B Does Destructive Interference Affect Starlight in a Double Slit Experiment?

[collinsmark]

@collinsmark can explain better than me but basically, [...]

@Devin-M, Sorry, I haven't been following this thread until now. But yes, I agree that ideally, if things are kept simple, I would expect the total ADU count of the double-slit should be pretty close to twice that of the single slit, all else being the same, if you were to add up all the (red) pixel values across a single line in each pattern. That's of course after dark frame subtraction. It's also assuming that your Nikon D800 isn't trying to modify the results in-camera.

Here's a few tips to help make sure your camera is not inadvertently messing up the results:

• If possible, store the images in RAW (NEF) format. When converting the Raw (NEF) file to some easily readable format, don't apply a white-balance, or if you have to, apply a neutral white balance (e.g., direct sunlight).
• If your workflow won't work with storing the data in RAW (NEF) format, then store the files in TIFF format. But don't store data as JPEGs.
• Do not use "Auto" white balance. Set it to something like "Direct sunlight," but the important thing is to make sure it's not set to auto.
• The Nikon D800 has selectable "Image Enhancements." Make sure your "Set Picture Control" option is set to "Neutral." See page 163 of your Nikon D800 manual for details.
• The Nikon D800 supports something called "Active D-Lighting." Make sure this is turned off. See page 174 of the manual for details.
• The Nikon D800 supports something called "Auto ISO Sensitivity Control." Make sure this is turned off. See page 111 of the manual for details.
• For that matter, try to make sure nothing is "auto," if you can think of anything else not mentioned here. Everything should be set to manual.

There's a lot of variables in what you're trying to do. A lot of stuff can go wrong, causing misleading results. So the above advice might serve as just a start.

 

[Devin-M]

• If possible, store the images in RAW (NEF) format. When converting the Raw (NEF) file to some easily readable format, don't apply a white-balance, or if you have to, apply a neutral white balance (e.g., direct sunlight).
• If your workflow won't work with storing the data in RAW (NEF) format, then store the files in TIFF format. But don't store data as JPEGs.
• Do not use "Auto" white balance. Set it to something like "Direct sunlight," but the important thing is to make sure it's not set to auto.
• The Nikon D800 has selectable "Image Enhancements." Make sure your "Set Picture Control" option is set to "Neutral." See page 163 of your Nikon D800 manual for details.
• The Nikon D800 supports something called "Active D-Lighting." Make sure this is turned off. See page 174 of the manual for details.
• The Nikon D800 supports something called "Auto ISO Sensitivity Control." Make sure this is turned off. See page 111 of the manual for details.
• For that matter, try to make sure nothing is "auto," if you can think of anything else not mentioned here. Everything should be set to manual.

I shot in RAW NEF format, white balance while shooting was set to sunlight, picture control was on standard instead of neutral (i'll retest), active d-lighting was off, auto ISO was off, everything was manual, all in camera noise reduction was turned off, RAWs were converted to 16 bit TIFs w/ all sharpening/noise reduction off and without any other modifications.

PS… according to this, picture control doesn’t affect the RAW files, only the JPGs or raw file conversion if you use the Nikon raw conversion software, which I didn’t do… I used adobe lightroom: https://www.dpreview.com/forums/thread/3600835

 

[PeroK]

I was expecting if I measure the # of photons N through each slit separately for X amount of time, and they are both the same quantity, that if I measure through both slits simultaneously for X time, that the # of measured photons would be 2N. But based on the previous unexpected results, when I do that test I now expect 1.22N, but I can’t explain the discrepency.

A classical probability argument would agree with you. Let's assume that with one slit open, the probability that a photon passes though the slit is $p$. With two slits open the probability must be $2p$. This probability translates to total intensity on the detector screen.

With that explanation, however, the double-slit experiment with one photon at a time (or one electron at a time) would yield no interference.

The QM explanation for double-slit interference calculates probabilities (hence intensities) as the square of intermediate probability amplitudes. In particular, the classical proposition regarding double the probability/number of photons fails in the case of the double-slit. The number of photons that are transmitted by a double-slit is not necessarily twice the number transmitted by each slit in isolation.

That is, in a way, the essence of QM in a slightly unfamiliar context.

The aperture formula, I suggest, is applicable only where the apertures are large enough and far enough apart for QM effects to be negligible.

