Understanding the Inverse Square Law for Laser/Coherent Light

[Glenn]

Does the inverse square law apply to laser/coherent light?

 

[mathman]

No at short distances, yes at long distances. The coherence keeps it from spreading, however, because the aperture is finite, it eventually starts spreading.

 

[Integral]

It is not a matter of starting to spread. The laser beam begins diverging as soon as it leaves the output mirror. The divergence is very small, so for small distances the net divergence is essentially zero. If you measure the spot size at various distances the divergence is noticeable. It is especially noticeable over any distance out side of a lab. So the spot size from a Earth based laser on the moon is appreciable. Beam divergence is not governed by r^-2 laws, it is a result of properties of the laser itself.

Now, if someone were to measure the intensity of the laser signal on the moon, and assumed that it was the result of a point source radiating equally in all directions (i.e. they applied r^-2 law they would arrive at an astronomically large number for the energy of the source. One of the big advantages of a laser is the ability to put a significant amount of energy into a very narrow beam.

I believe that r^-2 laws only apply to sources which emit energy into a significant portion of a sphere.

 

[kishtik]

Originally posted by Integral
I believe that r^-2 laws only apply to sources which emit energy into a significant portion of a sphere.

Then how can we determine the radius of a laser beam with a given distance and starting radius?

 

[chroot]

Originally posted by kishtik
Then how can we determine the radius of a laser beam with a given distance and starting radius?

You have to also be given the divergence angle.

And Integral, any source emits light into some non-zero solid angle -- lasers included -- and so 1/r² is always relevant.

- Warren

 

[kishtik]

Originally posted by chroot
You have to also be given the divergence angle.

I had no idea about that but now I think that shouldn't be hard with the div. angle. We can solve it analitically or using direct proportion. Am I wrong?

 

[chroot]

It's just middle-school geometry.

- Warren