# The double-slit experiment with a pit in the screen

• I
• Imya
Imya
TL;DR Summary
How does the distribution of light power between the pit and the rest of the flat screen depend on opening two slits or closing one slit? Five predictable experiments were previously described to formulate a question for the experiment #6.
The main question is contained below in the experiment #6. But first, 5 simple experiments are described to describe the designations that are used in the question of the experiment #6.

In the double-slit experiment, the slits are labeled α and β. A region C is designated on the screen. The power passes through the open slit: ##{P_\alpha} = {P_\beta} = {P_{\rm{sl}}}##

Experiment #1: Both slits are open, and the screen is a flat solid surface. This is well described and shown in the diagram (which is taken from here). The total power of light on the screen is ##{P_{\rm{scr 1}}} = {P_\alpha} + {P_\beta} = 2 \cdot {P_{\rm{sl}}}##. The power of light outside the region C is ##{P_{{\rm{scr}} - {\rm{C }}1}} = {P_{{\rm{scr }}1}} - {P_{{\rm{C }}1}} = 2 \cdot {P_{{\rm{sl}}}} - {P_{{\rm{C }}1}}##.

Experiment #2: Slit α is open. Slit β is closed. The total power of light on the screen is ##{P_{{\rm{scr 2}}}} = {P_\alpha } = {P_{{\rm{sl}}}}##. The power of light outside the region C is ##{P_{{\rm{scr}} - {\rm{C }}2}} = {P_{{\rm{scr }}2}} - {P_{{\rm{C }}2}} = {P_{{\rm{sl}}}} - {P_{{\rm{C }}2}}##.

Experiment #3: Slit β is open. Slit α is closed. The total power of light on the screen is ##{P_{{\rm{scr 3}}}} = {P_\beta } = {P_{{\rm{sl}}}}##. The power of light outside the region C is ##{P_{{\rm{scr}} - {\rm{C }}3}} = {P_{{\rm{scr }}3}} - {P_{{\rm{C }}3}} = {P_{{\rm{sl}}}} - {P_{{\rm{C }}3}}##.

Considering symmetry, the index 23 is added:
$${P_{{\rm{C }}2}} = {P_{{\rm{C }}3}} = {P_{{\rm{C 2}}3}} \Rightarrow {P_{{\rm{scr}} - {\rm{C }}2}} = {P_{{\rm{scr}} - {\rm{C }}3}} = {P_{{\rm{scr}} - {\rm{C 2}}3}}$$

In Experiment #1, the power of light in the region C is greater than the sum in Experiments #2 and #3. Therefore, the power of the rest of the screen is less:
$${P_{{\rm{C }}1}} > 2{P_{{\rm{C 2}}3}} \Rightarrow {P_{{\rm{scr}} - {\rm{C }}1}} < 2{P_{{\rm{scr}} - {\rm{C 2}}3}}$$

Next, the conical pit with A and B is made in the screen within the region C, as shown in the next diagram. The region A sees slit α and does not see slit β. The region B sees slit β and does not see slit α. This means that a straight ray from A falls on α, but not on β. A straight ray from B falls on β, but not on α. Because the angle between A and B is less than the angle between the straight rays from the slit α and β to the screen centre, as shown in the next diagram. Diffraction at the edges of the transition from regions A and B to the flat screen is neglected, assuming the region C size is much larger than the wavelength.

The experiments #4 and #5 are repeated experiments #2 and #3 with the new screen. The power in the regions A and B can be easily determined:
$${P_{{\rm{A }}4}} = {P_{{\rm{{C} }}2}} = {P_{{\rm{{C} }}23}}$$
$${P_{{\rm{B }}4}} = 0$$
$${P_{{\rm{A }}5}} = 0$$
$${P_{{\rm{B }}5}} = {P_{{\rm{{C} }}3}} = {P_{{\rm{{C} }}23}}$$
$${P_{{\rm{A }}4}} + {P_{{\rm{B }}4}} + {P_{{\rm{A }}5}} + {P_{{\rm{B }}5}} < {P_{{\rm{C }}1}}$$
$${P_{{\rm{scr}} - \left( {{\rm{A}} + {\rm{B}}} \right){\rm{ }}4}} = {P_{{\rm{scr}} - {\rm{{C} }}2}} = {P_{{\rm{scr}} - {\rm{{C} }}23}}$$
$${P_{{\rm{scr}} - \left( {{\rm{A}} + {\rm{B}}} \right){\rm{ }}5}} = {P_{{\rm{scr}} - {\rm{{C} }}3}} = {P_{{\rm{scr}} - {\rm{{C} }}23}}$$

Experiment #6: Both slits are open, and there is the conical pit in the screen.
What will be the power in the regions A and B? What will be the power in the the rest of the screen? Is there the right (1) or (2) or something else?