 

[PeroK]

@Devin-M what we are talking about here is the transmission coefficient for the double-slit relative to twice the transmission coefficient for the single slit. Your data might be quite instructive in this respect. You might try researching this online. E.g.

https://www.nature.com/articles/s41598-020-76512-5

The answer to your original question is that in the double-slit experiment almost as much light gets absorbed by the barrier as for a single slit!

 

[Devin-M]

The non-parallel slits seem to have a higher transmission coefficient than the parallel slits in the fringes, because with the non-parallel slits you have twice the area of fringes that have the same intensity as a single slit, but with the parallel slits, you have the same area of fringes as a single slit, and those fringes have less than twice the transmission coefficient of a single individual slit.

 

[Devin-M]

The usual explanation for the dark minima in the case of Fraunhofer diffraction is that the path length difference between the edges of the slits differs by half a wavelength or an odd multiple of a half wavelength path difference at the minima location, so the arriving waves are out of phase at the detection screen at the locations of these minima.

Transmission coefficient on the other hand implies that the energy which doesn’t arrive at the detector (the 25% deficit per slit in the 2 slit case) is somehow blocked by the slit.

Is one of these interpretations more valid than the other? It’s hard to understand how opening a second slit would decrease the energy transmitted via the other slit.

“ When the two waves are in phase, i.e. the path difference is equal to an integral number of wavelengths, the summed amplitude, and therefore the summed intensity is maximal, and when they are in anti-phase, i.e. the path difference is equal to half a wavelength, one and a half wavelengths, etc., then the two waves cancel, and the summed intensity is zero. This effect is known as interference.”

https://en.m.wikipedia.org/wiki/Fraunhofer_diffraction

 

[PeroK]

The usual explanation for the dark minima in the case of Fraunhofer diffraction is that the path length difference between the edges of the slits differs by half a wavelength or an odd multiple of a half wavelength path difference at the minima location, so the arriving waves are out of phase at the detection screen at the locations of these minima.

Transmission coefficient on the other hand implies that the energy which doesn’t arrive at the detector (the deficit per slit in the 2 slit case) is somehow blocked by the slit.

Is one of these interpretations more valid than the other? It’s hard to understand how opening a second slit would decrease the energy transmitted via the other slit.

They are two different aspects of the experiment. The diffraction pattern is the pattern. But, the intensity also varies with slit width and distance apart. If you do the double-slit experiment with very narrow slits, then very little light gets through.

The additional factor affecting the intensity is how the light interacts with a double-slit as opposed to a single slit. Classically, this would be simple. The double-slit would allow twice radiation through. But, in QM it's not so simple and it appears that much less than twice the light gets through.

 

[Devin-M]

So in QM, the explanation for the minima isn’t that 2 waves arrive out of phase and cancel out, because of the half wavelength path length difference between the slit edges and minima location?

 

[PeroK]

So in QM, the explanation for the minima isn’t that 2 waves arrive out of phase and cancel out, because of the half wavelength path length difference between the slit edges and minima location?

The QM explanation involves the components of the wavefunction arriving out of phase. You should try Feynman's QED book:

https://en.wikipedia.org/wiki/QED:_The_Strange_Theory_of_Light_and_Matter

 

[hutchphd]

The non-parallel slits seem to have a higher transmission coefficient than the parallel slits in the fringes, because with the non-parallel slits you have twice the area of fringes that have the same intensity as a single slit, but with the parallel slits, you have the same area of fringes as a single slit, and those fringes have less than twice the transmission coefficient of a single individual slit.

Simple Question:
Are you collecting 4π radians of scattering from the slits?
Then why do you expect these conservation laws?
Simple answer:
The energy spreads out variously and your camera catches only a forward part of it. This all can be calculated and there is no mystery here.

 

[collinsmark]

I'm slowly working on reproducing the experiment. But my setup isn't quite ready yet. But I'm working on it. (As I write this, my secondary [i.e., old] astronomical camera is busy gathering temperature controlled dark frames for the experiment.) I have a new mount, new control system, new hardware, etc, that should be close to ideal for this experiment (in a quick-and-dirty, on-the-cheap, sort of way). But it's not all set up yet and calibrated.

I would be really surprised though if the energy measured by the sensor, after dark frame subtraction, by having both slits open is anything other than roughly twice the energy of a single slit (all else being equal [e.g., equal slit sizes, equal exposure times, etc.]). And by "roughly" I mean within my own experimental error.

Let's make sure to keep this thread open for replies; and give me a week or so of time, and some good weather, and I plan to have some results to post then.

 

[Dale]

It’s hard to understand how opening a second slit would decrease the energy transmitted via the other slit.

