Question:
Cosmetic Tattoo Removal info
What if the car in the infomercial was yellow, or what ever is the complement
of the blue laser? That is, it was a pigment that absorbed the particular
wavelength of the laser very well? Would only the paint be damaged?
I'm interested in the tattoo removal scenario as well. Its reds that can't be
removed too well with laser surgery, correct? Because in that situation, more
than just the pigment would be removed, the red meat under it would receive
more damage. And I imagine a green laser would be needed to remove red, and a
red laser to remove green.
I've only seen lasers used on skin once, for cosmetic wrinkle removal (burning
away the outer layer).
Since I've learned that the shorter higher wavelengths have more energy, I tend
to think of them as pssoibly being able to do the most damage, while the red
lasers would need higher wattage to do the same 'damage'. Is this the case?
Maybe AFCA will chew on this more than the science fiction newsgroup did.
A whole chunk of questions about lasers follow, concerning what in particular
is possibly damaging about a laser, and how reflection and abosorption may be
involved.
There is a late nite infomercial that shows off their car wax by shining a
laser at a car. This blue laser has just cut through a grey slab of metal, but
reflects off of the white car (white reflects all colors). Assuming no other
change of the setup (the same beam that cut the metal is hitting the car), what
is possibly misleading about the car wax commercial? Could a simple UV coating
provide enough protection? If the car were blue like the laser, would you see
damage? What if it were the complimentary Yellow or Orange?
And tattoos of certain colors are difficult to remove with laser surgery (red
being too close to skin tone?)
Can a laser of any wave length cut through an appropiately colored object if
the wattage is boosted and there are enough photons involved? Is the actual
cutting done by absorption of that frequency creating heat, or breaking bonds
whatever? Is the wavelength more inportant than the wattage?
Are most industrial cutting lasers using Infrared frequencies> Visible> UV xray
etc? is the frequency itself doing the work, or does the cutting happen due to
thermodynamic loss, (if UV or X-ray lasers were used for cutting, would the
material
get hot?)
Larry Nivens Science Fiction has two particular examples of laser
absorption and reflection dealing with
mirrors and black superconducting cloth as protective sheilds. The
effectiveness of
these particular armors tend to be limited as there is no perfect reflector or
absorber. And if absorption has to do with cutting power wouldnt black be one
of the least safe colors (assuming the cloth isnt superconducting.
Answer:
- I worked with various high-power lasers during the two years I was
part of an Atomic/Molecular/Optical Physics research group (I switched to
Physics Education), just so you know I'm not completely pulling these answers
out of my butt.
The main trick is the reflectivity of the substance. As another poster
in the thread has pointed out, matte or dull substances absorb the energy
more easily and therefore are vaporized (lasers cut by heating the surface so
quickly that it vaporizes, there's rarely any molten material). The
misleading part is that the car probably could have bounced that beam with
just the factory-standard shiny paint job. A clean car is a laser-reflective
car. So, in that the wax job or whatever cleans off all impurities from the
surface, it's at least reasonably worthwhile.
If the car were blue like the laser, would you see
No real effect, since it's the metallic sheen of the paint which does
the job. The color we see on the car is just a matter of which color is
reflected better, they're all reflected pretty well. Mind you, being a blue
surface means it reflects blue better, so it might have been more iffy on an
orange car.
One side note: complementary colors are simply an artifact of the way
our eyes perceive light, with the red, blue and green receptors. The only
relevant detail for purposes of reflection is frequency difference, in which
case red is even worse than orange, since it's farther from blue.
The way lasers remove tattoos is by having the subcutaneous ink absorb
most of the energy of the beam and thus vaporize. The closer the ink color
is to skin color, the harder it is to do this without also having the skin
and blood absorb energy and, well, vaporize.
Actually, impurities are the most important. Anything which can absorb
the frequency of the laser without re-emitting it at the same frequency
(i.e. anything other than being totally transparent or 100% reflective) will
start to burn through as energy is dumped into the impurities. The stronger
the laser, the better the object has to be at reflecting or transmitting to
not be cut.
To give an example, in the AMO lab we'd pay a few hundred bucks for
really good mirrors, which reflected nearly 100% without any surface
impurities. If a speck of dust landed on the mirror and the beam hit it,
BANG, hole in the mirror's surface. And even a clean mirror would fry if
the power was too high for too long. Color is no protection.
Depends on the application. IR is easy to make at high energies, but
the fact that it's invisible can lead to safety issues, so they're often
frequency-doubled (and thus intensity-halved) into visible wavelengths. In
cases where color doesn't matter, like cutting, it's probably left IR or
green. In cases where color does matter (such as to drive a specific
chemical reaction), they either use a tunable diode laser or (less often
these days) a dye laser which has the incoming beam cause a laser emission
from a dye suspension. Dye lasers are messy as hell when they go wrong, so
they're generally not used in industrial applications when the laser can be
built to the frequency needed (dye lasers can be adjusted more easily, making
them more popular in research).
