Re: Michelson Interferometer and conservation of energy
- From: "Timo A. Nieminen" <timo@xxxxxxxxxxxxxxxxx>
- Date: Wed, 7 Feb 2007 05:15:54 +1000
On Wed, 6 Feb 2007, Mapsread wrote:
On Feb 6, 10:24 am, "PD" <TheDraperFam...@xxxxxxxxx> wrote:[cut]On Feb 6, 9:10 am, "Mapsread" <cw_...@xxxxxxxxx> wrote:
If you move the movable mirror to just the correct location, it seems
that the light interferes destructively and "disappears". This appears
to lead to a loss violating the conservation of energy. Searching
around the web some shows that is not the case and energy is
conserved. I'm trying to understand what principle is at work in
conserving energy here.
To me it sounds like they're glossing over some pretty heavy duty
physics with the statement "because there is a re-distribution of
energy at the detector...." Can somebody clarify this statement or
point me to a good article on this subject? Is there some quantum
activity at work here?
As far as energy conservation is concerned, rest assured that for
every place where the waves destructive interfere, there is another
place where the waves constructively interfere, and that's where the
energy goes.
Thank you for your kind response. I guess I'm trying to understand
your statement of "where the waves constructively interfere". I
believe you, but where? From the literature I've found, it appears
that all the light heads back to the source. Why? Why doesn't head off
towards Jupiter? I don't see what principle is at work directing the
light back to the source. Also, I cannot seem to draw a picture to
show where the waves might be constructively interfering in the
particular case I point out. That is okay if quantum mechanics is
involved because the picture can't possibIy capture the quantum
nuances. But this is my problem: If quantum mechanics is causing this
behavior, what principle is at work and why is it sending it back to
the source? If it's the far simpler case of the amplitudes of sine
waves (light waves) interfering, what does the picture look like with
a single monochromatic light beam of say, 3 wavelengths? I cannot draw
it. Thanks again!
You don't need any quantum stuff for this. The exact details will depend on the exact setup, but the basic idea can be seen in the simplest cases.
Take a single interface between 2 materials, say air and glass. Have two beams of light incident on the same spot on the interface so that the transmitted portion of each overlaps the reflected portion of the other. Depending on the phase difference between the two, you'll get constructive and destructive interference in the two outgoing beams.
Set the phase difference so that you get maximum destructive interference in the outgoing beam in the glass - the two incident beams are 1/2-wave out of phase. What happened to the energy? Look at the other outgoing beam, and there it is. Why are the two beams in phase here, and out of phase in the other? The beam incident from the air undergoes a 1/2-wave phase shift on reflection.
Usually, you will see a 1/2 wave phase shift on reflection when a beam goes from a "fast" medium to a "slow" medium, and no phase shift on reflection from slow to fast, and no phase shift on transmission, for any waves.
--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
E-prints: http://eprint.uq.edu.au/view/person/Nieminen,_Timo_A..html
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html
.
- References:
- Michelson Interferometer and conservation of energy
- From: Mapsread
- Re: Michelson Interferometer and conservation of energy
- From: PD
- Re: Michelson Interferometer and conservation of energy
- From: Mapsread
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