Attenuation - Revision history

Lwcamp at 18:58, 7 March 2026

2026-03-07T18:58:29Z

← Older revision		Revision as of 11:58, 7 March 2026
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	==References==		==References==

	[[Category:Lasers]][[Category:Physics & Engineering]]		[[Category:Lasers]][[Category:Physics & Engineering]][[Category:Physics & Math & Engineering]]

Lwcamp: /* Shooting through the atmosphere */

2025-04-08T22:48:38Z

Shooting through the atmosphere

← Older revision		Revision as of 15:48, 8 April 2025
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	If you are in orbit around a planet and trying to shoot at ground targets, or if you are on the ground trying to shoot at spacecraft in orbit, it helps to be able to estimate the effective thickness of the air. To go from Earth to space or vice versa, you need to shoot through an effective distance of 8.5 km of air at the density of the shooter or target in the air. This distance is called the <i>scale height</i>. For other planets, you can use		If you are in orbit around a planet and trying to shoot at ground targets, or if you are on the ground trying to shoot at spacecraft in orbit, it helps to be able to estimate the effective thickness of the air. To go from Earth to space or vice versa, you need to shoot through an effective distance of 8.5 km of air at the density of the shooter or target in the air. This distance is called the <i>scale height</i>. For other planets, you can use
	<div class="center" style="width: auto; margin-left: auto; margin-right: auto;">		<div class="center" style="width: auto; margin-left: auto; margin-right: auto;">
	~~<math>~~		scale height = 8.5 km × [ T / (288.15 K) ] × [ (9.80 m/s²) / g] × [ (28.96 g/mole) / W]
	~~\mbox{~~scale height} = 8.5 ~~\, \mbox{~~km~~} \, \frac{~~T}{288.15 ~~\, \mbox{~~K~~}} \, \frac{~~9.80 ~~\, \mbox{~~m}/~~\mbox{~~s~~}^2}{~~g~~} \, \frac{~~28.96 ~~\, \mbox{~~g}/~~\mbox{~~mole~~}}{~~W}		</div>
	~~</math>~~</div>		where T is the average temperature, g is the acceleration due to gravity, and W is the average molecular weight of the gases in the air. If you are shooting through the atmosphere at an angle, the distance of all that atmospheric junk you need to go through can be approximated as the scale height divided by the cosine of the angle you are shooting at. The amount of air you have to get through along your beam path is called the [https://en.wikipedia.org/wiki/Air_mass_(astronomy) air mass], and if you follow the link you will find more accurate methods of approximating it if you need to.
	where ~~<math>~~T~~</math>~~ is the average temperature, ~~<math>~~g~~</math>~~ is the acceleration due to gravity, and ~~<math>~~W~~</math>~~ is the average molecular weight of the gases in the air. If you are shooting through the atmosphere at an angle, the distance of all that atmospheric junk you need to go through can be approximated as the scale height divided by the cosine of the angle you are shooting at. The amount of air you have to get through along your beam path is called the [https://en.wikipedia.org/wiki/Air_mass_(astronomy) air mass], and if you follow the link you will find more accurate methods of approximating it if you need to.

