Double slit experiment (one slit closed)

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  • #1
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Hi,

In young experiment, say one slit is completely closed, what observed is looks like a single band light. But why is not a diffraction pattern? I wonder whether the reason is, the slit width(don't mean distance between slits) more narrow than single slit diffraction in Young experiment?
 

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  • #2
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Having one slit completely closed off is equivalent to the single-slit diffraction experiment. Yes, there should be diffraction if the slit width is of the appropriate size relative to wavelength.

Many introductory physics discuss Young's interference experiment before diffraction effects, and they are usually in separate chapters. So if you are somewhere between these chapters then the presentation may only consider interference effects and this would explain the single-band intensity pattern. Both effects are present and there should be a section in your text that discusses it.
 
  • #3
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Hmm, you say there is diffraction too. Can you suggest me a web site which explain diffraction in Young experiment? I couldn't find a good source. BTW, my books aren't explaining diffraction effect in Young. Thanks.
 
  • #5
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Great. Good illustrations. So we can not see the diffraction "directly" in multiple slit interference but the envelop is the evidence of diffraction. I will study this, good start point for me.

But an early question(guest); if one slit closed in Young experiment, we will see a diffraction but the center of diffraction will shift to the side of open slit and has less intensity(approx. 1/4 of double slit), is it?
 
  • #6
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And one more question; What will we observe if one of the slit narrow?

1. Intensity will lessen
2. The center of peak will not change(not sure this)
3. more diffrence... mmm... :P sorry
 
  • #7
548
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And one more question; What will we observe if one of the slit narrow?

1. Intensity will lessen
2. The center of peak will not change(not sure this)
3. more diffrence... mmm... :P sorry

I'll assume your initial scenario (Young's experiment with two slits open and then we close one off) when giving the answers.

1. Two things are going here: (1) A narrower slit implies greater diffraction, so you will see the light-bands have a larger spread. (2) You can imagine the slit to be a point source emitter of light. So at larger radial distances from the point source, the intensity falls off. Your physics text may make the assumption that the light intensity on viewing screen is constant for all points (ignoring 2 then), and therefore, the resulting intensity pattern is due solely to diffraction effects.

2. If you compare the central peak from Young's interference experiment (two slits open) to the scenario where you close one slit of the previous setup, you will see a shift in the central peak. In the former scenario, the central point of the peak will occur exactly between the two slits, and in the latter, the peak will occur directly across the open slit. If you only perform the single-slit experiment and gradually narrower the slit, the location of the central peak will not change.
 
  • #8
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A little explanation for my first question:

http://tr.wikipedia.org/wiki/Çift_yarık_deneyi

(left side: Sun light ; right side: monocromatic light)

above link has a picture. And if one of slit closed(for monocromatic), observed pattern is not look like a diffraction. So asked the first question. Now I belive that, there is diffraction while a slit closed in Young experiment. The picture incapable. Don't know where is the original place of that picture.

1.
If you compare the central peak from Young's interference experiment (two slits open) to the scenario where you close one slit of the previous setup, you will see a shift in the central peak.
Agree. This is ok.

2.
If you only perform the single-slit experiment and gradually narrower the slit, the location of the central peak will not change
No, I compare for double slit. One of them is narrower.

For the second question(one of them narrower): I think the central peak will stay in place. Central peak(and rest) intencity will lessen. Finally light bands will have a larger spread but not sure about the symmetry. I guess, will symmetrical too.

Summarize: I need help about the symmetry and band spread.
 
  • #9
548
2
above link has a picture. And if one of slit closed(for monocromatic), observed pattern is not look like a diffraction. So asked the first question. Now I belive that, there is diffraction while a slit closed in Young experiment. The picture incapable. Don't know where is the original place of that picture.

I see in the right-most pictures (monochromatic light) that there is only one band. There can be other bands, but the intensity of these bands will be some small fraction of the main, central band. Are you familiar with the equation that gives the intensity as function of angle from the central axis for single-slit diffraction:

[tex] I\left(\theta\right) = I_{max}\left(\frac{sin\left(\alpha\left(\theta\right)\right)}{\alpha\left(\theta\right)\right)}\right)^{2}[/tex]

where

[tex] \alpha\left(\theta\right)\right) = \frac{\pi a}{\lambda}sin\left(\theta\right) [/tex]

"a" is the slit width and everything else should be obvious. With these equations, you can determine the fractional intensity of the other bands relative to the main band. The picture may not be accurate, or the modeled parameters may produce those results. If you have Mathcad, I can send you some simulations that model single-slit diffraction and double-slit interference.
 
