0
$\begingroup$

Suppose I am mathematically modeling a visible green laser pointer beam which starts from the ground and ends up a half mile or full mile in the sky, projected upon an aircraft window 2 inches thick. How much will the ground-borne visible green laser pointer beam spread out in the final 2 inches between the front and back pane of the aircraft window? Please compare this result with that for an incoherent visible light green beam originating from the ground.

The reason I ask this question is to solve one of society's problems where commercial aircraft are grounded due to laser pointers sent from the ground near airports?

I just read the following 2 Stack Overflow articles which discuss some of the physics and equations necessary to solve my question:Physics of Focusing a Laser

Can radio waves be formed into a pencil beam? Thank you.

$\endgroup$
  • $\begingroup$ Society's solution against these kinds of problems is called "the law". The minimal divergence of a laser at the distance you mention is negligible for your "problem". If somebody wants to do some real harm with these things, they can do so with very cheap, readily available technology, but I do not think that this is a good forum to explain how to "do it right", especially not since your "effort to solve it" is doomed to fail. $\endgroup$ – CuriousOne Feb 16 '16 at 8:03
  • $\begingroup$ @CuriousOne, Thank you for your comment. What is the minimal divergence of an incoherent visible light green beam at the distance of 1/2 to 1 mile over the inch of the aircraft window's thickness? $\endgroup$ – Frank Feb 16 '16 at 8:13
  • $\begingroup$ Like I said, it's small enough. There is no technical solution to the problem from that end. Aircraft manufacturers could (and eventually probably will) make windows that go blind when hit by a beam, but if you were one of the guys designing such a window, you wouldn't be here, right? $\endgroup$ – CuriousOne Feb 16 '16 at 8:18
  • $\begingroup$ @CuriousOne,May I ask you to scan the following article which was patented as a software algorithm, discovery.ucl.ac.uk/2670/,Detection of Coherent Light in an Incoherent Background wriiten by a University of London professor? Thank you. $\endgroup$ – Frank Feb 16 '16 at 8:31
  • $\begingroup$ While the link is broken I think I know which paper you are referring to, but that's for a different scenario. When a pilot gets blinded by a laser from the ground, detecting the source is not a problem, at all. Indeed, one could build a relatively trivial camera system that can take an image of the area and show law enforcement with pinpoint accuracy where the perpetrators were. Now if you can only get police there within a minute, then you are onto something... it's the last part that doesn't work. $\endgroup$ – CuriousOne Feb 16 '16 at 8:43
0
$\begingroup$

Assume your incoherent light comes from a source smaller than the window and spreads out to cover the window. Then its angular divergence, defined as the angle between the light on the left edge and the light at the right edge, or the width of the window over the distance to the plane.

$$ \Delta \phi = w_{\rm{window}}/d_{\rm{plane}}$$

Any beam in these conditions will keep its overall beam divergence. A good laser might be smaller than the window at that distance. If so, the divergence in the window is given by that beam divergence. If larger, see formula above.

But I agree with the commentators: in any case, the effect of divergence as the beam goes through the glass is negligible.

$\endgroup$
-1
$\begingroup$

@anna v, a theoretical physicist from Greece, wrote on Physics Stack Exchanges 3 years ago, "Coherence matters because it retains the angles , there is spacial coherence. With incoherent light there is large divergence.One could not shine incoherent light on the moon, for example. The incoherent spot on the unit sphere will diverge and enlarge proportional to r^2 as the radius grows. A laser beam spot due to coherence enlarges slowly as can be found in various links". – anna v Jul 3 '12 at 18:47 in this URL, How can a laser pointer have range of several kilometers in atmosphere?

The answer to my question is, The incoherent spot on the unit sphere will diverge and enlarge proportional to r^2 as the radius grows. Therefore: 1/2 mile = 804.5 meters 2 inches thickness = 5 centimeters So, (804.55 meters) * (804.55 meters) - (804.5 meters) * (804.5 meters) = 80 square meters difference which equates to the radius of the incoherent spot on the unit sphere growing by 5 meters over the final 2 inches of the aircraft cockpit window. I must admit that this result may needs to be corrected because it is relative to the unit sphere. Please correct me if you have the time.

$\endgroup$
  • $\begingroup$ @anna v, Could you please critique and correct my calculation? I have a great part-time opportunity which you might be interested in. Thank you. $\endgroup$ – Frank Feb 16 '16 at 11:38
  • $\begingroup$ FYI, even if you delete one of your posts, the site retains it and I and other users with high enough reputation can still see it. In case you ever get into a legal fight about the validity of your patents, that's perfectly enough to invalidate whatever you have disclosed in them. Should you work for a company that has such patents at stake, you have just royally screwed your employer and you would be legally liable. $\endgroup$ – CuriousOne May 29 '16 at 8:26
  • $\begingroup$ I think @annav may know a bit more now than she did in 2012. PhysicsStackExchange should not be considered the source of physics truth. Rather, it should be a place where you can listen to what other physics-interested people have to say. There is a LOT of "fake truth" expressed here. If you find a statement here that makes sense to you, it's a good idea to do serious literature search (i.e., textbooks and peer-reviewed papers) to make sure it makes sense to the professional physics community as well. $\endgroup$ – S. McGrew May 12 at 2:50
  • $\begingroup$ The area of the beam grows as the square. Each dimension grows linearly. That’s why divergence, a linear function of distance not a quadratic one, is useful. $\endgroup$ – Bob Jacobsen Sep 11 at 2:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.