Method for determining the difference between two similar optical wavelengths

[lichen]

Hi optical engineers/physicists,

I'm designing an optical sensor system, and it would be very beneficial if there was a method for determining the difference between two very close (down to tens of femtometers) optical wavelengths.

The difference measured can be relative, although absolute would be preferred. The wavelengths will be around the 1550nm area.

I thought there would be accurate interferometric methods already in place for this, but as far as I can see the only method used is to measure the beat wavelength created from adding the signals (completely impractical in this case, where the beat wavelength would be of the order of metres).

Am I missing an obvious or common technique? The question sounds simple, but I cannot seem to find a simple solution at all.

Any advice is greatly appreciated of course.

Thanks,

LichenEdit:

One way of course is to measure the wavelengths separately and then subtract electronically, but the system is then subject to twice the measurement error. I would rather the differential measurement is performed optically and only then is photodetection performed.

 

[Redbelly98]

You wouldn't measure the beat wavelength, rather the beat frequency.

Let's say the difference is Δλ=15 fm for λ=1500nm light, for the sake of argument.

Δλ / λ = Δf / f

so

Δf = f Δλ / λ = c Δλ / λ²
= (3x10⁸ m/s) (15x10^-15 m) / (1.5x10^-6 m)²
= 2 MHz

There are photodiodes fast enough to measure GHz frequencies, so this would not be difficult.

 

[naresh]

Just to add to Redbelly98's reply, a photodetector is all you need to generate the beat frequency. Assuming you align the two beams and shine them on a detector, the photocurrent is proportional to the incident optical power, i.e the square of the amplitude of the electric field. This automatically gives you a mixing term that generates the beat frequency.

 

[lichen]

Thank you both very much.

I will try and summarise how I plan to use your advice. Hopefully this makes good sense.

This will be a fiber optical system, with the two wavelengths on separate fibers. I will couple the signals into a single fiber, and then into the photodetector. The photodetector output will have a frequency equal to the beat frequency of the optical signals. I can then analyse the photodetector/amplifier output to determine the electrical frequency, and from this frequency derive ∆λ.

I hope that I have a correct understanding of the method.

Thanks again,

Lichen

 

[Redbelly98]

I'm not familiar enough with fibers to know about combining two into one, but I'll leave that up to you.

I'm guessing you'll need a polarizer between the fiber output and the photodetector, but I could be wrong.

 

[lichen]

Combining fiber signals is trivial enough. I can't see why a polarizer would be required prior to photodetection.

In fact, the two signals will be polarized on orthogonal axes of a single highly birefringent fiber, so taking this fiber straight to the photodiode should be sufficient for the detector to sum the signals on both axes and produce the beat frequency.

 

[lichen]

I'd like to ask another question if that's ok.

Regarding the beat frequency created by two wavelengths: What if the two optical signals were not one single perfect wavelength each, but were instead comprised of certain known linewidths. Would this destroy the beat frequency detection method? In effect, I would want to detect the difference between the two peak wavelengths.

I'm worried that this may undermine the idea.

Edit: Would this simply produce a distribution in the electrical frequency domain, with peak corresponding to the beat frequency of the peak optical wavelengths?

 

[Redbelly98]

Edit: Would this simply produce a distribution in the electrical frequency domain, with peak corresponding to the beat frequency of the peak optical wavelengths?

Yes, exactly.

Combining fiber signals is trivial enough. I can't see why a polarizer would be required prior to photodetection.

In fact, the two signals will be polarized on orthogonal axes of a single highly birefringent fiber, so taking this fiber straight to the photodiode should be sufficient for the detector to sum the signals on both axes and produce the beat frequency.

The beam polarizations can't be orthogonal (coming out of the last fiber) if you want to get beating. Orthogonal polarizations do not interfere.

I was thinking the optical fibers would "scramble" the polarization, making for a randomly polarized output. I don't know if that would eliminate the beat signal or just reduce it. That's why I suggested the polarizer.

It sounds like you plan to use polarization-preserving fibers (which I've never used, only heard about). In that case then the polarizations will have to be the same.

A test you can do is to split one of the beams, and send it into the two input fibers. At the detector end you should then observe good fringes (through an IR viewer). The detector size should be smaller than the fringe spacing if you are to detect beat frequencies with the 2 beams.