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  • #10
548
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2. No, I compare for double slit. One of them is narrower.

For the second question(one of them narrower): I think the central peak will stay in place. Central peak(and rest) intencity will lessen. Finally light bands will have a larger spread but not sure about the symmetry. I guess, will symmetrical too.

I never fully analyzed this situation, but it's an interesting question nonetheless. My thoughts are that the central peak would remain in the same location, and also, the intensity pattern would be skewed left or right depending on which slit width is narrowed. To make this last point clearer, the distances from the maximum point of the central band to the points associated with the first minimum on each side will not be the same distance. If your familiar with the derivations, you may be able to create a simulation that models this.

EDIT: See link for derivation of double-slit diffraction: http://cnx.org/content/m12916/latest/
 
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  • #11
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Thank you buffordboy23, I don't have Mathcad but if you suggest, I can try. Anyway, I know those formulas and don't think have a problem about the intensity as function of angle from the central axis for diffraction. I can not imagine the effect of different width slits.

My thoughts are that the central peak would remain in the same location, and also, the intensity pattern would be skewed left or right depending on which slit width is narrowed.
Agree this.

I looked the link which you show. But It is for calculation mostly and much imported is for identical widths. After your words, thought again and reach such an approach: let's narrow one slit continuously. What will we observe? Initially, will observe a known Young interference+diffraction then finally only the diffraction of single slit(other slit almost closed). In time, we will observe a pattern between two of them. Thus, if I'm not wrong, only difference will the distribution symmetry of brightness. It will be favour of widest slit. Is it?
 
  • #12
548
2
I looked the link which you show. But It is for calculation mostly and much imported is for identical widths.

It appears from a simple glance that all you have to do is make some new substitutions for the two different slit widths. The intensity is then proportional to the square of the electric field at that point. It is an interesting question to ask (although I can't think of any useful applications for the result), so when I have the time I will develop a simulation to model the behavior and upload the results (it probably won't happen for a day or so).
 
  • #13
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Thank you again for to share your time and knowledge. Greets
 
  • #14
548
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I put together a quick and simple simulation and here are the results for slit widths of two different sizes (remember that intensity is proportional to the square of the electric field). As one slit is made larger while the other is held constant, the intensity of the central peak increases (because one slit has less spread in diffraction), and there appears to be no effect regarding the symmetry of the intensity plot with respect to the central axis or maxima.

I can post the full-derivation used in constructing the simulation. Just let me know and I will post it later.
 

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  • #15
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Thank you very much. Impressed two times. What a good software. But how did you use two different slit widths above formula? I would like to see the background of this graph, this is your skill, great.

Your graphs opposite of what we talk about. One of slit width is increasing opposite of decreasing. Thus, this help to comment on the opposite too. I suppose, we may say: Intencity will decrease and diffraction envelop will spread(I think).

On of impressive thing is, certain, hard symmetry of diffraction envelop. Honestly, this not exactly what I expected. Glad too see the graphs. This helped too much.

buffordboy23, don't know what will I say; how can I thank to you. My mind relaxed :) Can you show me the calculations and the thing what you used with that software? I liked it.
 
  • #16
Doc Al
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I put together a quick and simple simulation and here are the results for slit widths of two different sizes (remember that intensity is proportional to the square of the electric field). As one slit is made larger while the other is held constant, the intensity of the central peak increases (because one slit has less spread in diffraction), and there appears to be no effect regarding the symmetry of the intensity plot with respect to the central axis or maxima.
Are you saying that the location of the central maxima remains fixed regardless of the relative width of the slits? That seems a bit counter-intuitive. What if you kept reducing the size of one slit--and thus the intensity of the light through it--until it was essentially gone?
 
  • #17
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Your graphs opposite of what we talk about. One of slit width is increasing opposite of decreasing. Thus, this help to comment on the opposite too. I suppose, we may say: Intencity will decrease and diffraction envelop will spread(I think).

Yes, the graphs are opposite to what we talked about (I should of paid more attention) but qualitatively the results are the same. I verified this again.

