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In a laser interferometry experiment, we project a pattern of interference fringes onto a CCD sensor. For best results, we want good contrast between the bright and dark fringes, and we carefully compensate for various sources of noise - for example, by taking camera images with no fringes present, and with the laser turned off, and subtracting these images in the proper sequence. We'd expect therefore that the remaining signal should be highly linear, with the CCD signal at each pixel in direct proportion to the number of photons reaching it during the shutter time.

What we actually find is that, as we vary the laser intensity and shutter time such that the average intensity across the image remains constant, with no pixels saturated, there is a definite "sweet spot" where the fringes are much more well-defined than at other settings. Either increasing or decreasing the laser intensity away from this point (with corresponding decreases or increases in shutter time) causes the fringe definition to deteriorate.

I can't think of any reason why this should be. I know that in cases where the process generating the pattern has a time-constant (for example, using laser speckle interferometry to measure Brownian motion), there is an optimum exposure setting, but that shouldn't be the case in our system which is entirely static. So, what am I missing? I assume it's some property of the CCD sensor that I've overlooked.

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    $\begingroup$ Is it possible that there is shot noise at low intensity? Are you changing the laser intensity directly (which would also change its stability) or are you using ND filters to adjust the intensity in this test? Does the fringe amplitude change or is it just noisier? $\endgroup$
    – Steve Byrnes
    Commented Jul 20, 2012 at 0:36
  • $\begingroup$ We're using a stabilised fibre-coupled laser and exposure times in the range 80-200ms, so we don't think shot noise is significant. The general noise level seems low (the black areas outside the beam area really are black), and there's no saturation. $\endgroup$
    – Eos Pengwern
    Commented Jul 20, 2012 at 6:54

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The CCD is not perfectly linear, and the electronics in the analog pathway are not perfectly linear either. You should map the response of the system over your experimental range before you diagnose further, i.e. sweep a known calibrated source and measure and plot the pixel response. Especially since you are forming data by subtracting, small non-linearities will be much more visible.

Usually you either find a region which is linear enough for the purpose, and you make sure you're always in that region, or you fit a curve to the response and use that to linearize any data before analysis.

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  • $\begingroup$ The curious thing is that we're satisfied with the linearity of the 'peaks' in the fringe pattern, but puzzled by the behaviour in the 'troughs'. If the peaks were saturating then we'd expect to see the troughs fill up so some extent as a result of blooming, but there's no saturation here so we can't think why else the peaks would appear to spread out so much. $\endgroup$
    – Eos Pengwern
    Commented Jul 20, 2012 at 7:03
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Without knowing a lot of details about your particular set-up, it's impossible to give an exact answer. It can depend on your laser, the exact CCD you are using and how it is being read, your optics, etc. In general though, I can think of a few things that could cause this.

  • When you vary intensity while compensating with exposure, you are affecting the balance of dark current noise, read noise, and signal noise. Read noise is roughly constant, so at a short enough exposure it becomes a more significant fraction of the total. With really long exposures you will have more dark current, and thus more dark current noise. If your detector is linear, then your signal noise will be roughly constant, since you would be adjusting the intensity to deliver the same number of photons in each case; but if the detector isn't linear w.r.t. exposure time, then your signal noise can change as well.

  • Detectors are not necessarily linear versus intensity, nor versus exposure time (although good ones are).

  • The way you are adjusting intensity can have an impact. If you're not careful, you can introduce unexpected reductions in the polarization or polarization purity.

  • If there is stray light in the room, then it becomes more significant when you dim your beam.

  • I've seen cheap or old laser produce "burps" of incoherent florescence light.

  • Vibration becomes more of an issue over long exposures. Even if you can't see it, it will reduce your measured fringe visibility surprisingly fast.

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  • $\begingroup$ This all assumes, by the way, that you've checked the obvious things, like saturation in the CCD or the ADC. If it's a cheap camera, the ADC can be another source of nonlinearity. $\endgroup$
    – Colin K
    Commented Jul 20, 2012 at 3:10
  • $\begingroup$ We're using an industrial camera with a Sony ICX655 CCD sensor, and the camera cost a couple of thousand Euros so is reasonably well-specified. There's no saturation (since we aim at a maximum intensity of about 90% of the camera's range) and, with the background corrections, the areas outside the laser beam are properly dark (<<1% of the camera's range). $\endgroup$
    – Eos Pengwern
    Commented Jul 20, 2012 at 7:01
  • $\begingroup$ Be very careful when you claim that there is no saturation. If you turn the analog gain way down, you can saturate the electron well long before you saturate the ADC. This is insidious if you are unprepared, especially if the CCD has anti-bloom. It will not be a sharp, obvious effect like ADC saturation; it will be a sudden, and but smooth drop off in sensitivity. $\endgroup$
    – Colin K
    Commented Jul 20, 2012 at 17:03
  • $\begingroup$ Normally I run the camera with gain set to 0dB, but I tried a sequence of measurements with gain at different values up to 9dB and it didn't seem to make any difference at all - my auto-exposure algorithm corrected for the different values of the gain but the contrast levels were indistinguishable. $\endgroup$
    – Eos Pengwern
    Commented Jul 21, 2012 at 14:58

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