It doesn’t. When you open the second slit more energy goes through than when the slit was closed. However, unless you actually measure which slit it goes through there is no sense in which it went through one. So you cannot say that the energy through the first slit has been reduced.

You are assuming that the energy that goes through each slit is definite and independent. Neither of those is true, and the second should be obviously untrue. Since the double slit experiment is usually portrayed as showing that one photon goes through both slits it is clear that the energy going through the slits is not independent.

 

[Cthugha]

I'm sorry. I meant ratioanle for the numbers will help. Equations, assumptions, etc. that sort of thing,

I would not add photons to the mix.

I can only second that. This setup is so macroscopic that QM will not add anything to understanding it. And: yes, it is expected that for such a setup the total intensity behind the double slit will be the sum of the intensities of the single slits. This is tested frequently in student labs all over the world. I would strongly suggest to thoroughly quantify the detector before looking for esoteric solutions. Numbers indeed will help here.

Yes, unless a pixel is saturated (ie reaches the maximum value) the red values should be directly proportional to intensity.

The first thing I tell my students is to never trust your equipment and to never assume anything you are not sure of. We already found some of the deviation between your expectation and your data by having a closer look at how you treat the background noise. I would now go on and thoroughly characterize your detector before jumping to conclusions.

Do you have any means to linearly vary the incoming intensity in a controlled manner, e.g., by using neutral density filters, over a really wide range? If so, use them and check whether the curve of the detected intensity versus the incoming intensity is really linear. To be honest, even with most image corrections of modern cameras switched off, I would be heavily surprised if the response you get is really a completely linear response without any offset across all light levels. Also, averaging over several images will help to get more reliable data and reduce the graininess.

Edit: The manual already mentions a lot of features, such as software and hardware binning. Do you use some kind of binning? Is there any other kind of processing taking place (center of gravity detection to avoid blooming at low light levels or something like that)? What is the gain setting you actually use?
Other sources seem to imply that the camera has some kind of non-linear behavior at low light levels:
http://www.arciereceleste.it/articoli/translations/115-test-asi6200-en

 

[Devin-M]

I downloaded a piece of software called RawDigger which lets me inspect the RAW pixel values without needing to first convert the file from RAW (NEF) to TIF. The camera is a Nikon D800 by the way.

I sampled the center most 23x51 (ignoring the green and blue pixels) portion of each diffraction spike and looked at the average red pixel value.

Single Slit Red Average: 2143
Double Slit Red Average: 3537
Same Area Noise Average: 45
Single Slit Red Average Noise Subtracted: 2098
Double Slit Red Average Noise Subtracted: 3492 (1.66x higher)

I would like to do another test with a lower gain setting as a few of the pixels in the center of the double slit pattern saturated (was at 6400iso, want to retest at 1600iso), however I don't expect a break in the clouds for the next 7 days.

 

[Drakkith]

I would like to do another test with a lower gain setting as a few of the pixels in the center of the double slit pattern saturated (was at 6400iso, want to retest at 1600iso), however I don't expect a break in the clouds for the next 7 days.

You can likely do the test inside. A flashlight shining through a tiny hole in a piece of cardboard or thick sheet of paper would probably serve just fine as a point-like source. Of just do away with the paper or cardboard if you don't think you need a point source and point it at the wall or something.

 

[Devin-M]

This paper tests the linearity of the sensor response to varying degrees of illumination, the results look quite linear...

Source: https://www.researchgate.net/publication/23699681_Camera_calibration_for_natural_image_studies_and_vision_research

 

[Cthugha]

This is for exposure times of 2 milliseconds and high luminances on the order of more than 1000 cd/m^2. I am not sure that these conditions are really that comparable to the conditions you use. At high light levels it is easy to have a linear response. At low light levels, where all kinds of noise source become prominent, it is a challenge to get something that resembles a linear response. I have been fighting with photodetectors at low light levels for long enough now to not trust them.

Possibly, it might help to reperform the measurement without saturation for the double slit compared to the sum of the spectra of the two individual slits. Possibly an artificial light source will already be sufficient as @Drakkith suggested. To me, it seems striking that for the green pixels, the dark count corrected ratio of the double slit intensity per pixel to the single slit intensity per pixel is on the order of 2.1 (454/216), which is quite well within the range of what is expected.

