# C Modulated Slit

1. Nov 15, 2013

### jkg0

Suppose a slit much wider than photons from a coherent source is placed in front of the source. Let this slit be opened and closed at a rate faster than one wavelength of the source.
Will light pass through the slit?

2. Nov 15, 2013

### TumblingDice

Interesting question that I'm eager to hear the experts weigh in on. An early thought would be how relativistic the slit speed would need to be for longest wavelength. But that's just blowing smoke into a good question. Hoping to learn a lot from replies!

3. Nov 15, 2013

### Simon Bridge

Photons don't have a width.

Do you mean the slit-width starts out large compared with the wavelength of the incoming light?

1st lets see if the experiment is even physically possible.

What you are talking about is an aperture that closes like a camera shutter.
The speed of the shutter can be nowhere near the speed of light. v<<c

If the aperture width is a = kλ: k>1
The time to completely close (at constant speed) would be kλ/2v where v is the speed of the aperture. One wavelength passes in time λ/c.
For the shutter to close in that time, $$\frac{k\lambda}{2v}<\frac{\lambda}{c}\implies \frac{v}{c} > \frac{k}{2}$$ - for v<<c then, k<<2 ... i.e. the aperture width must be very close to the wavelength of the light for the experiment to be physically possible.

If you were relying on the slit being a lot wider than the wavelength, perhaps you didn't want to start with an interference pattern, then all bets are off now - it is not possible to set up an experiment as described.

You described it in terms of photons.
Photons don't have dimensions the way you'd normally think of such things.
Instead they are described by probability.

Your setup has a light-source, does not need to be coherent, and a detector, and a barrier between them with a slit in it whose width d cycles rapidly between wide open d=a and shut d=0, say: $$d(t)=a\cos(2\pi v t/a)$$. We can ask what the probability is that the detector at position (r,θ) would detect a photon at some time.

A formal calculation would involve Feynman's sum-over-many-paths and will be annoying to do because of the time variance. However, we can use the known solutions to similar problems to cheat it a little.

The result is an interference pattern in the probability distribution that varies rapidly with time.

Something like: $$p(\theta)=\text{sinc}^2\left(\frac{\pi a}{\lambda}\cos\big(\frac{2\pi v t}{a}\big) \sin\theta \right)$$... I have yet to check this but you get the idea.

Therefore, we can expect that there should be no position you can put the detector where it will not have some probability of detecting a photon sometime.

Looking at the probability as a function of time, staying in the same position, then we can expect there will be some instants where the probability of detecting a photon is zero.

The exact effect will depend on the function d(t).

Conclude: photons will get through the slit - yea even though it be opening and closing very fast.

You can intuitively figure how this would work - at any time there is a probability that a photon will arrive at the slit while the slit is either opening or closing. You needn't worry about the photon getting chopped off by the shutter closing while it is half-way through, photons don't have dimensions like regular objects do.

There's all kinds of discussions:
http://www.ecse.rpi.edu/~schubert/Course-Teaching-modules/A012-Photon-length.pdf [Broken]

What gets treated as the dimension of a photon depends on what you want to do with it.

Last edited by a moderator: May 6, 2017
4. Nov 15, 2013

### TumblingDice

If the aperture can move rapidly enough for the experiment, wouldn't this simplify to the light passing through a constantly expanding and contracting slit width? The smaller the slit, the larger the diffraction, and varying at a constant rate with time. How much of an interference pattern would occur? I'm thinking center always has a band no matter how slim, but a lot of consistent overlap on each side as the width changes. Might this just create a bright center that gradually fades off left and right, or are there mathematical patterns still at play?

5. Nov 15, 2013

### Simon Bridge

That is exactly what I did.

The pattern would blur out because of limitations of the detector - does not have to be the good-old mark 1 eyeball btw.

The pattern is still there in the math - the rest depends on the detector, which needs a seperate model. I don't think the experiment described approaches HUP limits.

I imagined a shutter that closed from both sides because it makes the math easier.
The common way to do this would be to use counter-rotating disks with slits cut in them.
Those would normally produce a non-zero length of time in which the slit is completely closed - making the pattern flicker.

We could imagine the disks rotate at relativistic speeds, allowing k→2 ... you only need the aperture of the order of magnitude of the wavelength to get noticeable interference. It would also change the shape of the slits.

6. Nov 15, 2013

### Simon Bridge

Looking at the brightness of the fringes:$$I(\theta,t) = I_0\cos(2\pi v t/a)p(\theta,t)$$ ... where $I_0$ is the total number of photons falling evenly over the entire slit and $p(\theta,t)$ is the probability density function already provided (someone check the normalization please?)