On of impressive thing is, certain, hard symmetry of diffraction envelop. Honestly, this not exactly what I expected. Glad too see the graphs. This helped too much.

Me neither. I used the Fraunhofer Diffraction model as in the link that I submitted earlier.

Can you show me the calculations and the thing what you used with that software? I liked it.

Yes, when I get the time. This will be a somewhat lengthy endeavor, but it should be done so that the posted data can be criticized accordingly.
 
  • #18
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If a goes to 0 then the expression just becomes that for Youngs double slit

the quoted sentence above is from the page you show me. What does it mean? When does it for Youngs double slit? Which slit width required for Young? Supposing that, some slit widths essential for Young.
 
  • #19
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the quoted sentence above is from the page you show me. What does it mean? When does it for Youngs double slit? Which slit width required for Young? Supposing that, some slit widths essential for Young.

The equation for the intensity is in a nice compact form. There are two key factors in that equation: (1) alpha is related to the interference, and (2) beta is related to diffraction. If d goes to zero, then the two slits become one as they overlap (interference effects no longer exist), and similarly, if a goes to zero, the slit widths are now points (diffraction effects are now absent).

When I made the graphs, I did not have such a nice equation to work with. I had to approximate a time-averaged value of the electric field. I will explain all when I get the time.

EDIT: Do you know how the author arrived at the very first equation in that link that I provided? This will help me with determining where I should start with the derivation. See single slit diffraction in the Related Material links along left-hand side of the webpage for more help: (http://cnx.org/content/m12915/latest/)
 
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  • #20
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I suppose alpha is about the place of selected point and beta is about path difference. At least so far understood this.

EDIT: Do you know how the author arrived at the very first equation in that link that I provided?
if you mean the first link you gave about first equation then I suppose it is a superposition equation for two slits. Of course there is path difference term. BTW, working already.
 
  • #21
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Here is a brief outline of the derivation that I used to produce the graphs. Sorry that it took so long, but I don't really have the time to go into full detail so consult a text, like "Optics" by Hecht, if you need more details. In fact, I used that text as my basis and accepted all of the approximations used in the derivation. Doc Al brought up a good point, but I haven't had a chance to really think about it.

From "Optics" by Hecht, Eq (10.22) states that

[tex] E = C \int^{\frac{b}{2}}_{-\frac{b}{2}} F\left(z\right) dz + C \int^{a+\frac{b}{2}}_{a-\frac{b}{2}} F\left(z\right) dz[/tex]

where

[tex] F\left(z\right) = sin \left[ \omega t - k\left(R-zsin \theta\right)\right] [/tex]

C is the secondary source strength per unit length along the z-axis (which is perpendicular to the slits) divided by R (which is measured from the origin to a point P on the screen and is assumed to be a constant in the Fraunhofer approximation, "far-field"), a is the slit separation distance, b is the width of both slits, and k is the angular wave number. If you look at Eq (10.22), notice that it is the sum of two individual electric fields (one for each slit). So, if we replace b with c, such that b does not equal c (permits two slits with different widths), in the limits of the second integrand, and integrate (used Maple) we get the electric field as a function of time and angle:

[tex] E\left(\theta, t\right) = E_{1}\left(\theta, t\right)+E_{2}\left(\theta, t\right) [/tex]

where the individual fields are (after simplification)

[tex] E_{1}\left(\theta, t\right) = \frac{bC}{\beta}sin\left(\beta\right)sin\left(\omega t - kR\right) [/tex]

[tex] E_{2}\left(\theta, t\right) = \frac{cC}{\gamma}sin\left(\gamma\right)sin\left(\omega t - kR\right+2\alpha) [/tex]

where

[tex] \alpha = \frac{ka}{2} sin \theta [/tex]

[tex] \beta = \frac{kb}{2} sin \theta [/tex]

[tex] \gamma = \frac{kc}{2} sin \theta [/tex]

Now, intensity is proportional to the square of the electric field and to get the intensity as a function of theta, we integrate the square of the electric field with respect to time. Depending on the computer power available, this may be difficult (it was in my case). So, what you can do is choose specific time values and take the average.
 
