Reducing loss of energy for Lasers

[Josiah]

TL;DR

Reducing loss of energy

Hi
I was told by someone that if you place the laser in a vacuum, then if would not loose energy. Is this true?

 

[anuttarasammyak]

Not only laser but light in general do not lose energy in vacuum.
Laser beam go in one direction. A beam receiver get the energy without the loss. Light, for an example sun light, goes all the direction whose energy density drops by r^-2 law. Total r^2-surface surrounding the light source get generated energy without loss.

 

[berkeman]

@Josiah -- Please take care not to try to reopen a closed thread. Your previous thread on this question was closed because you veered off into non-physical territory with your laser power questions. I will let this thread here proceed, but only if you stick to known Physics.

Here is how your previous thread was tied off:

The OP question has been sufficiently addressed. Thread closed. Thanks to all who participated.

 

[PeterDonis]

I was told by someone

This is not a valid reference. Can you find a valid reference (textbook or peer-reviewed paper would be best) that says what you're asking about? If not, we can't discuss it here.

 

[DaveE]

TL;DR: Reducing loss of energy

Hi
I was told by someone that if you place the laser in a vacuum, then if would not loose energy. Is this true?

Do you mean the laser beam, or the entire laser? What's in the vacuum?

There are some laser applications that require a vacuum, but they are pretty unusual. One is very, very short wavelengths, often called "vacuum uv" because of the absorption by O₂ and N₂. Another application is in vacuum spatial filters, where the beam is focused so small that the energy breaks down (ionizes) the air molecules. This can also be a problem with ultra fast laser pulses.

Gas lasers aren't vacuums in the gain media (laser tube), but they are quite low pressure.

But basically, no. Putting a laser in a vacuum won't make it better. In the above examples, it's just part or all of the beam that needs a vacuum, not the entire machine. You're likely to break it because of some reliance on air convection for cooling of various parts. I recall having to significantly overdesign a 5KW forced air cooled power supply so we could sell it in Denver et. al.

As in my reply to your previous post, you can't easily make a laser more powerful with simple tricks. They have already made it as powerful as they can given constraints like how much money you can spend. From laser pointers to fusion research, they are all optimized by people that know more than us.

PS: "not losing energy" is laughable in the laser world. They are all very inefficient devices, every single one.

 

[berkeman]

Another application is in vacuum spatial filters, where the beam is focused so small that the energy breaks down (ionizes) the air molecules.

I was at a photonics trade show many years ago, and as I walked down one of the aisles, I heard a "snap, snap" noise every few seconds. As I got closer, I saw that it was a large laser (about the size of a small car, including the power source), and they had it set up to focus the pulsing beam in the air behind some clear shields. You could see the ionization at the focus point with each laser pulse. Pretty impressive.

 

[Matterwave]

https://www.edmundoptics.com/knowle...es/lasers/gaussian-beam-propagation/?hl=en-US

I am no expert in laser technology, but I found the above article helpful!

A laser in a vacuum (I'm thinking here, laser light traveling in a vacuum) does not lose energy since there's nothing to lose energy to (with the exception of the cosmological red shift effect), but no laser is perfectly collimated. Diffraction effects at the edges will always spread the laser out and the spread for a perfect Gaussian beam (which no real laser achieves, but can be used as an estimate based on the above article) is given by eqn 3 in the article:

$$
\theta=\frac{\lambda}{\pi w_0}
$$

Where $w_0$ is the initial beam width (see fig. 1).

If you plug in some reasonable numbers for that you'll see how your beam spreads out. Once you see how your beam spreads out you can calculate how much energy a certain observer some distance away would receive from your laser.

I was thinking you're interested in finding out if you can send a light signal very far away. But say you shine a laser at Alpha Centauri, 4 light years away, you'll find that even if you had a gigantic laser (of order meters wide in lens shape) the laser would still get spread apart quite a bit by the time it goes 4ly.

 

[DaveE]

...a perfect Gaussian beam (which no real laser achieves...

Yes, they do. Single frequency TEM₀₀ propagation is a thing people have been using for nearly 50 years. About as perfect as anything in modern technology. That's how the computer you're looking at was made. Also, for many lasers it's where most of the energy is in the far field even if they have higher order modes too.

 

[Matterwave]

Yes, they do. Single frequency TEM₀₀ propagation is a thing people have been using for nearly 50 years. About as perfect as anything in modern technology. That's how the computer you're looking at was made. Also, for many lasers it's where most of the energy is in the far field even if they have higher order modes too.

Ah gotcha. I don't know this field that well, I was simply parroting back the article:

In many laser optics applications, the laser beam is assumed to be Gaussian with an irradiance profile that follows an ideal Gaussian distribution. All actual laser beams will have some deviation from ideal Gaussian behavior.

 

[berkeman]

That's how the computer you're looking at was made.

Can you elaborate? Do you mean photolithography used in IC fabrication?