All I got was two parallel bright fringes instead, like the ones you would get by shining a torch through two very thick slits.
It means the separation between slit is not close and the slits size is not small enough! Those two light beam must overlap to have interference. Small slit size is required to have large diffraction, the optimal slit size is certainly small than wavelength $\lesssim\lambda\approx0.5\mu m$ which gives you large diffraction. However, larger slits size is ok, but you have to (a) Make two slits as close as possible (b) move the setup far from the screen. You will know that it is enough when the light beam can overlap.
For the slit, you need better tools than a knife as well as a better material. First, you should use a shape cutter. Second, you need a material that can have a sharper edge such as film. I believe that film was used in the first few experiments of this kind. You have mentioned a hair is enough so $10\mu m$ should probably be ok, you just need to move the screen further away.
For the light source, you should always use a laser, since a high coherent light is required. Any laser out there is ok, it just cost 1 dollar and I can sure you can borrow a laser pointer near you. As I remember when I was doing Michelson Morley experiment, a tungsten light only gives interference pattern for $<0.1m$ with short coherent length, but a laser can have coherent length $>2m$. It means your life can be easier as you can use a 20 times larger slit with a laser!
Edit: Additional info on the methods Young used for this experiment.
The wiki about Young' interference experiment has quoted his paper on "On the nature of light and colours" (Also around page p.140 in the book Method and Appraisal in the Physical Sciences). The relevant excerpt is:
In order that the effects of two portions of light may be thus combined, it is necessary that they be derived from the same origin, and that they arrive at the same point by different paths, in directions not much deviating from each other. This deviation may be produced in one or both of the portions by diffraction, by reflection, by refraction, or by any of these effects combined; but the simplest case appears to be, when a beam of homogeneous light falls on a screen in which there are two very small holes or slits, which may be considered as centres of divergence, from whence the light is diffracted in every direction.
So, I guess the experiments were carried out as follow:
- Light source: In a room with all windows covered with thick curtains so that the inside is completely dark. Then let a small beam of sunlight go in.
- Monochromatic light: Use prism to split light into different color (This is known method back to Netwon). To get high quality single frequency light, a slit is required in front of prism to get a narrow sunlight beam.
- Point source of monochromatic light: Add another slit to get the required color (S1 in Fig. 1), the output monochromatic light is therefore from a single point source.
- Interference: Add another double slits (S2 in Fig. 1) so that the light can have two different path. Make sure that light from S1 falls on the slits S2. To ease observations, the screen should be far away.
Since his results cover all color, so it is very likely that he used sunlight rather than other light source such as candle (There was no light bulk at that time). Also, there is no diffraction grating, so it is likely that he was just using a simple prism.
For home experiments carried out these day, we can use LED as a monochromatic light source so that step 1 and 2 can be skipped. If you use a torch, you still need the step 2.
I think the experiment you are proposing is not possible in the way you want it.
Let us say we produce two photons in an electron-positron-annihilation with total momentum zero. (Since I don't see an easy way to produce entangled electrons I will talk about photons here, but I think it is not important for the argument). Those two photons are of course entangled in momentum: if one has momentum $\vec p$ the other one has momentum $-\vec p$.
But in order to make this statement you have to make a moemntum measurement on the initial state, i.e. know that the total momentum is zero with a certain $\Delta p$. But then, by means of the uncertainty relation, you only know the position where the photons were emitted with an uncertainty $\Delta x \propto (\Delta p)^{-1}$.
Now you can have two scenarios:
Either your double-slit is small enough and far enough away that due to the uncertainties $\Delta p$ and $\Delta x$ you do not know through which slit your photon goes. Or you still can tell (with some certainty).
In the second case there will never be an interference pattern. So no need for entanglement to destroy it.
But in the first case, due to the uncertainty $\Delta x$, measuring the position (by determining which slit your photon takes) does not give you an answer about the entangled photons position that is certain enough to tell which slit it will go through. Therefore you will see interference on both sides.
So an EPR like measurement is not possible in the experimental setup you propose.
I would assume that in general you need commuting observables, like spin and position in the Stern-Gerlach experiment, in order to measure EPR. But I didn't think that through yet.
addendum, 03-19-2014:
Forget about the second photon for a while. The first photon starts in a position state which is a Gaussian distribution around $\vec x_0$ and a momentum state which is a Gaussian around $\vec p_0$. After some time $t$ its position has evolved into a Gaussian of $\mu$ times the width around $\vec x_0 + \vec p_0 t$ (mass set equal to 1) while the momentum state is now $1/\mu$ times the width around $\vec p_0$. So while your spatial superposition gets larger - and thus better to measure with a double slit - the superposition in the momentum state, in which you have entanglement, gets smaller. You don't gain anything from entanglement, since your momentum wave-function is so narrow, that you know the momentum anyways.
It is actually not important to have space and momentum for this. Just take any non-commuting observables A and B, say with eigenstates A+, A-, B+, B-, and take two states S1 and S2 that are entangled in A. So measuring S1 in A+ implies S2 in A- and vice versa. But what you want is measure if S1 is in B+ or B- and from this conclude if S2 is in B+ or B-. And since A and B do not commute, measuring B with some certainty gives you a high uncertainty on A, meaning, for knowing if S1 is in B+ or B- you completely loose the information if it is in A+ or A-. So you cannot say anything about S2. On the other hand, as long as you are still in an eigenstate of A and know what to expect for the A measurement of S2, you don't know anything about the result of the B measurement.
So in order to do an EPR experiment you need entanglement in the observable you measure or an observable that commutes with it.
Please tell me if my thoughts are wrong.
Best Answer
After you put a detector, the basis of states becomes a tensor product of the states of the particle (with the two states meaning: came from this or that slit) and the ones of the detector (went through this slit or not).
So the cross terms that appear when you evaluate the probability of having reached a given position at the final screen coming from one or another slit, are now orthogonal (there is no overlapping of the detector states, otherwise the detector would be useless) and don't contribute to the probability, so you recover the classic result.
So basically, you'd get a no-fringe pattern whether or not you let multiple electrons through at a time, as long as you lace a detector
And this gives rise to a very interesting question: because, if you put the detector after the screen (for example a very fine, ideal, bubble chamber that could give you the direction the electron had) the result would be the same, and that is deeply counterintuitive (the classical physicist would scratch his/her head wondering: does the partice see into the future too?)