Drider_it

Ok was tossing this around at work

now keep in mind lol this was a joke that got on a tangent and ended up with a thought behind it

we use a rotating laser on our jobs doin concrete.. sits on top of a tripod useing a reciever on a 6 foot metal stick for measurin..

we got to thinking..

if you had a laser with a semi powerful beam.. that shot into a .. say for lack of better words..

a round container of undefigned size.. in the middle is a crystal with 50 facets each facet is off-centered to a mirror that will reflect the beam back to another facet .. thus each facet hits each mirror in order.. now say that it goes back onto number 1 facet and begins the process over again.

in doing so.. could you shut the light off and the beam would keep going with out a source?

i havent found any info on a subject like this yet for my time is limited so was wondering the same.

i mean light is there right.. is constant but fades.. but if it is refected constantly would it fade.. the obvious answer is yes.. but still seems there is a way to keep is going with out the original laser source..


if it does fade.. this come to the second part.. could this set up intensify the beam in any way? say tweak the facets so each one is stronger magnfication then the last then at the end it the last facet would hit the last mirror and instead of hitting the number one facet.. instead shoots out of a predetermined hole?

we read alot of scifi books and play several scifi tabletop rpgs..

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fckm

the light would fade, as mirrors are not perfectly reflective and the medium that the light is traveling in will also absorb energy. There are dielectric structures called photonic crystals which can theoretically have a 100% reflectivity, and you'd have to run everything in a vaccum so you don't get absorption in your medium. But, you still have to inject the laser beam somehow, which means you have to have a hole and light will eventually leak out.

Intensity is power/area, so to intensify a laser, you can simply pass it through a single lens, it's not really necessary to use multiple lenses. (although the intensity is only increase at the focal point) Keep in mind that increasing the intensity doesn't mean that you have necesarily increased the power of the beam. Using a lens simply means that the spot size has been decreased.

You're probably thinking of amplification. That is, a way of putting more power/energy into the beam itself. This can be achieved using an optical amplifier. Basically, you make an optical cavity and fill it with a material that you can pump to a high energy level such that the transition energy corresponds to your lasing frequency. But this is aways an active process, it consumes energy. There's no way to maintain the power of the light inside your container unless you add energy into it.

Basically, the object that you're thinking of is a microwave. The sides of a microwave are metallic and are pretty highly reflective at microwave frequencies. One side as a portion cut out so that you can pump more microwaves into the cavity. Of course, there is still significant loss, especially if you're cooking something

I suggest you take a look at howstuffworks.com, they should have lots of info on lasers, optical amplifiers, and microwaves

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The short answer is no, because of attenuation

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shakran

but there are things called light traps which do this - not forever, but for quite a long time.

Slavakion

Yeah, there's no way to keep the light reflecting forever. But that concept is what's used in "light clocks" in theoretical physics to explain relativity. But that's getting off topic

Sapper

Your premise would be true in what I like to call "physics land" where you can define "annoying variables".

If there were a perfect vacuum and the reflective surfaces were 100% efficient, then your premise is correct - except for the "magnification" part. I'll briefly explain the magnification: this notion is derived from classic geometrical optics where you can use refraction to increase or decrease the size of an image that a lens subtends. This means using the fact that light traveling through media of different refractive indicies (things in which light travels at different speeds), you can cause the rays to "bend" and thus an object appears larger.

As noted, no reflective surface is even near 100% This goes as far as atomic structure - as a photon hits the reflective surface, some of the energy is lost in the collision with an atomic body (electron, nucleus) and according to the law of conservation of energy, some energy is lost from the photon. This means that as the photon is reflected from the atom, it has lost some of its energy. Loss of energy continues to infinity whereby all energy is converted from light to heat.

Cool.

It is possible to surmise that if you were to shine blue light into your theoretical apparatus, with an efficiency of "x" you would be able to watch the light turn from blue to red and onward to the lower frequencies. Observing the light of course would introduce an exponential decay in efficiency.

Wow - come to think of it .. this sounds a lot like Schrodinger's cat.

Oh god, I think I've gone cross-eyed.

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tehhappyboy

Bringing moon beams home in a jar.....


(remember, light is really fast, around the earth 8 times in a second (provided it could curve), so even if it was almost perfectly reflective, the light would degrade too quickly. No, no perfect mirrors. Yet.)

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fckm

Sapper

not quite. Photons don't lose energy when they interact with matter. A photon is a quantization of the electromagnetic field. When light interacts with matter, a photon is either absorbed, or it is not. A photon cannot lose some portion of energy to an atom, it must be either absorbed fully, or not at all. The energy of a photon is completely determined by it's frequency, so E=plank's constant*nu.

That said, it is possible for the atom to remit a photon of different energy/frequency after it absorbes a photon of one frequency.

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Sapper

Fckm.

Well, as you said, a photon is a quanta of EM energy. When light interacts with matter, photons certainly can lose a portion of their energy.

Plank's equation E = H * nu only tells you how much energy a certain frequency (nu) has - nothing about conservation of energy. Light hits an object, the frequency of reflected light is slightly lower than the incident light. The question is asked, why?

As you dilligently mention, an atom will fully adsorb the photon and re-emit the photon with a lower frequency (and hence less energy). This means that the atom which emits the new photon does so with greater energy (in terms of the atom) and the photon which is emitted has less energy. Conservation of energy is observed - that is until you consider quantum mechanics (which we rightfully should). The question then is what can we measure?

I didn't want to over complicate my initial response (as I often do) and as such did not explain the adsorption and emission of photons to save time and effort. Only someone with some degree of background in chemistry or physics (such as in your case) would be able to follow such a conversation anyways.

What I find more intriguing though is that all moving objects exhibit a wave-like motion. This was derived by Schrodinger and is known as the Schrodinger wave equation:

lamda (wavelength) = h (Planck's constant) / m (mass) * v (velocity)

So anybody on this board could calculate the theoretical wavelength of themselves walking. All you need to know aside from your velocity and mass is Planck's constant h = 6.626e-34 kg m^2 s^-1

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