laser energy dissipatation rate

"lucy" wrote in
:

Hi!
Can someone help me? I'm not a math major or anything, so please answer
SLOOOWLY.

Can someone tell me the rate that light or energy dissipates over
distance?

Presuming it's an omnidirectional light source, like the Sun or
a light bulb (no reflectors or lenses) it follows the Inverse Square
Law. The intensity falls off as the square of the distance.

If at one meter the light intensity is X, then at 2 meters the light
intensity is 1/(X^2).


Specific Example:
A beam of energy (light, coherent light, like a laser) leaves the end of
the emitter at 1 MegaWatt.
It hits a target 1,000 meters distant.

How much energy is lost, and how much impacts the target?

We can assume Vacuum, so air doesn't factor in.

Ahhh, a laser is different. As you indicate, presuming a vacuum, then
there is nothing in the path of the light to absorb any of it, so if
it leaves the laser at 1 megawatt, then after 1000 meters, there's still
1 megawatt. However.....

Even laser beams spread out. This is called divergence. It is NOT
possible to make a perfectly straight and narrow laser beam that
stays straight and narrow (or even non-laser light).

On can make the beam spread out less using collimating optics, but this
is at the expense of beam diameter. In order to make a given laser beam
spread out less, you must make it larger in diameter.

For example, I have several HeNe lasers, mostly just bare tubes, but
I also have a stabilized collimated HeNe for interferometer use.

On the bare tubes the laser beam starts out about 1 mm in diameter, but
it spreads out fast. After a hundred meters the beam has spread out to
about 25 centimeters.

On the collimated laser the beam starts out at about 10 mm, but
after the same 100 meters the beam isn't but 15 mm.

So yes, there is still a megawatt of laser energy after a thousand
meters but it's going to be spread out over a larger area than when
it had left.

If this megawatt laser is, say, a weapon, then the important part is
how much laser energy per surface area there is at the target. This is
usually rated in watts/cm^2. For a weapon, this number must be higher
than the damage threshhold of the material to be effective. Otherwise
it's just a bright light.


And if the target was 1,000,000 meters distant?

Same story as above but divergence will have spread it out even more.


Just so I won't have to ask in the future, how much does a standard
atmosphere absorb of the energy, in addition to the distance
dissipation?

It depends on the wavelength of the light. Most light in the visible
part of the spectrum will not be absorbed much. However, there are
some ranges in the infrared where our atmosphere is the equivilant
of black. Actually, scatter is probably more of a problem in the
visible rather than absorption.


Would this fomula work with all or most forms of energy? Light, X-Rays,
IR, UV, etc?

For the most part, especially in a vacuum. Remember, atmospheric
attenuation is very dependent on frequency.

thanking you...

lucy

lucyh at dhimaging dot com dot au


Brian
--
http://home.earthlink.net/~skywise71...ics/laser.html
"Great heavens! That's a laser!"
"Yes, Dr. Scott. A laser capable of emitting a beam of pure antimatter."

Sed quis custodiet ipsos Custodes?

"Skywise" wrote in message
link.net...
"lucy" wrote in
:

Hi!
Can someone help me? I'm not a math major or anything, so please answer
SLOOOWLY.

Can someone tell me the rate that light or energy dissipates over
distance?

Presuming it's an omnidirectional light source, like the Sun or
a light bulb (no reflectors or lenses) it follows the Inverse Square
Law. The intensity falls off as the square of the distance.

If at one meter the light intensity is X, then at 2 meters the light
intensity is 1/(X^2).


Specific Example:
A beam of energy (light, coherent light, like a laser) leaves the end of
the emitter at 1 MegaWatt.
It hits a target 1,000 meters distant.

How much energy is lost, and how much impacts the target?

We can assume Vacuum, so air doesn't factor in.

Ahhh, a laser is different. As you indicate, presuming a vacuum, then
there is nothing in the path of the light to absorb any of it, so if
it leaves the laser at 1 megawatt, then after 1000 meters, there's still
1 megawatt. However.....

Even laser beams spread out. This is called divergence. It is NOT
possible to make a perfectly straight and narrow laser beam that
stays straight and narrow (or even non-laser light).

On can make the beam spread out less using collimating optics, but this
is at the expense of beam diameter. In order to make a given laser beam
spread out less, you must make it larger in diameter.

For example, I have several HeNe lasers, mostly just bare tubes, but
I also have a stabilized collimated HeNe for interferometer use.

On the bare tubes the laser beam starts out about 1 mm in diameter, but
it spreads out fast. After a hundred meters the beam has spread out to
about 25 centimeters.

On the collimated laser the beam starts out at about 10 mm, but
after the same 100 meters the beam isn't but 15 mm.

So yes, there is still a megawatt of laser energy after a thousand
meters but it's going to be spread out over a larger area than when
it had left.

If this megawatt laser is, say, a weapon, then the important part is
how much laser energy per surface area there is at the target. This is
usually rated in watts/cm^2. For a weapon, this number must be higher
than the damage threshhold of the material to be effective. Otherwise
it's just a bright light.


And if the target was 1,000,000 meters distant?

Same story as above but divergence will have spread it out even more.


Just so I won't have to ask in the future, how much does a standard
atmosphere absorb of the energy, in addition to the distance
dissipation?