	==Credit==		==Credit==

Lwcamp: /* Scatter */

2025-04-08T22:44:40Z

Scatter

← Older revision		Revision as of 15:44, 8 April 2025
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	The next thing to worry about is scatter. Rather than simply making the light in your beam go away, scatter is what makes the light go in a different direction. Particulates or aerosols in the air are good at scattering light. This is why we can’t see through fog or clouds. But even perfectly clean air will scatter light to some extent. The electric field in the light will make the electrons in air molecules slosh back and forth at the light’s frequency. And these electrons then act like antennas to radiate light away in different directions (while simultaneously taking that energy away from the beam). This is called <i>Rayleigh scattering</i>, and it is more effective for higher frequency light than lower frequency light. This is why, for example, the sky looks blue - more high frequency blue light from the sun is being scattered into our eyes than lower frequency light of other colors.		The next thing to worry about is scatter. Rather than simply making the light in your beam go away, scatter is what makes the light go in a different direction. Particulates or aerosols in the air are good at scattering light. This is why we can’t see through fog or clouds. But even perfectly clean air will scatter light to some extent. The electric field in the light will make the electrons in air molecules slosh back and forth at the light’s frequency. And these electrons then act like antennas to radiate light away in different directions (while simultaneously taking that energy away from the beam). This is called <i>Rayleigh scattering</i>, and it is more effective for higher frequency light than lower frequency light. This is why, for example, the sky looks blue - more high frequency blue light from the sun is being scattered into our eyes than lower frequency light of other colors.
	The Rayleigh scattering length for clean sea level air is <ref name=CRC>Calculated using atomic and molecular polarizabilities taken from David R. Lide, "CRC Handbook of Chemistry and Physics: 71st Edition 1990-1991", CRC Press (1990)</ref>		The Rayleigh scattering length for clean sea level air is <ref name=CRC>Calculated using atomic and molecular polarizabilities taken from David R. Lide, "CRC Handbook of Chemistry and Physics: 71st Edition 1990-1991", CRC Press (1990)</ref>
	<div class="center" style="width: auto; margin-left: auto; margin-right: auto;"><~~math~~>		<div class="center" style="width: auto; margin-left: auto; margin-right: auto;">
	~~R_{~~Rayleigh} = 948 ~~\, \mbox{~~km~~} \left~~[ ~~\frac{\lambda}{~~1 ~~\, \mu\mbox{m}} \right~~]^4		R<sub>Rayleigh</sub> = 948 km [ λ / (1 μm)]<sup>4</sup>
	</~~math~~></div>		</div>
	for wavelength ~~<math>\lambda</math>~~.		for wavelength λ.

	Note that scatter makes it so you can see the laser beam. Even a 1 watt blue laser shows up clearly in clean air from its Rayleigh scattering. A very powerful visible light laser like you would use for a weapon will give an obvious trace through the air.		Note that scatter makes it so you can see the laser beam. Even a 1 watt blue laser shows up clearly in clean air from its Rayleigh scattering. A very powerful visible light laser like you would use for a weapon will give an obvious trace through the air.

Lwcamp: /* The Beer-Lambert law */

2025-04-08T22:41:35Z

The Beer-Lambert law

← Older revision		Revision as of 15:41, 8 April 2025
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	==The Beer-Lambert law==		==The Beer-Lambert law==

