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we take long, narrow slits for diffraction, I am convinced that the slit should be narrow for the diffraction to occur as the waves spread out vertically,

but I can not understand why do we take long slits,the book sears and zemansky's says it is to avoid the horizontal spreading as the horizontal dimension is relatively large,

how does the diffraction pattern vary with respect to the length of the slit, thanks.

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    $\begingroup$ I don't understand certain bits of your post, but if you have short slits (e.g. diminute squares), you would end up with diffraction in both dimensions and the output image would be a little messy for academic purposes. Here you can see the expression in the Fraunhofer approximation for a rectangular slit among other examples. $\endgroup$
    – Lith
    May 25 '19 at 17:10
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The answer to your question is spatial coherence. It describes the correlation between waves at different points in space, lateral or longitudinal.

Now as light's partial waves pass through both slits, and interfere, you get an interference pattern. But to have this, you need very small relative path difference between the partial waves that pass through both slits.

There are a few main reasons why your interference pattern would disappear:

  1. Now in this case you need spatial decoherence, and if you have a wide slit, partial waves can pass many ways, and to the screen many path lengths

  2. To have an interference pattern, you need your slits relatively far apart. If your slits get close, the pattern will disappear. As your slits would get wider, the (far edges of the) slits would get naturally farther apart, since you cannot get wide slits and have their far edges close at the same time. If the far edge of the slits gets farther, and you need to keep the close edges far too (remember to have interference, you need the slits far apart), what you will get is two wide slits really far apart, which will cause the light to become spatially coherent again like natural Sunlight (remember you need spatial decoherence to have interference).

  3. The very prerequisite for the double slit experiment is to have a point like lightsource in front of the slits, this can be done with laser or a single slit before the double slits. Of you have wide slits, far apart, you will destroy what you achieved with the laser or the single slit in front of the double slits.

  4. This is why double slit experiments do not work with direct Sunlight, without a single slit in front of the double slits, or without a laser.

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This is not quite correct:

... the slit should be narrow for the diffraction to occur as the waves spread out vertically,

In fact, the slits are narrow to reduce horizontal blur, and thereby to increase fringe contrast. The amount of blur is approximately proportional to the width of the slits, but the spacing of the fringes is approximately inversely proportional to the spacing between the slits.

but I can not understand why do we take long slits,the book sears and zemansky's says it is to avoid the horizontal spreading as the horizontal dimension is relatively large, how does the diffraction pattern vary with respect to the length of the slit, thanks.

Two pinholes will produce essentially the same set of fringes produced by two slits, but the fringes would be very dim. Two more pinholes just above the first two pinholes will produce the same fringes added to the first set of fringes, but offset vertically by the vertical spacing of the two sets of pinholes. The fringes become brighter as more such pinhole sets are added vertically.

Adding pairs of pinholes like that, with very small vertical spacing, is precisely equivalent to simply stretching the first two pinholes vertically to make long vertical slits.

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  • $\begingroup$ With a bright source (easy in this age of lasers) the two-pinholes pattern is plenty bright enough. But it also has a complicated 2D structure. Slits give you an effectively 1D experiment which corresponds well to the 1D treatment seen in introductory books like S&Z. Likewise, with a monochromatic source (lasers, again) you get passable fringe contrast over a wide range of slit widths. $\endgroup$ May 25 '19 at 17:49
  • $\begingroup$ I wouldn't call the structure of a two-pinhole interference pattern complicated. It is composed of slightly curved fringes, which closely approximate linear fringes. $\endgroup$
    – S. McGrew
    May 25 '19 at 18:05
  • $\begingroup$ Use a brighter source so that you can see the part of the pattern that lies far from the line joining the two pinholes. Not that the explanation of that complexity is itself complex, but if you in the context of a introductory text you are trying to simplify as much as possible. $\endgroup$ May 25 '19 at 18:17

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