diffractiondouble-slit-experimentopticsvisible-light
My professor asked me (in my viva exam), "If, in the Fraunhofer single slit diffraction experiment, if we have 2 slits instead of one (at very short distances), What would happen?"
I answered with "Young's double slit experiment intensity distribution". He said my answer was wrong. Out of curiosity, what is the right answer then?
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:
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.
The answer I gave before was wrong, as was kindly pointed out by CuriousOne. The fading of the intensity isn't because of the $1/r^2$ fall-off of light, since the diffraction formulas are only for small angles anyway.
First of all, it's clear from the second figure in the question that the only relevant thing is the diffraction and not the interference; that is, we might as well consider a single slit. Also, since this is only a wave phenomenon, the wave-particle duality only comes into play if you think light is made of particles (which it is, with a certain definition of "particle"), but we don't need that here; the classical wave behavior of light can explain this perfectly well.
The main idea is that as we get farther away from the center of the diffraction pattern, the phase difference between all the rays coming from different parts of the slit gets bigger and bigger. When we are very near the center, the rays are almost in phase, and if we move a little bit, they are still almost in phase. But if we keep moving, they get a bit out of phase, and the interference between them all makes the intensity go down. As we get farther and farther away from the center, the phase difference gets bigger; and since the dependence of the phase with the distance from the center (or the angle) is different for all the rays, they won't get in phase again (as would happen if we had two very narrow slits). Simply put, the reason that away from the center the intensity goes down is that the waves coming from the slit are all out of phase with each other.
Best Answer
The intensity distribution will remain same as fraunhofer one only , however intensity will increase at each point . Means supposedly earlier there was Amplitude 2A , at a point of maxima , now it will become 4A , since 2 waves will come from the new fraunhofer slit also as it is not very distant from the previous slit , hence , you can assume the maximas and minimas of the new slit are located at same positions as of the previous slit .
So intensity will get 4 times at each place , but distribution will remain same .