The conservation of energy requires:
$${P_{{\rm{A }}6}} + {P_{{\rm{B }}6}} + {P_{{\rm{scr}} - {\rm{{C} }}6}} = {P_{{\rm{scr 6}}}} = {P_\alpha } + {P_\beta } = 2 \cdot {P_{{\rm{sl}}}}$$
If the power outside the region C corresponds to Experiment #1, then the power in the regions A and B must change and become greater than in Experiments #4 and #5.
$${P_{{\rm{scr}} - {\rm{{C} }}6}} = {P_{{\rm{scr}} - {\rm{{C} }}1}} \Rightarrow {P_{{\rm{A }}6}} + {P_{{\rm{B }}6}} = {P_{{\rm{scr 6}}}} - {P_{{\rm{scr}} - {\rm{{C} }}6}} = 2 \cdot {P_{{\rm{sl}}}} - {P_{{\rm{scr}} - {\rm{{C} }}1}} = {P_{{\rm{C }}1}} > 2{P_{{\rm{C 2}}3}} = {P_{{\rm{A }}4}} + {P_{{\rm{B }}4}} + {P_{{\rm{A }}5}} + {P_{{\rm{B }}5}}$$
$$\left\{ \begin{array}{l} {P_{{\rm{scr}} - {\rm{{C} }}6}} = {P_{{\rm{scr}} - {\rm{{C} }}1}}\\ {P_{{\rm{A }}6}} + {P_{{\rm{B }}6}} > 2{P_{{\rm{C 2}}3}} \end{array} \right. \qquad(1)$$
But straight ray from each of the slit falls on only one corresponding region. Each of the regions sees only one corresponding slit. How can the power on it depend on whether the second slit is open or closed?

And if the power in the regions A and B remains the same as in Experiments #4 and #5, then the power of the rest of the screen must change and become greater than in Experiment #1:
$${P_{{\rm{A }}6}} + {P_{{\rm{B }}6}} = {P_{{\rm{A }}4}} + {P_{{\rm{B }}4}} + {P_{{\rm{A }}5}} + {P_{{\rm{B }}5}} = 2{P_{{\rm{C 2}}3}}$$
$${P_{{\rm{scr}} - {\rm{{C} }}6}} = {P_{{\rm{scr 6}}}} - \left( {{P_{{\rm{A }}6}} + {P_{{\rm{B }}6}}} \right) = 2 \cdot {P_{{\rm{sl}}}} - 2{P_{{\rm{C 2}}3}} = 2{P_{{\rm{scr}} - {\rm{{C} }}23}} > {P_{{\rm{scr}} - {\rm{{C} }}1}}\\$$
$$\left\{ \begin{array}{l} {P_{{\rm{scr}} - {\rm{{C} }}6}} > {P_{{\rm{scr}} - {\rm{{C} }}1}}\\ {P_{{\rm{A }}6}} + {P_{{\rm{B }}6}} = 2{P_{{\rm{C 2}}3}} \end{array} \right. \qquad(2)$$
But how can the power on the rest of the screen change due to the appearance of pits in it?

Imya said:
How does the distribution of light power between the pit and the rest of the flat screen depend on opening two slits or closing one slit? Five predictable experiments were previously described to formulate a question for the experiment #6.
This isn't a quantum physics question, it's a question about classical wave optics. But when you say
The region A sees slit α and does not see slit β. The region B sees slit β and does not see slit α. This means that a straight ray from A falls on α, but not on β. A straight ray from B falls on β, but not on α.
you are switching to ray optics and this inconsistency is causing the apparent contradiction.
To approach this problem correctly you will have to use wave methods consistently throughout: calculate the phase and amplitude at every point on the "mouth" of the pit (which is easy because it's just what we would find if the pit weren't there and we've already calculated that as C in case #1). This is the light source that is illuminating the walls A and B of the pit, and we can use that to calculate the intensity at every point along A and B.