 

[Devin-M]

Yes the green channel is very close to double, however I don't trust the green (495–570 nm) detections are mostly monochrome light which might be necessary for destructive interference as that light appears to be leaking through the 656nm H-Alpha narrowband filter in front of the sensor.

https://www.astronomik.com/en/clip-...alpha-6nm-ccd-maxfr-clip-filter-nikon-xl.html

https://www.astronomik.com/en/fotog...nfilter/schmalbandfilter-h-alpha-ccd-6nm.html

 

[Drakkith]

Yes the green channel is very close to double, however I don't trust the green (495–570 nm) detections are mostly monochrome light which might be necessary for destructive interference as that light appears to be leaking through the 656nm H-Alpha narrowband filter in front of the sensor.

I think it's more likely that the green pixels are merely picking up red light. I'd bet that the bayer filter on the CCD is letting some of that HA light through the green filters. But I'd have to check the data sheet to be sure.

 

[Devin-M]

The blue channel double slit is also showing 2.05x the single slit after noise subtraction. But the blue values could also be non-monochromatic leakage through the narrowband filter rather than 656nm HA light activating the blue pixels… It’s very possible the red detections would be much closer to monochromatic overall than the green and blue values.

 

[hutchphd]

This paper tests the linearity of the sensor response to varying degrees of illumination, the results look quite linear...

I did decades of work doing spectrophotometry/colorimetry at low light levels for precision medical instruments. This has involved fluorescence, reflectance, and transmittance using Photomultipliers, Avalanche photodiodes, photodiodes, and (sometimes photon counting) cameras.
I guarantee the photocurrent from silicon is rigorously linear over a tremendous range. In my experience a color camera is fraught with problems, most of which have been mentioned. Incidentally I have always liked ImageJ software (free from NIH) for analysis.
I am also still unsure as to your exact setup and a concise reiteration of it and your particular expectations would be useful.
I feel the explanation here is very likely that your expectations for the diffraction signal are incorrect but I really don't understand your setup (a diagram would be good) Also details of the the CCD color filtration can be an issue. This should not be too difficult with facts in hand.

 

[Devin-M]

I’m investigating whether situations with high contrast destructive interference give twice the total photon detections with double slit compared to individual slit, or is it less due to the destructive interference.

Here’s my setup Nikon D800 on equatorial mount with H-Alpha narrowband filter in front of sensor, 300mm f/4.5 lens and TC-301 teleconverter for 600mm f/9 doing 5min exposures on Polaris at 6400iso/gain setting:

 

[Drakkith]

But the blue values could also be non-monochromatic leakage through the narrowband filter rather than 656nm HA light activating the blue pixels…

I'm not so sure. Typical HA filters are nearly 100% effective at blocking wavelengths outside of a small range around the central wavelength. However, I believe the green and blue parts of the bayer filter on the Nikon D800 transmit about 5-10% of light at 656 nm.

 

[hutchphd]

Here’s my setup

Thanks. Sorry if I am behind the curve here but what is the lens cap with slits?

 

[Devin-M]

It’s a bahtinov mask, and I’ve blocked off most of the slits with masking tape and then took exposures through each slit separately and then both at the same time in hopes of showing some destructive interference since it’s a monochromatic point source.

 

[collinsmark]

As promised (from post #41), here's my efforts to reproduce the experiment myself.

My Reproduction, Part I: Equipment, Data Acquisition, and Pre-processing

Equipment:

• Orion Bahtinov mask
• Blue, painter's tape (essentially masking tape)
• Explore Scientific ED80-FCD100 telescope
• Orion Field Flattener for Short Refractors
• ZWO Electronic Filter Wheel (EFW)
• An old Astronomik Hα filter (I think it's a ~7nm bandwidth)
• ZWO ASI1600MM-Pro main camera (monochrome, cooled)
• Electronic focusing system (Optec and Starlight Instruments mishmash)
• Orion 50mm Guidescope with Helical focuser
• ZWO ASI290MM-Mini camera for guidescope
• Pegasus Astro UPBv2 for power distribution, USB hub, and dew heater control
• Various dew straps (the good ones are from AstroZap)
• Various mechanical dovetails, clamps, rings as such from PrimaLuce and ADM.
• Minisforums (not to be confused with Physics Forums) mini PC
• Cheapy, Nexigo webcam, just to keep a wide eye on things.
• Sky-Watcher EQ6-R Pro mount

Figure 1. Equipment setup

Figure 2. Bahtinov mask. Although difficult to see, the entire inside of the Bahtinov mask is blocked out with blue, painters tape except for two of the slits (see vertical slits at the top).

