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I am trying to understand how a real world beam of laser actually reflects the physics description of oscillating electromagnetic waves.

So say we are looking side on at a vertically polarized laser beam, and this is section of it propagating through free space:

enter image description here

Ive cut down the opacity and zoomed in now to illustrate my question on what would the waves look like? Something like...

enter image description here

But how can this form a Gaussian intensity profile? Maybe there are more of these waves dispersed through out it, and their amplitude denotes the intensity e.g:

enter image description here

Where the waves closer to the edge of the beam are the same wavelength but smaller amplitude than the main section...? But now we have only considered a horizontal cross section, what would it be like if you looked at it from above?

Thanks

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    $\begingroup$ You must always label your pictures! What kind of waves are you drawing? Why do you have the laser beam still 2D when the y-axis must clearly represent something like current amplitude and not position? Please make more precise what you are trying to do/ask here, since zooming in is not what you did in the pictures (you can't zoom in in any meaningful sense since EM waves are only visible when they hit your eye, the light we see from laser beams is scattered off particles in the beam's way) $\endgroup$ – ACuriousMind Aug 27 '14 at 20:46
  • $\begingroup$ @ACuriousMind Hi, This is meant to be a hypothetical snapshot of a cross section looking perpendicular to a laser beam. I understand that all we see is the scattering from the particles in the air, but it was for illustration. And it is meant to be frozen in time, so a particular instant. $\endgroup$ – Steve Hatcher Aug 28 '14 at 3:01
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Your picture is ok, so long as you realise that this is a snapshot in time.

If you looked "from above" the electric field of the waves (by your definition of the polarisation direction) would be pointing towards or away from you - so this becomes rather difficult to draw in the same way.

It might be better to think of the E-field as a field of arrows, where the length of the arrow is proportional to the instantaneous E-field magnitude and the direction, well, shows the direction.

In which case, from above, you would be looking down on a field of arrow-heads and arrow-tails. If the laser had a Gaussian profile in both x and y directions (assuming the beam is heading towards z), then the maximum height of the arrows (in either direction) would diminish towards the edge of the beam.

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  • $\begingroup$ Thanks, yes it is meant to be frozen in time. I think I understand what you mean with the arrow heads. SO looking from above, you could maybe use a color to map out the 'height' of the arrow head, say red if its pointing directly up at you and blue if its pointing as far down as possible, and then map it out. I think I get it thanks. So is the single plane wave equation just kind of an average value for all these vectors? $\endgroup$ – Steve Hatcher Aug 28 '14 at 3:04

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