Any one here who loves Electromagnetic Theory?

[mixedgas]

I've got a tough one.

Two planar light waves, one at a wavelength of 2.1 microns and one at a wavelength of 600 nanometers are launched at a target. The oscillators generating in the waves are free running lasers.

What percentage of the time will the postive peaks of the two light levels hit a surface with the tops of the peaks within a 1 nanometer coincidence window of each other? Within a 10 nanometer window?

I know its a tiny number.

Any one feel like some math? I have a guesstimate I'd like to verify.

Steve

[logsquared]

I probably don't understand what you mean... but what the hell .00119% for 1 nm window

[steve-o]

My answer is "42"

[mixedgas]

I probably don't understand what you mean... but what the hell .00119% for 1 nm window

I have a client who wants to develop a non-linear effect in a thin film that only occurs when the wave peaks of the two femtosecond lasers hit the target at the same time. The lasers run at 1 Khz, and I dont have control over the relative phases, so I just want a idea of how horribly inefficient the process will be.

This does not mean when the pulses hit. I mean to determine how often th peeks line up.

Forget that the lasers are pulsed.

Just assume they are continious wave.

If you have a 600 Hz sine wave on channel A on a dual trace scope, and a 2200 hertz sine wave on channel B, every once in while the wave crests will line up. The question is how often does that happen, mathematically, assuming the scope is perfect and not sampling.

Steve

[steve-o]

600/2200=.272 or 1/x =3.66 s ? i dunno , I'm home and it's toddy-time and why am i doing this ? Am I close ? Did i win ? ..

[planters]

I'll bite. If you assume that the 2100nm wave passes through a plane that is 1nm thick, the "peak" of the wave will spend 1/2,100 of it's time within the plane. The peak of the other wave will similarly will spend 1/600 of it's time within this plane. If there is no coupling then I would guess that the random coincidence of these waves will be 1/2,100 x 1/600 = 1/1,260,000. You might quadruple this if you allow that the negative peaks will also lead to positive interference. I think that this is similar to the principle of two photon fluorescence where two long wavelength photons are simultaneously ( or nearly so ) present at one location to additively stimulate a molecule to fluoresce that otherwise would not respond to such long wavelengths. The cross section for this process is also very low and that is why such high intensities are sought.

[dnar]

I'll bite. If you assume that the 2100nm wave passes through a plane that is 1nm thick, the "peak" of the wave will spend 1/2,100 of it's time within the plane. The peak of the other wave will similarly will spend 1/600 of it's time within this plane. If there is no coupling then I would guess that the random coincidence of these waves will be 1/2,100 x 1/600 = 1/1,260,000. You might quadruple this if you allow that the negative peaks will also lead to positive interference. I think that this is similar to the principle of two photon fluorescence where two long wavelength photons are simultaneously ( or nearly so ) present at one location to additively stimulate a molecule to fluoresce that otherwise would not respond to such long wavelengths. The cross section for this process is also very low and that is why such high intensities are sought.

That is not how I understood the question.

Now you have me confused, I need a drawing of the wave, the surface and the coincidence window. Not to scale is fine

[-bart-]

I don't think it is as bad as planters predicts. I think it all boils down to the longitudinal mode spacing of the two lasers.
Some modes will have the desired "summing"condition relatively all of the time, some modes hardly ever.

For the condition :
sin(t*2pi/2100.xxx) + sin(t*2pi/600.zzz) > 1,999

In my layman's excel simulation roughly 25nm out of 65000nm will meet this condition.

Someone more experienced in methlab can surely empirically simulate all modes and plot them against the coincidence chance.
(Not me, today, thank you)

[planters]

I too may have misunderstood the initial set up, but I had assumed the simplest scenario as in single mode.

The waves had exited each of the two laser cavities and were propagating in free space and converging at a small enough angle (or from opposite sides) that they overlap within the 1nm plane.

The effect of increasing the number of longitudinal modes would be to proportionally increase the fraction of time the two beams' "peaks " interacted, but to simultaneously and proportionally decrease the energy in each interaction. Nevertheless, I think it is reasonable to assume that the overlap , when it occurs, will be random-the modes will not be coupled.
A lot of simplifications are being made. We don't know the dimensions of the lasers, the medium, the bandwidth and hence even the maximum allowable number of modes. We also don't know the application;as in the possibility of beam interaction at anything other than the precisely defined phase given in the set up. Also, the initial femtosecond at 1Khz would be a 10 to the minus 12 duty cycle which itself would create a very low interaction cross section.

[mixedgas]

I've talked with some world class experts on this. No one has given me a straight forward rate equation, and guestimates are all over the place. Nor can I find a rate equation in papers on the subject. PLs educated guesses jive very well with the experts.

Sources are roughly 8 nanometer wide, 1 Khz rate, ~60 fS pulses. Each optic they bounce off of, lengthens the pulse, and can change if the red, or the blue part of the pulse gets there first.
Going through glass lenses results in various nasty types of temporal dispersion and mode changes across the face of the pulse with respect to time.

Interaction angle with the target changes things drastically as well.

What I intend to do to ensure maximum interaction rate, I cannot discuss, but you may assume its active.


Thanks Guys,

Steve

[Doc]

The way understand beat frequency is: The beat frequency will occur at the mean of the difference of the two superimposed frequencies.

[planters]

Wouldn't it just be the difference? If one wave is at 1200Hz and is interfering with a wave at 1201Hz then you should hear one extra beat every second or in other words one extra beat every 1200 waves of the first note.

[-bart-]

I guess beatfrequency is not the tool to describe the effect that Steve wants. The peaks have to line up pretty exactly.
Theoretical speaking : 2100nm stands to 600nm as 7:2. So, if the wavelengths of the two lasers are exactly that, coupled, and in phase, they line up every 4100nm. However when a phase-shift occurs the peaks never line up

What happens if the wavelengths are 599,99nm and 1100,17nm, two primes ?

[planters]

I have not yet been able to access any related papers, but my guess is optical switching/modulation. Anyway, the sparseness of these pulsed beams will probably be a significant if not the overwhelming factor controlling interaction frequency. Beat frequency becomes interesting as the difference between two frequencies is small relative to the frequencies themselves. As I understand it the two frequencies don't have to have a common divisor and so prime should not be relevant, only the difference. So, in your example take 59,999nm and 110,017nm convert to frequency and subtract. Because the lasers are not coupled they will only interact randomly, but the average frequency (of this interaction) depends on the parameters of this interaction as in the 1 nm slice in time (3x 10 minus 18 sec) and the bandwidth ( 8nm and stretching through out the optics) which broadens the definition of the "peak" and hence the 1nm to at best 8nm and likely more. This is an interesting thought experiment. I don't think I'd relish trying to determine a good answer.

[mixedgas]

Where it gets worse is the process is angle sensitive. My reason for asking the question was to figure out how long I need to dwell at a given angle to see any possible phase matching.

Steve