Figure 3. Method for covering left or right slit. It was a simple matter of blocking out a slit by carefully placing a doubled-up strip of blue, painters tape over the top of Bahtinov mask. Or, in the case of both slits open, I just removed the strip and placed it on the table.

Acquisition Details:

• Target: Aldebaran, because it's bright and was near the zenith at my location (less atmosphere to worry about)
• Skies were clear
• Seeing was average
• ASI1600MM-Pro camera cooled to -10 deg C
• Camera gain: 139 (offset 50) placing the camera right around unity gain (i.e., ~1 e-/ADU)
• N.I.N.A. software was used for acquisition along PHD2 for autoguiding
• Autofocus routine was performed immediately before inserting the Bahtinov mask and acquiring data
• Left slit was covered and 40, 1.5 sec subframes were taken
• Both slits were left open and 40, 1.5 sec subframes were taken
• Right slit was covered and 40, 1.5 sec subframes were taken
• Bahtinov mask was removed, and lens cover was placed over objective of ES ED80-FCD100 telescope, and 100, 1.5 sec "dark" frames were taken (camera still at -10 deg C)
• Lens cover removed again, and using a flat panel, 30 "flat" frames were taken. With the lens cover back on again, another 30 "darkflat" frames were taken. However, I didn't end up using flat field calibration on this project (see below)

Preprocessing:

• The 100 dark frames were stacked using PixInsight, creating a master dark frame.
• In all cases (for this particular project), stacking was done using averaging (arithmetic mean), no weights (all weights = 1), and no normalization. However, Winsorized Sigma Clipping was in the stacking to eliminate statistical outliers (e.g., bad pixels caused by hot-pixels, cosmic rays, etc.)

Let me stop here and talk about flat frames for a moment. In the end, I decided not use flat frames in the calibration.

Case against flat frame calibration: Flat frame calibration is used to compensate for any a anomalies in the optics. However, in this experiment, the optics -- anomalies included -- are what we're trying to characterize in the first place. So flat frame calibration doesn't make sense in this experiment.​

​

Case for flat frame calibration: Optics aren't the only thing flat frames are good for. They can also compensate for any dust motes that may exist on my Hα filter for sensor.​

Well, I didn't notice any particularly significant dust mote signatures in the data, so I chose not to perform flat frame calibration (there's really no point). I have the flat frame data if I ever want to go back to it, but for now, flat frame calibration is skipped. Okay, moving on...

• PixInsight was used to perform "dark" frame calibration on the "light" frames. This is done by subtracting the master dark frame from each of the "light" frames (i.e., the subframes containing the diffraction/interference pattern) on a pixel by pixel basis. This basically calibrates out the arithmetic mean of the in-camera noise such as thermal noise and read noise.
• The 40 calibrated frames in each scenario (left slit covered, both slits open, right slit covered) were stacked, using PixInsight.

Figure 4. Example data after dark frame calibration and stacking. Shown here is the image with the left slit covered. Obviously some cropping needs to be done before continuing. (The above image was resized and "stretched" for display purposes here, but the data used in the actual calculations/evaluations did not use resized or stretched images.)

Figs. 5 and 6 show the data after appropriate cropping. In all cases, the crop was 519 pixels wide by 10 pixels high.

Figure 5. Data after cropping. From top to bottom: Left slit covered, Right slit covered, Both slits open.

Figure 6. Same data as Fig. 5, but with a "stretch" so it's easier to see the details. (Note: Stretching was done for display purposes only: forthcoming calculations were done based on unstreched data.) From top to bottom: Left slit covered, Right slit covered, Both slits open.

To be continued (spoiler: results are pretty much exactly what you should expect) ...

 

[collinsmark]

... continued from previous post.

My Reproduction, Part II: Results and Conclusions

Results and Conclusions in a nutshell:

Now that we have nice crops, all we need to do is add up or average the pixels values in each scenario. PixInsight has an easy way to do this using the Statistics process, where it gives you the arithmetic mean of the pixel values in a given image. So that is what I used.

Arithmetic mean pixel value, Left Slit Covered: 190.468
Arithmetic mean pixel value, Right Slit Covered: 189.908
Arithmetic mean pixel value, Both Slits Open: 371.243

Figure 7: Results

According to ideal theory, the average pixel value with both slits open should be quite close to twice the average value of a single slit. Or if the slits are of slightly different size, the value with both slits open should be about the same as the single slit values added together.