	Numerically, the attenuation is represented by an <i>attenuation length</i> <~~math~~>~~R_0~~</~~math~~>, which is the distance at which the intensity of the light (or beam power, or pulse energy) has been reduced to approximately 37% of its original value. If you start off with an intensity <~~math~~>~~I_0~~</~~math~~>, or a power <~~math~~>~~P_0~~</~~math~~>, or an energy <~~math~~>~~E_0~~</~~math~~>, then the intensity ~~<math>~~I~~</math>~~, power ~~<math>~~P~~</math>~~, and energy ~~<math>~~E~~</math>~~ after a given range ~~<math>~~R~~</math>~~ will be		Numerically, the attenuation is represented by an <i>attenuation length</i> R<sub>0</sub>, which is the distance at which the intensity of the light (or beam power, or pulse energy) has been reduced to approximately 37% of its original value. If you start off with an intensity I<sub>0</sub>, or a power P<sub>0</sub>, or an energy E<sub>0</sub>, then the intensity I, power P, and energy E after a given range R will be
	<div class="center" style="width: auto; margin-left: auto; margin-right: auto;"><~~math~~>		<div class="center" style="width: auto; margin-left: auto; margin-right: auto;">
	~~\begin{matrix}~~		<table>
	I & = ~~& I_0 \, \~~exp[-R/~~R_0~~] \\		<tr><td>I <td> = <td> I<sub>0</sub> exp[-R/R<sub>0</sub>]
	P & = ~~& P_0 \, \~~exp[-R/~~R_0~~] \\		<tr><td>P <td> = <td> P<sub>0</sub> exp[-R/R<sub>0</sub>]
	E & = ~~& E_0 \, \~~exp[-R/~~R_0~~].		<tr><td>E <td> = <td> E<sub>0</sub> exp[-R/R<sub>0</sub>].
	~~\end{matrix}~~		</table>
	</~~math~~></div>		</div>
	This is called the <i>Beer-Lambert law</i>.		This is called the <i>Beer-Lambert law</i>.
	<ref name=Bouguer_1729>https://archive.org/details/UFIE003101_TO0324_PNI-2703_000000 Bouguer, Pierre (1729). Essai d'optique sur la gradation de la lumière [Optics essay on the attenuation of light] (in French). Paris, France: Claude Jombert. pp. 16–22.</ref>		<ref name=Bouguer_1729>https://archive.org/details/UFIE003101_TO0324_PNI-2703_000000 Bouguer, Pierre (1729). Essai d'optique sur la gradation de la lumière [Optics essay on the attenuation of light] (in French). Paris, France: Claude Jombert. pp. 16–22.</ref>
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	<ref name=Beer_1852>https://books.google.com/books?id=PNmXAAAAIAAJ&pg=PA78 Beer (1852). "Bestimmung der Absorption des rothen Lichts in farbigen Flüssigkeiten" [Determination of the absorption of red light in colored liquids]. Annalen der Physik und Chemie (in German). 162 (5): 78–88. Bibcode:1852AnP...162...78B. doi:10.1002/andp.18521620505.</ref>		<ref name=Beer_1852>https://books.google.com/books?id=PNmXAAAAIAAJ&pg=PA78 Beer (1852). "Bestimmung der Absorption des rothen Lichts in farbigen Flüssigkeiten" [Determination of the absorption of red light in colored liquids]. Annalen der Physik und Chemie (in German). 162 (5): 78–88. Bibcode:1852AnP...162...78B. doi:10.1002/andp.18521620505.</ref>

	If you have more than one physical phenomenon contributing to attenuation (for example, absorption and scattering, or absorption off of oxygen and absorption off of nitrogen), the inverses of the attenuation lengths of each phenomena add together. In particular, if you have scattering with a characteristic scattering length <~~math~~>~~R_s~~</~~math~~> and absorption with a characteristic absorption length <~~math~~>~~R_a~~</~~math~~>, then		If you have more than one physical phenomenon contributing to attenuation (for example, absorption and scattering, or absorption off of oxygen and absorption off of nitrogen), the inverses of the attenuation lengths of each phenomena add together. In particular, if you have scattering with a characteristic scattering length R<sub>s</sub> and absorption with a characteristic absorption length R<sub>a</sub>, then
	<div class="center" style="width: auto; margin-left: auto; margin-right: auto;"><~~math~~>		<div class="center" style="width: auto; margin-left: auto; margin-right: auto;">
	~~\frac{1}{R_0}~~ = ~~\frac{~~1~~}{R_a}~~ + ~~\frac{~~1~~}{R_s}.~~		1 / R<sub>0</sub> = 1 / R<sub>a</sub> + 1 / R<sub>s</sub>.
	</~~math~~></div>		</div>

	==Absorption==		==Absorption==

Tshhmon at 21:28, 23 April 2024

2024-04-23T21:28:03Z

← Older revision		Revision as of 14:28, 23 April 2024
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	==References==		==References==

	[[Category:Lasers]]		[[Category:Lasers]][[Category:Physics & Engineering]]

Lwcamp: /* Different atmospheres */

2024-03-31T04:11:07Z

Different atmospheres

← Older revision		Revision as of 21:11, 30 March 2024
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	==Different atmospheres==		==Different atmospheres==