It depends on the wavelength of the light. Most light in the visible
part of the spectrum will not be absorbed much. However, there are
some ranges in the infrared where our atmosphere is the equivilant
of black. Actually, scatter is probably more of a problem in the
visible rather than absorption.


Would this fomula work with all or most forms of energy? Light, X-Rays,
IR, UV, etc?

For the most part, especially in a vacuum. Remember, atmospheric
attenuation is very dependent on frequency.

thanking you...

lucy

lucyh at dhimaging dot com dot au


Brian
--
http://home.earthlink.net/~skywise71...ics/laser.html
"Great heavens! That's a laser!"
"Yes, Dr. Scott. A laser capable of emitting a beam of pure antimatter."

Sed quis custodiet ipsos Custodes?


Brian,

THANK YOU so very much for an explanation that I can understand!

Hmm... now to get a list of metals and their melting points...

-Lucy

"lucy" wrote in message
...

"Skywise" wrote in message
link.net...
"lucy" wrote in
:

Hi!
Can someone help me? I'm not a math major or anything, so please
answer
SLOOOWLY.

Can someone tell me the rate that light or energy dissipates over
distance?

Presuming it's an omnidirectional light source, like the Sun or
a light bulb (no reflectors or lenses) it follows the Inverse Square
Law. The intensity falls off as the square of the distance.

If at one meter the light intensity is X, then at 2 meters the light
intensity is 1/(X^2).


Specific Example:
A beam of energy (light, coherent light, like a laser) leaves the end
of
the emitter at 1 MegaWatt.
It hits a target 1,000 meters distant.

How much energy is lost, and how much impacts the target?

We can assume Vacuum, so air doesn't factor in.

Ahhh, a laser is different. As you indicate, presuming a vacuum, then
there is nothing in the path of the light to absorb any of it, so if
it leaves the laser at 1 megawatt, then after 1000 meters, there's still
1 megawatt. However.....

Even laser beams spread out. This is called divergence. It is NOT
possible to make a perfectly straight and narrow laser beam that
stays straight and narrow (or even non-laser light).

On can make the beam spread out less using collimating optics, but this
is at the expense of beam diameter. In order to make a given laser beam
spread out less, you must make it larger in diameter.

For example, I have several HeNe lasers, mostly just bare tubes, but
I also have a stabilized collimated HeNe for interferometer use.

On the bare tubes the laser beam starts out about 1 mm in diameter, but
it spreads out fast. After a hundred meters the beam has spread out to
about 25 centimeters.

On the collimated laser the beam starts out at about 10 mm, but
after the same 100 meters the beam isn't but 15 mm.

So yes, there is still a megawatt of laser energy after a thousand
meters but it's going to be spread out over a larger area than when
it had left.

If this megawatt laser is, say, a weapon, then the important part is
how much laser energy per surface area there is at the target. This is
usually rated in watts/cm^2. For a weapon, this number must be higher
than the damage threshhold of the material to be effective. Otherwise
it's just a bright light.


And if the target was 1,000,000 meters distant?

Same story as above but divergence will have spread it out even more.


Just so I won't have to ask in the future, how much does a standard
atmosphere absorb of the energy, in addition to the distance
dissipation?

It depends on the wavelength of the light. Most light in the visible
part of the spectrum will not be absorbed much. However, there are
some ranges in the infrared where our atmosphere is the equivilant
of black. Actually, scatter is probably more of a problem in the
visible rather than absorption.


Would this fomula work with all or most forms of energy? Light,
X-Rays,
IR, UV, etc?

For the most part, especially in a vacuum. Remember, atmospheric
attenuation is very dependent on frequency.

thanking you...

lucy

lucyh at dhimaging dot com dot au


Brian
--
http://home.earthlink.net/~skywise71...ics/laser.html
"Great heavens! That's a laser!"
"Yes, Dr. Scott. A laser capable of emitting a beam of pure antimatter."

Sed quis custodiet ipsos Custodes?


Brian,

THANK YOU so very much for an explanation that I can understand!

Hmm... now to get a list of metals and their melting points...

-Lucy

Here is one list:
http://www.physics.ohio-state.edu/~kelch/melting.html

Alex

SNIP


Here is one list:
http://www.physics.ohio-state.edu/~kelch/melting.html

Alex

Thanks Alex!

ok - last question - how many joules to raise metal 1 degree Celcius?
I'd guess it would vary from metal to metal, but I was thinking of an alloy
of Titanium, but with TUNGSTEN having a melting point of 6150F and 3399C,
I'm having second thoughts...

-Lucy

ps: just double-checked and Titanium isn't on that chart. Is that because
it's an alloy?

"lucy" wrote:

SNIP


Here is one list:
http://www.physics.ohio-state.edu/~kelch/melting.html

Alex

Thanks Alex!

ok - last question - how many joules to raise metal 1 degree Celcius?
I'd guess it would vary from metal to metal, but I was thinking of an alloy
of Titanium, but with TUNGSTEN having a melting point of 6150F and 3399C,
I'm having second thoughts...

-Lucy

ps: just double-checked and Titanium isn't on that chart. Is that because
it's an alloy?

It's an element.

BTW if your aim is to melt metal with a laser that you can buy/make at home, I
suggest you investigate mercury.

--
Patrick Hamlyn posting from Perth, Western Australia
Windsurfing capital of the Southern Hemisphere
Moderator: polyforms group )