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  • #22
Born2bwire
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Another way to get the results is to do a FDTD simulation using the Yee algorithm. It will have the advantage that it is a full wave solver that does not need to make the Fraunhofer approximation. Depending on the wavelengths involved, the Fraunhofer approximations are usually only valid within a cone. If you try to use it for solutions that are greatly displaced horizontally to the main axis of the radiation then the solution can break down. This usually isn't a problem as long as you are mindful of the conditions required for the approximation to be accurate.

The nice thing about the FDTD is that you can make the code easily in Matlab and even create movies of the wave's propagation. So it makes a nice visual representation.
 
  • #23
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buffordboy23, I am a bit late but want to thank you again. Unfortunatly my math is not well enough. Anyway, tried to remember and also work to calculate, at least an approx. value for the same angle but opposite sides of central bright line. And wanted to tell "eureka" but not yet. By the way, recipe is in my hand :) and you gave it. I'm really grateful.

I hope will see what i expected as mentioned previously.

Born2bwire, your words sounds nice but heard FDTD simulation the first time. I mean don't know how to use it. But liked it too(as mathlab) however didn't much use both.
 
  • #24
Born2bwire
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I actually uploaded some examples of single slit diffraction that I did using FDTD (finite-differenc time-domain).



This took maybe a minute or less to simulate and then another minute or two for Matlab to process the data and generate the movie. Doing double slit test would just require a quick modification of two lines of code.
 
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  • #25
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Born2bwire, thank you for sharing. That program and algo are already a good toy :) But a good educational surely.

At "Single Slit Diffraction 7" look like there are very close tree wave source between slit and suppose this is reason of almost a line toward slit. There are both diffraction and interference in this simulation.

Well, I wonder what is the difficulty or possibility of seeing the results of equations above what buffordboy23 shown. Or generally, can I see the what I am hunting? After all, your simulations are not for optic(light). They are plane waves in pool.
 
  • #26
Born2bwire
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The last two simulations have a sheet of current source that will produce a plane wave. It is not actually a true plane wave because I stopped the source at the border of the PML (Perfectly Matched Layer). A PML is a lossy layer on the boundary that absorbs any waves that transmit into it but at the same time minimizes the reflections off the boundary of the layer. The idea is that it will allow your simulation to mimic being in infinite space. Otherwise, all the waves will bounce off the boundaries of the FDTD grid and cause all sorts of problems (unless you want to simulate cavity). Since I truncated the source the plane wave is a little distorted near the edges.

The point of doing a plane wave excitation is to show that the sub-wavelength slit will always reproduce a wave as if it was a point source regardless of the excitation. In addition, the plane wave more explicitly shows the edge diffraction in the wide slit.

I could try doing the double slit and reproduce the same results. I have a time-domain code so I need think about how I will process the data but I think I can just take a snapshot at the point of maximum amplitude after the waves have settled in. I'll see what I can come up with.
 
  • #27
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The last two simulations have a sheet of current source that will produce a plane wave. It is not actually a true plane wave because I stopped the source at the border of the PML (Perfectly Matched Layer). A PML is a lossy layer on the boundary that absorbs any waves that transmit into it but at the same time minimizes the reflections off the boundary of the layer. The idea is that it will allow your simulation to mimic being in infinite space. Otherwise, all the waves will bounce off the boundaries of the FDTD grid and cause all sorts of problems (unless you want to simulate cavity). Since I truncated the source the plane wave is a little distorted near the edges.
It is looking good already. You're right.

The point of doing a plane wave excitation is to show that the sub-wavelength slit will always reproduce a wave as if it was a point source regardless of the excitation. In addition, the plane wave more explicitly shows the edge diffraction in the wide slit.
Agree.

I could try doing the double slit and reproduce the same results. I have a time-domain code so I need think about how I will process the data but I think I can just take a snapshot at the point of maximum amplitude after the waves have settled in. I'll see what I can come up with.
I will be grateful. BTW, http://physics.neiu.edu/vpl/optics/diffraction.html this is a good one and there is the source code too. Source code is in link at top of page.
 