	The attenuation of atmospheres of greatly different composition than the Earth – especially in the infrared – goes well beyond the scope of this document. You can make a quick estimate, however, by dividing the attenuation length without aerosols by the planet's atmospheric pressure relative to Earth and then figure the effect of aerosols separately. This method is more accurate the closer the planet's atmospheric composition is to Earth. Unless the atmosphere contains chlorine, nitrogen dioxide, far more ozone than Earth's air, or approaching an atmospheric pressure or more of methane, the air will be nearly transparent to visible light; although the scattering may vary by a factor of two or so depending on the polarizabilities of the air molecules.		The attenuation of atmospheres of greatly different composition than the Earth – especially in the infrared – goes well beyond the scope of this document. You can make a quick estimate, however, by dividing the attenuation length without aerosols by the planet's atmospheric pressure relative to Earth and then figure the effect of aerosols separately. This method is more accurate the closer the planet's atmospheric composition is to Earth. Unless the atmosphere contains chlorine, fluorine, nitrogen dioxide, far more ozone than Earth's air, or approaching an atmospheric pressure or more of methane, the air will be nearly transparent to visible light; although the scattering may vary by a factor of two or so depending on the polarizabilities of the air molecules.

	This can also be used to estimate the attenuation of air on Earth at different altitudes.		This can also be used to estimate the attenuation of air on Earth at different altitudes.

	==Shooting through the atmosphere==		==Shooting through the atmosphere==

Lwcamp: /* Shooting through the atmosphere */

2022-07-27T18:21:42Z

Shooting through the atmosphere

← Older revision		Revision as of 11:21, 27 July 2022
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	<div class="center" style="width: auto; margin-left: auto; margin-right: auto;">		<div class="center" style="width: auto; margin-left: auto; margin-right: auto;">
	<math>		<math>
	\mbox{scale height} = 8.5 \, \mbox{km} \, \frac{T}{288.15 \, \mbox{K}} \, \frac{9.80 \, \mbox{m}/\mbox{s}^2}{g} \, \frac{28.96 \, \mbox{g}/\mbox{mole}^3}{W}		\mbox{scale height} = 8.5 \, \mbox{km} \, \frac{T}{288.15 \, \mbox{K}} \, \frac{9.80 \, \mbox{m}/\mbox{s}^2}{g} \, \frac{28.96 \, \mbox{g}/\mbox{mole}}{W}
	</math></div>		</math></div>
	where <math>T</math> is the average temperature, <math>g</math> is the acceleration due to gravity, and <math>W</math> is the average molecular weight of the gases in the air. If you are shooting through the atmosphere at an angle, the distance of all that atmospheric junk you need to go through can be approximated as the scale height divided by the cosine of the angle you are shooting at. The amount of air you have to get through along your beam path is called the [https://en.wikipedia.org/wiki/Air_mass_(astronomy) air mass], and if you follow the link you will find more accurate methods of approximating it if you need to.		where <math>T</math> is the average temperature, <math>g</math> is the acceleration due to gravity, and <math>W</math> is the average molecular weight of the gases in the air. If you are shooting through the atmosphere at an angle, the distance of all that atmospheric junk you need to go through can be approximated as the scale height divided by the cosine of the angle you are shooting at. The amount of air you have to get through along your beam path is called the [https://en.wikipedia.org/wiki/Air_mass_(astronomy) air mass], and if you follow the link you will find more accurate methods of approximating it if you need to.

Lwcamp: /* Shooting through the atmosphere */

2022-07-27T18:21:27Z

Shooting through the atmosphere

← Older revision		Revision as of 11:21, 27 July 2022
Line 62:		Line 62:
	<div class="center" style="width: auto; margin-left: auto; margin-right: auto;">		<div class="center" style="width: auto; margin-left: auto; margin-right: auto;">
	<math>		<math>
	\mbox{scale height} = 8.5 \, \mbox{km} \, \frac{T}{288.15 \, \mbox{K}} \, \frac{9.80 \, \mbox{m}/\mbox{s}^2}{g} \, \frac{28.96 \, \mbox{g}/\mbox{cm}^3}{W}		\mbox{scale height} = 8.5 \, \mbox{km} \, \frac{T}{288.15 \, \mbox{K}} \, \frac{9.80 \, \mbox{m}/\mbox{s}^2}{g} \, \frac{28.96 \, \mbox{g}/\mbox{mole}^3}{W}
	</math></div>		</math></div>
	where <math>T</math> is the average temperature, <math>g</math> is the acceleration due to gravity, and <math>W</math> is the average molecular weight of the gases in the air. If you are shooting through the atmosphere at an angle, the distance of all that atmospheric junk you need to go through can be approximated as the scale height divided by the cosine of the angle you are shooting at. The amount of air you have to get through along your beam path is called the [https://en.wikipedia.org/wiki/Air_mass_(astronomy) air mass], and if you follow the link you will find more accurate methods of approximating it if you need to.		where <math>T</math> is the average temperature, <math>g</math> is the acceleration due to gravity, and <math>W</math> is the average molecular weight of the gases in the air. If you are shooting through the atmosphere at an angle, the distance of all that atmospheric junk you need to go through can be approximated as the scale height divided by the cosine of the angle you are shooting at. The amount of air you have to get through along your beam path is called the [https://en.wikipedia.org/wiki/Air_mass_(astronomy) air mass], and if you follow the link you will find more accurate methods of approximating it if you need to.