With a sluggish response from the detectors, we'd expect the accumulated effect to follow the time-rms value of the intensity function - like measuring an AC voltage with a moving-coil voltmeter - you can see by inspection that some variation of intensity with angle will remain.

So there would be a fuzzy interference pattern corresponding to the rms value of the aperture width.
To get the actual pattern, someone should do the calculus - thank you :)

7. Nov 15, 2013

### TumblingDice

I also kept the slit 'centered' in thought. Wasn't interested in bringing up tangents to the question.

What detector limitations are you speaking of?

I was imagining a nice beam of light at consistent wavelength passing through a slit and producing a visible interference pattern as in normal demonstrations of single slit diffraction. Except slit varies in width. Interference 'bands' get further apart and then closer together as slit closes and opens. Is that correct?

If so, this is why I'm wondering if result over time would just be a smooth fade from center to edges. I don't want to belabor this if my thinking is off-base or it would be easier to perform the experiment than the calculations.

EDIT: Thank you for the follow-up, Simon. You think and type more quickly than I...!

8. Nov 16, 2013

### Simon Bridge

Using the eye as an example - there are artifacts like:
- persistence of a detection event after the source has gone
- a time delay for the full intensity to register
- post processing artifacts as the detection gets interpreted
- dead time - during which nothing gets detected even if it is present
- efficiency - it may take many photons to trigger a detection

You get this to some extent in all detectors.
The first two are the reason a rapidly blinking light can look like a continuous one.
Also why you can appear to see through a wooden fence that you walk past rapidly.
With the third, you get the illusion of motion that we use for so much of our entertainment.
... and so on.

That's the picture I was painting - yes.

If you had an array of perfect detectors, then the detectors would simply show the rapid rise and fall in the photon count-rate at the place they are sitting.

If they keep a record, we can play it back and see the diffraction pattern expanding and contracting.

But what if they are not perfect?

In an imperfect detector, turning on a light spot produces a slow-increase in response (voltage say) until enough light has fallen on the detector that a maximum has been reached. The max voltage reached is usually dependent on the intensity of the light. Less light, a lower max voltage.

Turn the pot off and the reverse process occurs... the voltage gradually dies away.

I say "gradually", in good detectors this can be nanoseconds. The shape of the voltage-time curve when a light is switched on is called the "signal response" of the detector.

What happens if we turn the light off before the max voltage is attained?
Well then the detector registers a lower intensity than was actually there.
If you repeat a series of pulses like that - it appears to be a flickering light.
As you bring the pulses closer together in time, the flickering smooths out, and you see a more continuous light. The fast-pulse light will be dimmer than actual continuous light from the same source.

The brightness that gets detected from the time-average of the pulses...

You should be able to fill in the rest.
It's the illusion-of-motion effect you get with cinema.

Alternatively you can do the math and disprove me.
This is all very back-of-envelope after all.

Note: cannot remember if I'm supposed to time-average the intensity or the electric field though - I suspect electric field ... which is a vector... so you are more likely to get interference patterns showing up.

9. Nov 16, 2013

### jkg0

Clarification

For clarification:
First, by width I mean λ, the wavelength of the light in question.

Second, for the slit imagine the intersection of two plates each with a narrow, slightly curved slit in them rotating past each other. The window will be the intersection of the two slits. This window can be made to move very fast as its movement is a beat frequency of the union of plates. In this case the plates can rotate at much less than the speed of light and still create a window that moves at the speed of light or even faster as it is composed of the intersection.

I understand that each plate will have some depth, however this depth can easily be made less than the wavelength of light, for example 800nm light, 10nm depth plates.

Jonathan

10. Nov 16, 2013

### Staff: Mentor

I have to ask... Where are you going with this thought experiment?

So far there's no QM involved; it's a problem in classical wave behavior. And without actually solving that (as yet incompletely specified) problem, I can say how it its solution will relate to the QM prediction: the intensity of light at at any point beyond the screen, as calculated in the classical solution, will be proportional to the probability of detecting a photon at that point.

11. Nov 16, 2013

### Staff: Mentor

How wide is a photon? (No, it's not the wavelength, if that's what you're thinking.)

This doesn't make any sense when interpreted literally. "Rate" implies a certain number of times per second. "Wavelength" is a distance. Do you mean "Let this slit be opened and closed at a rate faster than the frequency of the light?"

12. Nov 16, 2013