I’m thinking of doing this experiment and I’d like to know if it’d work first. YouTube doesn’t seem to have any videos of this.
So if I had 2 parallel mirrors, I could angle the laser to get any path length of light I want. If the angle were close enough to perpendicular, could I see light traveling? Or would the laser fade away first?
No matter how many mirrors or what angles you use, the photons are still traveling at 186,200 miles per second. How do you expect to see them?
Also, they are only going to be visible if they scatter off of dust in the air, so the intensity is going to fade rapidly.
Oddly enough, this is how a laser beam is generated - two parallel mirrors and light, with a hole in one of the mirrors. How a Laser Works - YouTube gives a wonderful description of this. Start 3:20 for the mirror thing
Your question is complicated. I’m not sure what it is you expect to see if the light is perpendicular. Even if you line up the mirrors to be precisely parallel and the beam perfectly perpendicular, so there is no walkoff, it’ll still just look like a beam of light travelling through the open air.
And you can’t see light in the open air unless it bounces or scatters off something. Hence those people in movies walking through crossed laser beams, making them visible by spraying stuff into them. So if you want to make a laser beam visible, just pass it through a spray, or a cloud of disdyt, or smoke, or fog from a gof machine.
Lasers, by the way, don’t often use two perfectly parallel flat mirrors – it’s too easy for the beam to “walk off”. It’s more comon to have one or both mirrors concave, and to place them an appropriate distance apart, so that a beam that tries to “walk off” will be re-=directed back towards the axis. This is also true if you’re not building a laser resonator, but a “cell” to contain the beam for a number of bounces. Such devices are called White Cells and Herriott Cells, and the combinatioons of curvature and spacing are the same as for a laser resonator.
Incidentally, the number of bounces that a beam will make is on the order of R/(1-R), where R is the reflrectivity. If your mirrors reflect 99.99% of the light, that’s approximately 10,000 bounces. If your mirrors are about 4" apart (10 vcm), you can expect an infinitesimally fast burst of light to be absorbed or scattered in about 0.3 milliseconds.
I understand the OP’s question but I think there are a few problems with turning it into practices. If I position the mirrors 1 foot apart (in a vacuum) it will take 1 second for the laser bounce off them 1,084,163,568 times. If I make the angle such that the light travels 1 foot along the mirrors in those 1 billion plus bounces then I will have the appearance of light traveling 1 foot per second.
Here are the problems.
The angle I need to set the laser is arccos (1/1084163568) which is 89.9999999471521 degrees. Assume precision to 1/10000 of a degree. Then you are anywhere from perpendicular to the light “traveling” along the mirrors at almost 2000 feet per second.
But that is not the biggest problem. Using CalMeacham’s numbers, you are orders of magnitudes short of having the laser complete the trip and that’s not even accounting for the absorption of the light by dust/smoke so you can see it travel.
You think he wants to see the light rattling around between the mirrors? That’s trivially easy – use a continuous wave laser, which the kind you’re likely to buy, and just let it rattle back and forth between the mirrors. Even if you get a pulsed laser, the pulse will likely be so long that you’ll see a lot of bounces (making really short pulses is an art in itself – and an expensive one).
In the 1999 Mummy movie, they illuminate the inside of the mummy’s tomb by bouncing sunlight from one mirror to another. There was a distinct flow of the beam of light from one mirror to the next. Hollywood science at its worst.
My interpretation of what he’s asking is that if you put the mirrors almost, but not quiiiite, perpendicular, you could maybe see a spot of light slowly “walking” up the mirror from the initial impact site to the edge. I think the answer to that, as has been said, is that it would take far more bounces than any real material could produce.
And the way I designed it, you would see a 1 foot wide light traveling 1 ft/sec.
Let’s suppose your device is 1 meter wide and you want the pulse to travel at 10 m/s, so that the duration is 1/10 s–short but certainly visible to the naked eye.
