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Inverse square law and lasers

  1. Jul 16, 2007 #1
    I've read from several sources that says the inverse square law doesn't apply to lasers, but I've been told that it does apply. Which is right?

    Also if it does apply in today's lasers what about in a perfect laser?

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  3. Jul 16, 2007 #2


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    In a perfect laser beam the [itex] \frac 1 {r^2} [/itex] law would not hold. The beam diameter would not change, you would have the same energy density at any point on along the beam.

    HOWEVER. Perfect laser beams do not exist. Every real laser has a bit of beam divergence. If the beam is divergent, then at a significantly large distance you would measure a drop off in energy density according to [itex] \frac 1 {r^2} [/itex].
  4. Jul 16, 2007 #3
    How long is a significant distance? 1 AU? I ask becuase I was discussing the application of lasers as weapons in space.
    Last edited: Jul 16, 2007
  5. Jul 16, 2007 #4
    Normally "perfect laser" doesn't require the wavelength to be so negligible compared to the beam waist..
  6. Jul 16, 2007 #5


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    Energy falling off "as 1/r2" is true for light expanding from the source in a sphere- the energy is spread across the entire surface of the sphere which area is increasing as r2.

    If by "perfect laser" you meant "no spread at all", then the energy would not drop (except of course by absorption by space dust or whetever material was in its way).

    If your laser has a spread angle of [itex]\theta[/itex] (That would be a three dimensional "dihedral" angle) then the area over which it is spread at distance r from the origin would be [itex]\theta r^2[/itex] and so the strength would still fall off at a rate proportional to r2 (but the proportionality would be small for very small angles).
  7. Jul 16, 2007 #6
    Whats Known About Very Small Return Waves Riding Back On Em Waves Or On Any Other Wave?
    Last edited: Jul 16, 2007
  8. Jul 16, 2007 #7


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    I have a really good book about science for SF writers, written by Ben Bova (who is both a real scientist and a novelist). The section on lasers in space indicates some counter-intuitive behaviour. Something about the beam remaining coherent for quite some distance, then diverging at some rate that doesn't follow the inverse law.
    I can't remember the details, and the book is back at my house. I'll try to find it after work and get back to you.
  9. Jul 16, 2007 #8

    Claude Bile

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    The inverse square law does not hold true because the inverse square law assumes the source is radiating isotropically. Lasers do however have a far-field divergence angle which can be used to calculate the irradiance profile some distance away. This far-field divergence angle depends on the wavelength of the laser and the minimum spot size of the beam.

  10. Jul 17, 2007 #9


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    On one of the Apollo missions, they left an array of mirrors, with each shaped like a cube sliced in half diagonally, so they would reflect from any direction. Earth based lasers can be pulsed at the mirrors, and it takes a bit before the reflection comes back.

    http://science.nasa.gov/headlines/y2004/21jul_llr.htm [Broken]

    One of the advantages of being an old guy (55 years old), is being able to remember this stuff. At the time, major networks included snippets of videos of the laser beams being bounced off the mirrors.
    Last edited by a moderator: May 3, 2017
  11. Jul 17, 2007 #10


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    That sounds familiar. I never got the chance to go back to the house today, but this might very well be what Bova was referring to.
  12. Jul 17, 2007 #11


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    Shouldn't you be dead by now? Bloody hell, but you're ancient!
  13. Jul 26, 2007 #12
    if you take the average on the whole spherical surface at a distance r, even for lasers, perfect ones, the inverse square law will hold. On average, though.

    best wishes

  14. Jul 26, 2007 #13


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    Interestingly, these are the same type of things used on the masts of sailboats to make them more visible on radar.

    Radar Reflector
  15. Jul 26, 2007 #14
    Huh? That can also be said of a single cannonball, which is exactly the case for which you would prefer to say that the inverse square law doesn't apply.
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