Purity of Interference Pattern in a Classical Double-slit Experiment

[Quarker]

In a classical double-slit experiment with a laser as the light source, is it possible to create an interference pattern where at least two fringes are completely devoid of light? Or will there always be some light contamination, even in the darkest fringes?

 

[phinds]

Have you done any research on this?

 

[Quarker]

Yes, I did searches of double-slit experiments, but the fringes are pretty much ignored in every one.

 

[Ibix]

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

 

[Quarker]

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

Thank you Ibix, but I’m not sure that this answers my question. Contrast is a measure of the detectability of light. I’m more interested in whether it’s possible to produce completely light free fringes in an interference pattern.

 

[PeroK]

I’m more interested in whether it’s possible to produce completely light free fringes in an interference pattern.

Given that light interacts with a detector screen in terms of a finite number of photon detections, then the answer must be yes, it's possible.

That said, if you run any experiment for long enough something is bound to go wrong eventually. There might be a small Earth tremor near the lab that disrupts your experiment. How long do you wait until you are satisified that the dark patches will remain dark forever?

 

[Quarker]

Given that light interacts with a detector screen in terms of a finite number of photon detections, then the answer must be yes, it's possible.

That said, if you run any experiment for long enough something is bound to go wrong eventually. There might be a small Earth tremor near the lab that disrupts your experiment. How long do you wait until you are satisified that the dark patches will remain dark forever?

Assuming no seismic activity ; )

 

[PeroK]

Assuming no seismic activity ; )

There's nothing fundamentally QM about your question. It's more about the limits of experimental error and capability.

 

[Quarker]

There's nothing fundamentally QM about your question. It's more about the limits of experimental error and capability.

I’m not sure what you mean by fundamentally QM.

 

[Ibix]

Thank you Ibix, but I’m not sure that this answers my question. Contrast is a measure of the detectability of light. I’m more interested in whether it’s possible to produce completely light free fringes in an interference pattern.

Look at the formula for visibility. If the minimum intensity is zero then $v=1$, otherwise $v<1$. The page also notes that the visibility is a measure of coherence, so you only get $v=1$ for slits illuminated with perfectly eternally coherent light. No real source fits the bill, only an ideal one.

As PeroK says, you can step the input beam brightness down far enough that quantisation means that the brightness of a dark fringe is probably zero for an arbitrarily long time, at the expense of the brightness of the bright fringes also being very low.

Note also the existence of detector noise, particularly the dark current.

 

[PeroK]

I’m not sure what you mean by fundamentally QM.

It applies to any experiment. It doesn't really tell you anything about QM.

 

[Quarker]

Look at the formula for visibility. If the minimum intensity is zero then $v=1$, otherwise $v<1$. The page also notes that the visibility is a measure of coherence, so you only get $v=1$ for slits illuminated with perfectly eternally coherent light. No real source fits the bill, only an ideal one.

As PeroK says, you can step the input beam brightness down far enough that quantisation means that the brightness of a dark fringe is probably zero for an arbitrarily long time, at the expense of the brightness of the bright fringes also being very low.

Note also the existence of detector noise, particularly the dark current.

I see.

 

[Quarker]

It applies to any experiment. It doesn't really tell you anything about QM.

This has been helpful. Thank you.

 

[sophiecentaur]

whether it’s possible to produce completely light free fringes in an interference pattern.

In the simplest model, waves from two slits have exactly the same amplitude which means that the slits must be equal width and are exactly 180 degrees out of phase. Then, at just one place in the minima, you can get almost complete cancellation. There will be many reasons to give some inequality in the the slot sizes and equal path lengths of light from the source and through both.

You will have looked at information about the Young's Slits experiment so you will recognise what I'm talking about. No need for me to draw a diagram (look at your favourite hit on Google); the null will always 'bottom out' and not go down to zero because the slits will have finite width and you are not actually getting 'point sources.

Of course, any detector will have a certain resolution (it has a finite pixel width) so it will be getting some non-zero resultant on either side of the null. Basically, the better the depth of the nulls (nearly zeros) you want, the more expensive your optics needs to be.

In addition, you can experience QM effects and statistical behaviour so ,for very low energy flux, you may get no response for a long time because there can be very low numbers of photons.

 

[tech99]

The two lobes are in antiphase, so even a detector with finite width will find a null.