  • #28
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I suppose the problem solved in my mind. While narrowing one slit, places of fringes will not change and intencity symmetry will conserve. The only difference will be the the total intencity for any place of bright fringe. This is the same result what buffordboy23 show. In fact, he used math to solve and show to me but there were still one point as Doc_Al mentioned. Seemed there was contradiction for me. Now I suppose understood at last with your(every contributer) helps :)

Well, didn't work with math. As I said, my math is not good. So I will explain myself without it. Surely if I can. I'll try my best :)

I tried to think with water waves and ask myself that, what is the equivalent of narrowing one slit. If I am not wrong it is intencity of light coming from slits. Since it is proportional to amplitude, the equivalent of narrowing one slit is lessen amplitude of one of wave source in water waves.

amplitude difference between both wave sources can not change the place and symmetry of interference pattern. Everytime frequency, velocity and wave length will be the same. Again; amplitude difference will not change the place or type of fringe. It will be totaly symmetric.

Now it's time to talk about what Doc_Al mentioned. After thinking with water waves, It is easy to compare. Single source is not the same as double slit. The slit width(amplitude) does not effect anything, except any bright fringe intencity. If there are two sources then there is interference else there is nothing to compare.

Am I right?
 
  • #29
Born2bwire
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I don't necessarily agree with the results of buffordboy23. In subwavelength slits, the reradiated fields are that of a point source. But once you expand the slit width above a wavelength, then the field no longer acts like a point source but will carry through along its primary axis as if the slit was not there. Only on the edges and at angles away from the primary axis will the wave become diffracted.

So expanding the slit will increase the amplitude of the transmitted wave and it will make the wave along the primary axis more like the original wave. I would expect that the measured intensity would become asymetrical. Attached are my results, but my results are against a flat plane of 100 wavelengths from the screen. The width of the plane encompasses the pi/4 radian beamwidth that buffordboy24's simulation did. Amplitude won't change the points of inteference, but it will change the amplitude of the total wave at that point. The fields will drop off at around 1/r2 in intensity. Ignoring phase and such like that, we know that at a given point the distance from the two slits (except along the perpendicular bisector) will be different. If we are 5 m to the left of the midpoint then the radiation from the left hand and right hand slits will scale as [tex]1/r_l[/tex] and [tex]1/r_r[/tex] where [tex]1/r_l > 1/r_r[/tex]. This situation is perfectly reversed 5 m to the right of the midpoint. So it's obvious that if the slits radiated at different amplitudes, that the combined amplitudes should be asymmetric.

The only difference is that he is assuming distances beyond the Frauhenhoffer limit. For a radiator this is something like [tex]\frac{2D^2}{\lambda}[/tex]. The question is what is D then... If it's the size of the largest slit, then my first two simulations of 3 and 6 wavelengths is in the far-field but then the third is in the region between (I can't remember what that is called). But given the fact that we have two radiators I think D is on the order of 100 wavelengths which is not a far-field simulation I can do.

EDIT: Here are some videos I made (URL= http://www.youtube.com/watch?v=FeCV27-mk0I&fmt=18 in case they don't load).

https://www.youtube.com/watch?v=<object width="425" height="344"><param name="movie" value="http://www.youtube.com/v/FeCV27-mk0I&hl=en&fs=1"></param><param [Broken] name="allowFullScreen" value="true"></param><param name="allowscriptaccess" value="always"></param><embed src="http://www.youtube.com/v/FeCV27-mk0I&hl=en&fs=1" type="application/x-shockwave-flash" allowscriptaccess="always" allowfullscreen="true" width="425" height="344"></embed></object>

https://www.youtube.com/watch?v=<object width="425" height="344"><param name="movie" value="http://www.youtube.com/v/Xx-_Me4Rock&hl=en&fs=1"></param><param [Broken] name="allowFullScreen" value="true"></param><param name="allowscriptaccess" value="always"></param><embed src="http://www.youtube.com/v/Xx-_Me4Rock&hl=en&fs=1" type="application/x-shockwave-flash" allowscriptaccess="always" allowfullscreen="true" width="425" height="344"></embed></object>

Sorry for taking so long, I got so caught up with things I went ahead and revamped my code. Now it uses OMP, among other optimizations I made to the code, and can take input files so that you can simulate any PEC structure and current sources that you want in free-space. Yay. That means it's easy to do something with three sources that turn on and off with different frequencies and amplitudes in multiple scatterer environment.

https://www.youtube.com/watch?v=<object width="425" height="344"><param name="movie" value="http://www.youtube.com/v/GT1aRoauKm8&hl=en&fs=1"></param><param [Broken] name="allowFullScreen" value="true"></param><param name="allowscriptaccess" value="always"></param><embed src="http://www.youtube.com/v/GT1aRoauKm8&hl=en&fs=1" type="application/x-shockwave-flash" allowscriptaccess="always" allowfullscreen="true" width="425" height="344"></embed></object>
 

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  • #30
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Born2bwire these are more interesting. I liked first two simulations but couldn't understand the third. Are there three source?
That means it's easy to do something with three sources that turn on and off with different frequencies and amplitudes in multiple scatterer environment.
this is very good too. Seemingly this is the third simulation.

