Double slit experiment (one slit closed)

1. Apr 27, 2009

Volcano

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?

2. Apr 27, 2009

buffordboy23

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. Apr 27, 2009

Volcano

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.

4. Apr 27, 2009

Staff: Mentor

Last edited by a moderator: Apr 24, 2017
5. Apr 27, 2009

Volcano

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. Apr 27, 2009

Volcano

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. Apr 28, 2009

buffordboy23

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. Apr 28, 2009

Volcano

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.
Agree. This is ok.

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.

9. Apr 28, 2009

buffordboy23

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:

$$I\left(\theta\right) = I_{max}\left(\frac{sin\left(\alpha\left(\theta\right)\right)}{\alpha\left(\theta\right)\right)}\right)^{2}$$

where

$$\alpha\left(\theta\right)\right) = \frac{\pi a}{\lambda}sin\left(\theta\right)$$

"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.

Last edited: Apr 28, 2009
10. Apr 28, 2009

buffordboy23

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/

Last edited: Apr 28, 2009
11. Apr 28, 2009

Volcano

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.

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. Apr 28, 2009

buffordboy23

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. Apr 29, 2009

Volcano

Thank you again for to share your time and knowledge. Greets

14. Apr 29, 2009

buffordboy23

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.

Attached Files:

• Slit Width Graphs.JPG
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15. Apr 30, 2009

Volcano

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. Apr 30, 2009

Staff: Mentor

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. May 1, 2009

buffordboy23

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.

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

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. May 3, 2009

Volcano

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. May 3, 2009

buffordboy23

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/)

Last edited: May 3, 2009
20. May 4, 2009

Volcano

I suppose alpha is about the place of selected point and beta is about path difference. At least so far understood this.

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. May 10, 2009

buffordboy23

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

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

where

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

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:

$$E\left(\theta, t\right) = E_{1}\left(\theta, t\right)+E_{2}\left(\theta, t\right)$$

where the individual fields are (after simplification)

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

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

where

$$\alpha = \frac{ka}{2} sin \theta$$

$$\beta = \frac{kb}{2} sin \theta$$

$$\gamma = \frac{kc}{2} sin \theta$$

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.

Last edited: May 10, 2009
22. May 10, 2009

Born2bwire

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. May 14, 2009

Volcano

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. May 15, 2009

Born2bwire

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.

Last edited by a moderator: Sep 25, 2014
25. May 15, 2009

Volcano

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.