propagation loss of laser light

The intensity (W/m^2) of an electromagnetic wave from an ordinary antenna decreases with the square of the distance from the emitter (in the far field.) Is the same true for a laser beam?

Yes.
As the laser beam diverges, the beam diameter is increasing while the power density decreases proportionally.
The inverse square law applies.

WHAT Ra? My laser's power decreases with the square in the distance travelled!!?
I don't quite think so.:crackup:

It does decrease like RA said. However no value was given for the proportion of the power decrease for the square of the distance traveled. That number depends on the divergence of the laser.. or the type of antenna used. The more directional the antenna, or the lower the divergence on the laser, the lower the power decrease ratio is going to be, but it's still always going to be a linear change with the square of the distance.

anselm said:

WHAT Ra? My laser's power decreases with the square in the distance travelled!!?

Click to expand...

power density. (See units in op)

It actually decreases even faster, because of scattering ans absorption in the air. But divergence is stronger effect most of the time IMHO.

So a laser behaves like an ordinary antenna but with an enormous amount of gain. Ya?

RA_pierce said:

Yes.
As the laser beam diverges, the beam diameter is increasing while the power density decreases proportionally.
The inverse square law applies.

Click to expand...

apchar said:

So a laser behaves like an ordinary antenna but with an enormous amount of gain. Ya?

Click to expand...

No, I am confusing the inverse square law with the beam divergence. qumefox is right. What I meant is that the power density decreases as the beam diverges. The beam divergence angle determines the ratio at which the W/m^2 is reduced.

Also, I forgot to take into account the beam profile. Typically the power distribution of a laser beam is not uniform. It is more or less Gaussian with a higher power density near the center. You can calculate the power density by dividing the Wattage by the area of the laser's beam at x distance (As you know, W/m^2).

Knowing the beam divergence will allow you to calculate the beam diameter, and therefore the area it covers at any given distance.
Off to class again... Maybe I shouldn't have avoided taking math classes for 3 years. I'm rusty.

What about scattered laser light. I would think that a laser beam reflected off a curved surface would behave like a directed point source. Wouldn't that be subject to the inverse square law?
FYI: My interest is in calculating the incident power on a vehicle from a police laser gun compared to that received at the guns detector.

That too would be subject to inverse square law. But someone said common laser beam is not ?

According to my calculations, A laser does seem to follow the law.

The inverse square law says that if you double the distance, you quadruple the area and therefore cut the intensity to ¼. Yes?

A 1W 1mm 1mrad laser at 50 meters is about 50mm wide. Power over area is 530µW/mm².
At double the distance of 100m, it is 100mm wide. Power over area is ~127µW/mm², which is ¼ of the intensity at 50m.

The same is true for all divergences. A 1W 1mm Xmrad laser will be Xmm at distance Xm. Double the distance will be 2Xmm at 2Xm.
If we compare power densities: k*P1/A1=P2/A2 where k is a constant.
Assuming the losses in the medium are negligible, P2 = P1, so k*P/A1=P/A2.
A1 = πr² = π(Xmm)² and A2 = πr² = π(2Xmm)² = 4π(Xmm)²
k*P/(π(Xmm)²)=P/(4π(Xmm)²) Power, π, and (Xmm)² cancels, and we're left with:
k=¼ Once again, doubling the distance cuts the power density by 75%.

If you use values at very short distances, it's not as close, but that is because the "point source" is a virtual point that's actually behind the laser. That is one reason we calculate divergence at the furthest distance possible - to make minimize the effect of the initial diameter.

Awesome.
That is what I was trying to articulate but you made it much more clear.
I was just scribbling down some numbers to explain it to myself.

If you use values at very short distances, it's not as close, but that is because the "point source" is a virtual point that's actually behind the laser. That is one reason we calculate divergence at the furthest distance possible - to make minimize the effect of the initial diameter.

Click to expand...

^That just occurred to me this morning as I was thinking about this problem. When I first replied I figured that the power was asking about far-field power density, but when others replied I was puzzled, because in the near-field, lasers don't seem to drop in power density as much.
Anyway, great explanation, it makes it much easier to understand.

Exactly.
It's a little more complicated for close distances (like when focusing for burning). But after some distance, the beam always diverges in practically linear way, which means inverse square law in power density.

Some more interesting details here:

Gaussian beam - Wikipedia, the free encyclopedia