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Lasers with visible beam at night time...

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Advanced Member level 2
Jan 29, 2004
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My green 532-535nm laser pointer shows a clear beam at night; my blue laser pointer probably a ~405 nm bluish/violet does not.
Do not own a 445-455nm. Do you know before I buy one that wavelength, if will show the beam from source to target ?

if will show the beam from source to target ?
The type and density of dust floating around may be what most determines the beam's 'visibility', which in theory should be barely perceptible for any visible light spectrum.

I used to live near an Air Force antenna range from which a beam could occasionally be seen pointing skyward. Looked like cyan (argon?).

Anything you see is reflection, and absorption varies with wavelength, perhaps this is the mechanism at play.

there are more things in the air that will reflect green than there are things that will reflect voilet - clean air gives no indication at all.

Your eye is considerably more sensitive to green light than blue, or red. The also eye loses sensitivity to blue light due to ageing, the lenses and fluid in the eye start to become more yellow in colour.
Green lasers are used for pointing out astronomical objects at star parties. Red lasers also work, for this but need much higher power.

as noted above, you are seeng light reflected out of the beam
the size of the dust particles determines what wavelengths get reflected
if the wavelength of light is greater than the size of the dust, there is reflection
if the wavelength of light is smaller than the size of the dust, it does not reflect

if the wavelength of light is greater than the size of the dust, there is reflection
if the wavelength of light is smaller than the size of the dust, it does not reflect
Really? I thought it´s the other way round.

I have a picture in my mind:
A (big) stone, half in water.
If the stone is small compared to the wavelength, the wave goes "around" the stone with no considerable irritation or reflection. (wavelength > object)

But if the stone is big compared to the wavelength, the wave becomes irritatated and partly reflected. (wavelength < object)


your analogy works for mechanical waves like sound and water that require a medium to move so the wave travels. it isn't clear if that analogy holds for EM waves, that do not require a medium

i asked a friend more knowledgeable than myself
this is his response:

What you have in mind is Rayleigh scattering. Basically Rayleigh scattering has an intensity proportional to 1/wavelength^4 and is independent of the size of the scatterer as long as the scattering particle size << wavelength and the << means really several orders of magnitude in ratio. In truth, when scatterer size ~ wavelength all bets are off because then the physical and geometric properties of the scatter really matters. For example refractive index or dielectric constant can have profound effects. That is why microwave waveguides are sometimes filled or sometimes not.

In air where the scattering is mainly single molecular dipole radiation, Rayleigh scattering is always present for visible or near visible E-M radiation. If dust, either natural or artificial, polluting the air is very microscopic (volcanic, etc.) then violet, yellow or red sunsets result. If the dust size produced approaches the visible wavelengths in diameter, for example, then the smoke can appear either black or white depending what it is made of and there is no Rayleigh scattering. This is what is common for pollution debris just coming out of a smoke stack or for a forest fire. How black or how white depends on the density of "large" particles present and their composition.

In the final analysis you had the wavelength direction of increased scattering correct, but you erred in your criteria relative to the wavelength to particle size.

end of response

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