Lwcamp: /* Shooting through the atmosphere */

2022-07-27T18:18:43Z

Shooting through the atmosphere

← Older revision		Revision as of 11:18, 27 July 2022
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	\mbox{scale height} = 8.5 \, \mbox{km} \, \frac{T}{288.15 \, \mbox{K}} \, \frac{9.80 \, \mbox{m}/\mbox{s}^2}{g} \, \frac{28.96 \, \mbox{g}/\mbox{cm}^3}{W}		\mbox{scale height} = 8.5 \, \mbox{km} \, \frac{T}{288.15 \, \mbox{K}} \, \frac{9.80 \, \mbox{m}/\mbox{s}^2}{g} \, \frac{28.96 \, \mbox{g}/\mbox{cm}^3}{W}
	</math></div>		</math></div>
	where <math>T</math> is the average temperature, <math>g</math> is the acceleration due to gravity, and <math>W</math> is the average molecular weight of the gases in the air. If you are shooting through the atmosphere at an angle, the distance of all that atmospheric junk you need to go through can be approximated as the scale height divided by the cosine of the angle you are shooting at. The amount of air you have to get through along your beam path is called the ~~<i>~~air mass~~</i>~~.		where <math>T</math> is the average temperature, <math>g</math> is the acceleration due to gravity, and <math>W</math> is the average molecular weight of the gases in the air. If you are shooting through the atmosphere at an angle, the distance of all that atmospheric junk you need to go through can be approximated as the scale height divided by the cosine of the angle you are shooting at. The amount of air you have to get through along your beam path is called the [https://en.wikipedia.org/wiki/Air_mass_(astronomy) air mass], and if you follow the link you will find more accurate methods of approximating it if you need to.

	==Credit==		==Credit==

Lwcamp: /* Shooting through the atmosphere */

2022-07-27T18:11:10Z

Shooting through the atmosphere

← Older revision		Revision as of 11:11, 27 July 2022
Line 64:		Line 64:
	\mbox{scale height} = 8.5 \, \mbox{km} \, \frac{T}{288.15 \, \mbox{K}} \, \frac{9.80 \, \mbox{m}/\mbox{s}^2}{g} \, \frac{28.96 \, \mbox{g}/\mbox{cm}^3}{W}		\mbox{scale height} = 8.5 \, \mbox{km} \, \frac{T}{288.15 \, \mbox{K}} \, \frac{9.80 \, \mbox{m}/\mbox{s}^2}{g} \, \frac{28.96 \, \mbox{g}/\mbox{cm}^3}{W}
	</math></div>		</math></div>
	where <math>T</math> is the average temperature, <math>g</math> is the acceleration due to gravity, and <math>W</math> is the average molecular weight of the gases in the air. If you are shooting through the atmosphere at an angle, the distance of all that atmospheric junk you need to go through is the scale height divided by the cosine of the angle you are shooting at.		where <math>T</math> is the average temperature, <math>g</math> is the acceleration due to gravity, and <math>W</math> is the average molecular weight of the gases in the air. If you are shooting through the atmosphere at an angle, the distance of all that atmospheric junk you need to go through can be approximated as the scale height divided by the cosine of the angle you are shooting at. The amount of air you have to get through along your beam path is called the <i>air mass</i>.

	==Credit==		==Credit==