\lambda} = 2.94 \approx 3.0$$

I'm okay with this value. I don't expect it to get a good laser. Next, the Rayleigh range, $z_R$ is a rough value beyond which the far lens approximation is valid. I think it might need a $M^2$ on it, but I really don't know.

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

I just wanted to throw that out there. The guide has equations as to how to calculate the lens equations as well as the magnification factor, but I can't figure out how the equations imply that you can't focus this to a finite dot with a sufficiently large lens. I know this is not possible, so I had no choice but to apply a high school level approach to the problem.

Below is my illustration of a beam of 100% collimated rays (I'm talking about Newtonian physics here), and the first vertical line is some hypothetical perfect lens, that would converge perfectly on the next vertical line. If the rays coming in were perfectly collimated then the rays would exactly converge on a single point, but we know they don't because of the divergence angle.

• Beam radius is 0.6 mm
• $2 \Theta = 0.001 rad$

Again, I'm going to act like a high school physics student, pretend I don't even know what wavelength is, and use geometry. I'll assume the angle the furthest beams are deflected at is 45 degrees, and denote the things about the area it focuses on as $dot$.

$$r_{dot} = (\sqrt{2} \omega_0) (\Theta) \sqrt{2} = 0.0011 mm = 1.1 \mu m$$ $$A_{dot} = 1.21 \mu m^2$$

Now, for power per unit area I'll use $\Phi$.

$$\Phi = \frac{P}{A} = 4.13 \frac{GW}{m^2}$$

What temperature could this reach? I'll just assume dissipation from blackbody radiation for now (back of the envelope).

$$\Phi = \sigma T^2$$

Where $\sigma = 5.67 \times 10^{-8} W m^{-2} K^{-4}$.

$$T = 16,500 K$$

I would also like to do this for the radial conduction equation. Maybe later.

The formula relating the beam waste and the numerical aperture of a Gaussian beam is:

$$w_0 \approx \frac{\lambda}{\pi \;\, \mathrm{NA}}$$

The highest numerical aperture you could dream of finding--with a good oil-immersion lens--is in the neighborhood of 1.5 (a bit more or less, I don't know). (Without oil-immersion, the numerical aperture is always less than 1.) So we get:

$$w_0 \gtrsim \frac{\lambda}{1.5 \pi}$$

So you can focus a 645nm beam (if ideal gaussian) to an area of roughly $\pi w_0^2 \gtrsim 0.05\mu \mathrm{m}^2$. If the power is 5 mW, I get $100 \,\mathrm{ GW}/\mathrm{m}^2$, or 100 million times brighter than daylight.

The real figure for a cheap laser pointer is lower, because $M^2>1$.

In practice, for a diode laser, your best bet is to first use a weak cylindrical lens to get the spot more round [all diode lasers have a built-in cylindrical lens, but in my experience the spot is still slightly elliptical], and then use a strong microscope objective lens (oil-immersion if possible) to focus as tight as possible. Also, put a two-lens telescope before the objective lens to either expand or shrink the beam so that it exactly fills the entrance aperture of the objective lens. Actually, maybe you shouldn't use an oil-immersion lens, you might burn the oil! Don't try this at home...