Sum of single slit values:
190.468 + 189.908 = 380.376

So, my results differ from theory by about

\frac{380.376 - 371.243}{380.376} = 2.4 \%

I'd say 2.4% ain't bad, and is within my own margin of experimental error. I expect I could have even done better with more careful cropping.

So I'm going to be so bold as to claim that my results pretty much agree with theory (more-or-less).

Additional analysis:

I also wrote a C# program that sums all the pixel values together in each column to be used for plotting the 1-dimensional spacial results. Here are the resulting plots:

Figure 8: Both slits open plot.

Figure 9: Left slit covered plot.

Figure 10: Right slit covered plot.

The shapes of the diffraction/interference patterns also agree with theory.

Notice the central peak of the both slits open plot is interestingly 4 times as high as either single slit plot. Yes, this agrees with theory and can be explained using classical theory or by quantum theory.

In the classical approach, the intensity of an electromagnetic wave is proportional to the square the amplitude of the electric field. So if you have two waves constructively interfering (doubling the electric field amplitude), the intensity increases by 2^2 = 4.

As others have pointed out, quantum theory is not necessary to explain the shapes. But you could use quantum theory if you really wanted to. If you did you'd find the probability density of a photon landing exactly on the central peak is approximately twice that for the double slit configuration than it is for the single slit configuration. So that's a factor of 2 right there from the probability. The other factor of 2 comes from the fact that there's two slits open in the double slit scenario, so twice as many photons arrive at the detector compared to only 1 slit open. And that's the other factor of 2. So 2 \times 2 = 4.

In any case, even though the central peak is 4 times as large in the both slits open scenario, the area under the curve is only twice as large, in part due to more frequent regions of destructive interference.

And that's about all I have on this.

 

[Devin-M]

I do have one question about your data. In the double slit case the data is dominated by the central peak and the 1st order diffraction spike. Suppose you sample only the central peak and 1st order spikes and compare to the same area single slit region. Is it double? So far so good. Now ignore this region and sample the region that only contains the much fainter 2nd, 3rd, 4th etc order spikes. Is it still double?

 

[collinsmark]

I do have one question about your data. In the double slit case the data is dominated by the central peak and the 1st order diffraction spike. Suppose you sample only the central peak and 1st order spikes compare and compare to the same area single slit region. Is it double? So far so good. Now ignore this region and sample the region that only contains the much fainter 2nd, 3rd, 4th etc order spikes. Is it still double?

It's not expected to be double unless the sample area of the detector is sufficiently large. Ideally, you'd want the detector size to be large enough such that the diffraction/interference pattern is vanishingly small at the detector's edges.

And just to be clear, it's the total area under the curve that is expected to be double. But that doesn't necessarily apply to the peak values on the curve.

For example, the peak value of the central lobe in the case of both slits open isn't just double that of a single slit, it's approximately 4 times as large. (That's applied to this particular experiment setup.)

As far as other lobes go, that depends. It depends on the width of the slits and also on the separation of the slits. Sometimes a double slit destructive interference location might coincide with where a peak would be in a single slit scenario. Or maybe not. It depends on the slit widths and slit separations.

Also, if light comes in at an angle (not perpendicular to the Bahtinov mask) or if there is a refractive element placed on one of the slits that differs from the other slit (causing an effective phase difference between the slits), that can affect things too, and effectively shift the central peak to a different location.

 

[Devin-M]

I was just wondering if your measurements agreed with my original measurements that comparisons of the pattern outside the central area show far less than double the average intensity with the second slit.

...because:

"The amount of light captured by a lens is proportional to the area of the aperture, equal to:

https://en.wikipedia.org/wiki/Aperture

...and the double slit has double the aperture.

I performed a new test, this time where I measured each slit individually, and then both at the same time:

Slit 1 ADUs: 7165
Slit 2 ADUs: 7076
Double Slit ADUs: 9339

Each test was targeting the star Polaris with a 5 minute exposure, 600mm, f/9, 6400iso, with a H-Alpha narrowband filter (656nm) in front of the sensor on a Nikon D800 DSLR, and the camera was mounted on an equatorial mount to compensate for the Earth's rotation.

View attachment 318927
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I've now (at your suggestion) looked at how much noise is in equal area with no signal...

Noise ADUs in area w/ no signal: 2634
Slit 1 ADUs w/ noise subtracted: 4531
Slit 2 ADUs w/ noise subtracted: 4442
Double Slit ADUs w/ noise subtracted: 6705 (1.47x higher than slit 1)

View attachment 318972
Noise Sample:
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