Or if we just want to resolve the apparent contradiction, we could recognize that the total power delivered to the walls of the pit must be equal to the total power that passes through the mouth of the pit.

Imya
Imya said:
TL;DR Summary: How does the distribution of light power between the pit and the rest of the flat screen depend on opening two slits or closing one slit? Five predictable experiments were previously described to formulate a question for the experiment #6.

But how can the power on the rest of the screen change due to the appearance of pits in it?
Nugatory said:
To approach this problem correctly you will have to use wave methods consistently throughout:
Exactly. In the same way that you calculate the basic interference fringe pattern you can calculate the resultant amplitude along the surface of the conical pit - Path difference between waves arriving from the two slits and landing (vectorially adding) on the conical surface - possibly including the cos factor for the relative amplitudes of the two arriving waves on the internal face of the pit. I use the term "waves" rather than your rays because rays would only be in one direction if the source slits were very wide and not your assumed omnidirectional from ideal zero width slits. You are mixing fish and fowl in your approach.

I assumed that the pit dimensions are greater than the wavelength because the width is the same sort of size as the fringe width. That would eliminate any problem of diffraction effects for (very short )Waves entering it.
If you want to calculate what would happen 'inside' a small cone, of intermediate dimensions of just a few wavelengths then upi'd have to do a complete analysis, including the diffraction pattern of such a pit and then correlate that patterns with the pattern of the two slits.

PS Only the energy arriving in the vicinity of the pit is involved; the modified fringe pattern will change a bit but not the total energy.

What type of pattern would appear on a second optical screen placed behind the pit?

Quarker said:
What type of pattern would appear on a second optical screen placed behind the pit?
Something indistinguishable from a single slit diffraction pattern, assuming the slit is small compared to the scale of the pattern, which is probably true.

Technically, what you do in every case is consider the complex amplitude (which is the amplitude times the phase as a complex exponential) of the incident light across the slits, set it to zero everywhere there is wall not slit, then take the Fourier transform. That'll often be a job for a numerical integrator, and in this case won't make any measurable difference from what I said in my first paragraph.

Ibix said:
Something indistinguishable from a single slit diffraction pattern
if the position of the screen is moved far enough then the fringes would be 'distinguishably' wider. In all of these "what if?" situations, it all depends on the actual numbers involved. And, because it's all in Fourier Land and things are inside out and bigger can produce smaller etc.. You have to do the sums.

If the slit in the first optical screen were located in a dark zone, where destructive interference predominates, would any type of pattern appear on the second screen?

sophiecentaur said:
You have to do the sums.
In general, sure. But if the slit width is small compared to the pattern from the first slit then the illumination is pretty much constant across the slit and you can skip to the answer I gave. And if you've got a naked-eye visible pattern then the pattern scale will be millimetres and the slit scale (if you're going to get much diffraction) is likely to be tenths or even hundredths of that.
Quarker said:
If the slit in the first optical screen were located in a dark zone, where destructive interference predominates, would any type of pattern appear on the second screen?
You mean, if no light goes in, will any light come out?

So if the original slit is kept open, the addition of a second slit in a dark zone won’t change the pattern appearing on the second screen?

Quarker said:
So if the original slit is kept open, the addition of a second slit in a dark zone won’t change the pattern appearing on the second screen?
No, for two reasons. First, there is no light coming through one of the slits. Second, the two slits would have to be separated by at least half the period of the pattern from the first pair of slits (typically a few millimeters) so they would be too far apart to expect any visible interference anyway.

As @sophiecentaur notes, all of this is only approximately true. The dark zones aren't completely black except in one zero-width place so there will actually be some light coming through, and there will be some interference effects between two slits no matter how far apart, and the illumination of the bright slit isn't completely uniform. Sufficiently precise lab equipment in the hands of a sufficiently capable experimentalist could, therefore, detect a difference from a single slit pattern. For reasonable assumptions about slit sizes, though, actually detecting those differences would be a very difficult and probably expensive undertaking.