 

[sophiecentaur]

The two lobes are in antiphase, so even a detector with finite width will find a null.

That implies any equipment would show a zero null depth. Any lack of symmetry or imperfect collimation will cause the nulls to 'fill in'. The phase between the two arriving waves has to be exactly antiphase when the amplitudes are equal. That's a big ask, imo. The detector will have a noise floor which will also affect the null depth.

 

[tech99]

If we assume a detector that has a finite width, it will receive contributions from both lobes. Even if the lobes are of unequal amplitude, as we move the detector from side-to-side we can find a point where the two contributions are equal. As the two lobes are in antiphase, we can find a null point.

 

[Gleb1964]

Only one point (or line) of detector would fulfil null, but not the other area of extended detector.

 

[Ibix]

I'd say it depends what the detector is detecting. If it's electric field amplitude then the integral across a finite size detector will cross zero at some displacement from the origin. If it's detecting intensity it won't. Optical experiments record intensity.

Note that the above is assuming a perfect fringe contrast. Depending on how bad a model that is of your experiment you may find that even if you're working in microwaves and can record amplitude you may not find a zero.

 

[sophiecentaur]

As the two lobes are in antiphase, we can find a null point.

The situation you are describing would be very idealised, with perfect symmetry and zero width slits for perfect cancellation. Clearly, you could make the setup better and better but there will always be something to fill in the nulls.

 

[BVirtual]

The null point you hope to find is of a width too small to be detected by current measuring devices. How wide is it? Look at the math equations. It would be infinitesimal in width.

Also, due to various interactions of the light beam going through the finite thickness of the slits, there will be many types of non collimated light exiting the slit inside your detector. There is refraction from the slits trailing edges that spread the 'wave' once inside the slit. And reflection from the slit thickness surface that create a non zero intensity across your detection device/screen, in a almost random variation across the screen. Thus, due to the finite geometry of your slit and the behavior of light going around edges, there will be no null, as you expect.

However, your experiment can still proceed. Now that you know these aberrant effects that change the null to a non zero value, you can estimate these effects, and subtract them from your measurements. You can then use a screen with a small horizontal slit in it, to let a finite width of the interference pattern though this slit, into a beam spreader, and broaden your 'null' onto another detector/screen. A concave lens or cylindrical mirror can work to spread the interference pattern.

It will be dimmer, and measured values will be smaller. Try a brighter source light. Try other adaptions and variations. A broader horizontal slit along with a convex lens to focus the light to increase it's intensity. A cylindrical convex lens would do nicely.

 

[Gezstarski]

I took the original question to be a theoretical one - is it possible in an idealised setup, (forget sesmic noise, detector dark current..,) to have two perfect nulls. The answer is yes. I can design the slits so that the intensity at any two arbitrary POINTS in the detector plane is exactly zero. By changing the width and position of the slits there are enough degrees of freedom to arrange that, at each of the two points, the wave amplitudes from the two slits are equal and opposite, In fact there are degrees of freedom to spare.

 

[sophiecentaur]

is it possible in an idealised setup, (forget sesmic noise, detector dark current..,) to have two perfect nulls. The answer is yes.

You need to include coherence in that statement. You are applying assumed perfection to one thing yet ignoring the inescapably finite coherence which the upstream part of the experiment will have. 'The maths' will give you a result but you always have to apply reality before drawing conclusion about your probable result. That's life.

 

[sophiecentaur]

at least two fringes are completely devoid of light?

No experiment will ever deliver a result of 'zero' (that can be relied on. That's because any measurement equipment will be affected by random noise and precision of manufacturing. If you bolt those effects into a mathematical model, the sums will not give you a zero minimum either.

I was at University in the mid 60s and they showed us a laser. It was a really big deal then and few of the teaching staff really got what was going on with the laser or in the equipment. (Fourier was well known but holography was not automatically associated with it - although it's obvs these days. We were shown some holograms and they were unbelievable!!!! The two slit experiment is the simplest version of holography and the fringe pattern is actually a hologram. The depth of a null is limited by coherence along the two (or multiple) paths. They told us that making holograms could only be achieved on a concrete base with a super stable optical arrangement. That would affect the achieved contrast of the resulting hologram, which is basically the depth of the nulls. There is no such thing as a perfect hologram for exactly the same the subject of this thread.