Re: What would happen if you had a tube and shot a laser into it?

Your question is an interesting one, and -- as with most interesting 
questions -- the answer is that "it all depends..."

For something to truly explode, most people believe (correctly) that it 
takes about 10,000 joules of laser energy to be applied to the surface per 
cubic centimeter of volume.  By explosion, this means to me  that the 
thing which absorbs the laser energy essentially instantly turns into a 
hot plasma (a gas of ions and electrons) and disassembles into an 
unrecognizable flash in something on the order of a millionth of a second.

Let's now think a bit about the laser...  To burn skin a little bit, you 
might start with a laser with pulsed energy output of 0.1 joules in a 
0.635 cm (that's 1/4 inch, if you're thinking English units) beam.  This 
laser will hurt a little bit when it hits your hand, but it won't leave a 
visible mark -- although it will blow the paint off the dial of your 
watch, which I know for a fact based on an experience I had with a very 
nice Omega Submariner which my wife gave me a few years ago.  If you scale 
up the laser energy a bit to 2 joules in a 1 cm beam, the results are 
quite different:  it causes flames (actually plasma) to blow off your skin 
and leaves a white mark of charred (but not penetrated) skin after a 
single pulse -- pretty ugly and hurts like the dickens, also based on 
personal experience.  For a first estimate, let's just say that something 
like 3 joules in a laser pulse can cause some sort of burn.

Now let's look at the tube you considered.  I'll guess that the wall 
thickness of the tube is about 0.020" -- which would be considered 
thinwall if it were stainless steel tubing.  0.020" is equal to about 
0.020 X 2.54 ~ .05 cm -- since I can only do sensible calculations in 
metric measurements, although I confess to thinking in English units 
sometimes.  I'll also guess that the tube might be about 1/4 inch in 
diameter -- which is about 0.6 cm -- pretty big for a capsule that you'd 
swallow, so I'll back off my guess and say that maybe it's 0.5 cm in 
diameter.  If the tube is 3/4" long -- that's about 1.8 cm or so -- the 
volume of the tubular section is equal to the cross sectional area of the 
tube multiplied by the length.  The cross sectional area of such a thin 
tube is exactly given by the expression (Pi) X ( Router*Router - 
Rinner*Rinner); Pi = 3.14159 (a constant), Router and Rinner refer to the 
inner and outer radii of the tube.  But I'm pretty lazy and, rather than 
do all this multiplication in a case where we don't need perfect accuracy, 
I use the fact that for a very thin-walled tube, the area is almost 
exactly equal to the circumference (Pi X Diameter X Wall Thickness), and 
for rough purposes this is just about 3 X 0.5 cm X 0.05 cm which we can do 
in our head to determine that the cross sectional area of the tube is 
about 0.075 square cm.  [Just for fun, I did the exact calculation, using 
0.5 cm as the outside diameter (which is twice the radius) of the tube and 
got the following:  3.14159 X(.50/2 X .50/2 -.45/2 X .45/2) = 0.0746 
square cm, which is very close to the answer using the much easier way of 
estimating.]  Multiplying this by the length of the tube to get the volume 
we find that this is 0.075 X 1.8 = 0.135 cubic centimeters -- and this is 
just for the cylindrical section.

Assuming that the ends of the tube are flat and of diameter 0.5 cm, the 
area of each end is calculated as -- using rough calculations again, since 
we don't have to be exact -- just Pi X Radius X Radius ~ 3 X .25 X .25 ~ 
0.4 square centimeters.  Since the ends are 0.05 cm thick, the volume of 
each is 0.4 X 0.05 = 0.02 cubic centimeters.  So the total volume of both 
ends is 0.04 cubic centimeters.

Adding the volumes of the tube and the ends we get 0.175 cubic 
centimeters -- and since we're making rough estimates and I'm very lazy 
and prefer to deal with calculations I can do in my head, we'll estimate 
this as 0.2 cubic centimeters.

If we take the 2 joules of laser energy we guessed above and put it into 
the capsule, the total energy deposition density is 2 Joules / 0.2 cubic 
centimeters or 10 joules per cubic centimeter...  Comparing this with the 
10,000 joules needed to turn something instantly into a flaming ball of 
plasma, it's pretty clear that we're a little shy in laser energy to 
create an explosion.