Light goes at 300,000,000 m/s, and the very best mirrors can handle about 1000000 bounces. If your total path length is 3e8 m/s * 0.1 s = 3e7 m, then your device needs to be 3e7 m / 1e6 = 30 meters long.
So the dimensions aren’t completely implausible. Aligning the thing sounds hard, though. And you’d probably need the whole thing to be in a near-vacuum, although not a perfect vacuum as you’d want some stray particles so that the beam is visible.
[quote=“Dr.Strangelove, post:10, topic:627225”]
…and the very best mirrors can handle about 1000000 bounces…
I went to the cite/site, and I still don’t understand this statement. Help?
Mirrors are rated in terms of their reflectivity, which according to the link can be 99.9999% or better.
To figure out the amount of light remaining after a bunch of bounces, we take reflectivity[sup]n[/sup], where n is the number of bounces. In this case, we have 0.999999[sup]1000000[/sup] = 0.368 = 36.8% of the light remaining. That’s enough to still see the pulse easily at the end.
We can go the other way with logarithms. If we have a very powerful laser and decide 1% remaining is still visible, we can do this:
0.999999[sup]n[/sup] = 0.01
log 0.999999[sup]n[/sup] = log 0.01
n*log 0.999999 = log 0.01
n = log 0.01 / log 0.999999 = 4605168
So if we’re willing to have a bit more degradation, 4.6 million bounces are possible.
Also, just as a bit of trivia: the reason I immediately knew that 1000000 bounces was ok is because of this handy formula: (1 - 1/n)[sup]n[/sup] = 1/e for large n. 99.9999% is one part-per-million short of 1, and so taking that to the millionth power is going to be close to 37%. 37% is an arbitrary cutoff point, but in this case it’s clearly enough so still see the laser pulse.
Terrific. Thanks. Clearly we can’t have 100% reflectivity, like carrying around an unquenchable lantern. Actually, it would be useless as a lantern, because it would not allow emission; the perfect reflectivity is as untenable as the infinite energy required.
I remember somewhere here a thread on this.
The opposite situation, in this case allowable(?) is a Black Box–is this correct?
Black-body emitter is the more common name. Despite that, it’s not really black, even in principle. Put simply, if you imagine a box that absorbs all incident radiation, it must nevertheless eventually heat up enough that it emits light all its own. This outgoing light will be essentially uncorrelated with the incoming light (well, this is probably arguable in the presence of quantum mechanics), but in the long run the energy transfer will exactly balance out. So yeah, in principle you can build a surface which will destroy any image incident on it, but it will still leak photons.
Apophrycal story: I read an account of a star party once, and someone decided to shoot their flash through the base of one of the big telescopes. People standing nearby said that the flash looked like a “photon torpedo” as it shot into the sky.
Yes, I think that’s what I’m trying to do. So, the problems are:
Mirror losses
Extremely small angle needed
Hey but if I just shone the beam straight, how far would it go before I couldn’t see it from the side?
Are you talking about this? http://www.youtube.com/watch?v=FtbEptrDpdg&feature=youtube_gdata_player
It’s something like what I’m trying to do, but I want the angle almost 90 and to be able to see the laser progress between the mirrors.
If I start with a 100 mW laser pointer, I think 5 mW would still be visible. Don’t think Ikea mirrors are 6N (6 nines) efficient though 
Yeah, definitely not.
The other problem is Rayleigh scattering. Let’s suppose we have a dust-free volume, but we need to know what level vacuum to pump to.
At 532 nm, the Rayleigh scattering cross-section for nitrogen is 5.1e-31 m^2. Suppose that we want the scattering losses to approximately equal the bounce losses (and ignore oxygen and other gases). That means we have:
5.1e-31 m^2 * [scattering cross-section]
2.686 7774e25 1/m^3 * [Loschmidt constant @ 1 atm]
p * [pressure in atmospheres]
30 m = [distance for one bounce]
0.000001 [proportion scattered for that distance]
So p = 0.0024 atm. Definitely a vacuum but not a hard one at all, and well within the range of existing equipment. In fact, you could put the entire experiment in this guy.