Sorry, one more question. If the first optical screen has two slits, both located in dark zones, any pattern that would appear on the second screen would be caused by light that hasn’t undergone destructive interference, correct? Because, as you say, the dark zones aren’t completely black.

Quarker said:
If the slit in the first optical screen were located in a dark zone, where destructive interference predominates, would any type of pattern appear on the second screen?
What ‘experiment’ are you actually describing? It makes no sense without a diagram.
If you draw a diagram then you may find the answer yourself. Random “what if” questions seldom help to improve understanding.
Quarker said:
If the first optical screen has two slits, both located in dark zones
Picture? A moment's thought wou d tell you that the Young's Slits need to be very close together in order to produce a visible (wide enough) fringe. A second pair of slits, separated by a fringe width would produce a fringe paterns with a fringe width with the spacing of the original pair of slits. Could you see it? Also, the energy arriving would be determined by which part of the original pattern you choose.
I seriously question whether you have actually grasped the principle of Young's experiment. Read the theory again, perhaps?

Quarker said:
Sorry, one more question. If the first optical screen has two slits, both located in dark zones, any pattern that would appear on the second screen would be caused by light that hasn’t undergone destructive interference, correct? Because, as you say, the dark zones aren’t completely black.
If you have two slits in a single minimum they will typically be illuminated very slightly so you could get something like a two slit pattern. Your eye is sensitive to approximately three orders of magnitude of brightness, so the dark regions are a factor of a thousand or more dimmer than the maxima, which will typically be a similar factor dimmer than the original incident laser (details depend on the setup, naturally). So it is no exaggeration to say the pattern would be around a million times dimmer than the first pattern. But the dark regions aren't technically black except at one point, so it's theoretically possible to use the little bit of light that's there - it's just light.

Ibix said:
so it's theoretically possible to use the little bit of light that's there - it's just light.
Agreed but I don't think the OP has grasped the fact that these two new slits could not be anywhere other than in a very small part of the first interference pattern.

Another piece of basic theory: the relative phases of the waves from the original slits would be changing quite rapidly and the two 'sources' would not be the same as the properly illuminated (colimated) original slits. So the resulting pattern would actually not be simple to predict. The questions have migrated into the realm of idle curiosity, not Physics, i think.

sophiecentaur said:
Agreed but I don't think the OP has grasped the fact that these two new slits could not be anywhere other than in a very small part of the first interference pattern.

Another piece of basic theory: the relative phases of the waves from the original slits would be changing quite rapidly and the two 'sources' would not be the same as the properly illuminated (colimated) original slits. So the resulting pattern would actually not be simple to predict. The questions have migrated into the realm of idle curiosity, not Physics, i think.
If you know of anyone who has actually conducted this experiment, I would love to read about it.

Quarker said:
If you know of anyone who has actually conducted this experiment, I would love to read about it.
It's probably intractable analytically, but you just need to take Fourier transforms of the aperture function. If you have basic programming skills and access to an FFT library it's fairly straightforward to work out for yourself. In python with scipy, if you can import a library, fill a list with ones and zeros, and call a function, that's about all you need.

The result won't be very interesting and likely dominated by rounding errors as you move the second slits into the minima, but it's doable.

Quarker said:
If you know of anyone who has actually conducted this experiment, I would love to read about it.

Why would they bother? This stuff was known about for hundreds of years. If you can understand even the most basic maths, the result of any of this kind of experiment can be roughly predicted.

If you use more advanced maths (Diffraction formulae etc.) you could look into the details but you'd need to involve actual dimensions and what approximations are involved.

Lord Jestocost

• Quantum Physics
Replies
9
Views
1K
• Quantum Physics
Replies
8
Views
1K
• Introductory Physics Homework Help
Replies
4
Views
1K
• Quantum Physics
Replies
7
Views
1K
Replies
2
Views
949
• Quantum Physics
Replies
9
Views
1K
• Quantum Physics
Replies
1
Views
1K
• High Energy, Nuclear, Particle Physics
Replies
6
Views
2K
• Introductory Physics Homework Help
Replies
6
Views
1K
• General Math
Replies
3
Views
1K