About the last three pictures, can you share the data(slits to screen, slit widths,...) while you get the graphs? And what is the horizontal axis there?

By the way, as you said above, if the slit widths wider than wavelenght then a diffraction pattern will occur with interference too. This is another difficulty to understanding something about observing interference. Maybe you give them to compare with what buffordboy23 give previously. I suppose, something will be easier to understand if we can rid of "single slit diffractions" from both.

About asymmetry, if we observe long distances from slit plane and if especially pay attention to more far distances from central fringe than we may observe an asymmetry about intencity. For my opinion, this is because of the angle between central line(perpendicular) and source-to-slit line. Those are not symmetric. Especially if slit widths wider(or close) than wavelength, "single slit diffraction" superpositions will not be symmetric for both sides.

So, I think taking slit widths much smaller than wavelength and observing for small angles(fraunhofer condition for Young exp.) is a good point to start. Nevertheless, I think I must think over again. Thank you Born2bwire for help me to understand.
 
  • #31
Born2bwire
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The images I posted are the maximum value of the electric field the x-z plane along changing x (now that I think about it, I really should have taken the maximum of the absolute value but since I have a single sinusoidal source this shouldn't matter). It was done emulating the same tests that buffordboy did. Two slits, 100 wavelengths apart. The only difference is that I needed to try and be in the far-field since he was using a Fraunhoffer (stupid optics people) approximation while my code is a full wave solver. So my plots are taken something like 133 wavelengths away from the slits. Not ideal but each time step only advances the wave by about 0.354 cells. So for a 1480 cell long problem space, it takes 8372 time steps for the wave to fully traverse the length and back again. So time is a bit of a problem, it took around 10-20 minutes to run 8000 time steps for a 1200x1480 problem space. I might be able to improve on this, I use field-splitting for the perfectly matched layer at the boundary but technically I only need to split the field in the PML layer, which is only 25 cells deep. So I could save a lot of memory and a one or two memory access and writes for each cell per iteration by fixing this. The only other difference is that his plots are respect to the angle where mine would be the equivalent of projecting his plots onto a flat plane.

The details of the movies are given in the description of the videos but, they are a 500x500 cell problem space with plane wave sources. Each cell is 0.25 wavelengths (125x125 wavelengths). The slits are separated by 200 cells (5 wavelengths) and are 1 cell (1/25 wavelength) and 75 cells (3 wavelengths) wide respectively. I wanted to show you how different a sub-wavelength and sup-wavelength slits are. These are also in the near-field but you can see how quickly the interference pattern sets up. You could print these out, draw and measure the angles and get a favorable comparison with what you would expect from theory.

I think though, that the primary reason for asymmetry is the fact that a larger slit will have a much larger amplitude sourcing from it. The first two videos I posted are on the same scale, so you can see how much more area the resulting waves from the larger slits clip the scale. If the field values are greater than the limits of the scale, then the field is just shown as red/blue.
 
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  • #32
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Born2bwire, thank you again. Actually, your last three graphs are fit to my previous expectation. I want to be sure. But I want to learn the amount of diffraction effects to results from large slits. I don't know the math you used to come this result(graphs). Can you try the same calculations with smaller slit widths? For example, left slit 0.05 wavelength and right one 0.1wavelength widths. Of course, a 0.1-0.1 wavelength couple slit widths graph would be better to compare. And every time slit spacing 100 wavelength. I select these because they are enough values to avoid diffractions. Again; I don't want simulations because these are very small values to draw easily. I will be contented with picture of graphs.

And finally, if possible I would like to learn the equation(s) especially which contains slit widths. I suppose you didn't use the same equations which buffordboy used(showed). Otherwise the graphs would be identical.
 

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