But what if you REALLY meant that the laser would be powerful enough to 
cause these light burns over your entire body -- and we'll consider my 
body to be a thin pancake that's 2 m high and 0.5m wide (I'm not 
particularly thin, but I am about that high and wide).  My surface area 
would then be 2 square meters and if I had 2 Joules per square centimeter 
distributed over my entire body, that would mean that our laser had 2 X 2 
X 10,000 Joules of total output, or 40,000 joules total.  If we stuff that 
through the small hole in the capsule, we'd be putting 40,000 joules onto 
a mass of .2 cubic centimeters -- so we'd have 200,000 joules of laser 
energy on a target that's just 0.2 cubic centimeters, resulting in an 
energy density of 1,000,000 joules per cubic centimeter.  This is well 
above our explosion limit, so the capsule would be expected to burst 
instantly into a plasma fireball and disassemble.

You may be concerned that I ignored the reflecting surface on the inside 
of the capsule.  It turns out that any reflector, even a pretty good one, 
leaks a little light through -- maybe a tenth of a percent or so of the 
enregy that hits it.  And any capsule or reflector material will absorb a 
little bit -- glass will absorb a tenth of a percent or so per centimeter 
of thickness, steel will absorb essentially all of it in a fraction of a 
centimeter.  Also, slight imperfections in the coatings and/or materials 
always exist which absorb a little more than the nominal values.  All of 
these factors, coupled with the hideously huge energy density of a MILLION 
JOULES PER CUBIC CENTIMETER would invariably cause a teeny localized area 
to be vaporized into a small plasma by part of the beam.  The vaporized 
volume -- since it's a gas of ions and electrons -- would absorb the beam 
much more heavily and propagate damage to adjacent areas of the tube, 
which would then vaporize and absorb more, and so on until the tube 
disassembles very rapidly.  

The reflector on the inside of the tube will help to keep the pulsed beam 
from leaking out of the tube.  This would probably allow the input pulse 
energy to be decreased somewhat -- it's hard to calculate exactly how 
much, but I'd guess that the energy needed to cause the explosion would be 
reduced by somewhere between a factor of 10 and 100, to something like 400-
4000 Joules.  This is pretty close to my personal experience, in which I 
put a big piece of exposed polaroid film in front of the world's biggest 
laser (at the time) and hit it with about 300 Joules of output energy.  
The paper didn't explode, exactly, but the top layers of the paper (maybe 
half the thickness of the capsule you postulate) were blasted off the 
surface in a fireball jet that deposited light ash over most of the laser 
lab.  In this case, the surface the beam hit was black (so that most of 
the beam was absorbed without reflection) but I hit it with so much lower 
beam energy that the situation is probably a reasonable, but rough, analog.

Last, but not least, comes the issue of whether any of this is practical 
or important...  From the standpoint of blowing up cylinders, it's pretty 
clear from the above that a lot of laser energy is needed -- of order 
thousands of Joule.  Since lasers nowadays cost something like $100,000 
for a 0.5 Joule lab laser, it's pretty apparent that things are likely to 
get pretty expensive pretty quickly.  Just the laser would cost a ton of 
money, and that doesn't consider anything that you might have to add to 
point the laser at what you wanted to shoot it at or into -- it takes a 
pretty good optical system to repeated shoot into a tiny hole, and the 
optics can cost as much or more than the laser.

You may also wonder why I keep talking about a pulsed laser.  It turns out 
that if you tried to achieve the same function with a laser that was not 
pulsed, but put out continuous power over a longer period of time, things 
would change.  If you used a low power laser and exposed the capsule for a 
long period of time (in this sense, any period greater than a few tens of 
microseconds probably qualifies as a long period of time) to deposit the 
energy slowly, a variety of things could happen.  If the time was long 
enough, the tube might just heat up as the heat absorbed was carried away 
by radiation or convection.  If the time was a little shorter, the tube 
might melt or burn.  Or, if the time wasn't too short, it could still 
explode.  What actually occurs depends on the laser power, the time, the 
hole size, the capsule size and material, and the reflectivity and surface 
quality of the inside of the capsule.

In other words, it all depends...