Suppose I have a Gaussian laser with a width $w$ and an intensity $I$ defined as $P/\pi w^2$, where $P$ is the laser power. If I shine the laser beam onto a convex surface with a radius of curvature $r$, would the effective intensity of the laser change due to this curved surface having a larger effective area under the beam than for a flat surface?
Optics – Does Shining a Laser on a Curved Surface Reduce Its Effective Intensity?
laseroptics
Related Solutions
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.
Intensity is energy per area per time. This means that if we integrate over the intensity profile of one pulse over time and beam profile we will get the pulse energy of that pulse. As formula $$ \begin{align}E_p &= \int ~\mathrm d^2r \int ~\mathrm dt \;I(r,t) \\ &= \int ~\mathrm d^2r \int~\mathrm dt \;I_\text{peak} \; \exp\left({-\frac{-4\text{ln2}t^2}{\tau^2}}\right)\exp\left({-\frac{-2r^2}{w_0^2}}\right)\\ &= 2\pi \int ~\mathrm d r \int~\mathrm dt \;I_\text{peak} \; r\exp\left({-\frac{-4\text{ln2}t^2}{\tau^2}}\right)\exp\left({-\frac{-2r^2}{w_0^2}}\right) \end{align} $$ The pulse energy $E_p$ is know to be 3 J from your problem. You only have to solve the integral and then solve for $I_\text{peak}$. Afterwards just compare $I_\text{peak}$ as a function of $w_0$ to $I_\text{threshold}$ and you are done. I don't really understand what you have done because you don't define $I_\text{total}$ but it looks wrong because the units don't add up (at least if $I_\text{total}$ is also an intensity).
EDIT: Full solution: When you solve the integral you find $$ E_p = \frac{\pi^{3/2} \tau\; w_0^2 }{4\sqrt{\ln2}}I_\text{peak}. $$ I just pluged the integrals into wolfram alpha. Now we find for the peak intensity $$ I_\text{peak} = \frac{ 4\sqrt{\ln2}}{\pi^{3/2} \tau\; w_0^2 }E_p $$ and we set $I_\text{peak} = I_\text{threshold}$ to find the smallest $w_0$ $$ \frac{ 4\sqrt{\ln2}}{\pi^{3/2} \tau\; w_0^2 }E_p = I_\text{threshold}\\ 2w_0 =2\sqrt{ \frac{ 4\sqrt{\ln(2)} E_p}{\pi^{3/2} \tau I_\text{threshold}}} $$
Best Answer
Intensity is defined as the power through a unit area. In other words, it is the amount of energy that arrives per unit area, per unit time.
Therefore, over a larger surface area, the intensity of the light (laser) will be reduced at any given point. This means if you were to shine the laser over a convex surface as oppose to a flat surface, the intensity at any point will be smaller since the amount of area is increased.
A good example of this generalized for any light source, is by considering the seasons on earth (earth is also a convex surface). Many people confuse the temperatures of the seasons being due to the distance of the earth from the sun. But in reality, because the earth has a tilt, when this tilt is pointing toward the sun, the amount of light strikes a smaller area and there is more direct sunlight per unit area making it warmer, whereas in winter the tilt is in the opposite direction and so light strikes a greater area making it cooler.