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I just wrote a test and there was a question on young's double slit experiment...we were given the width of the slits and the sepration between their centres
My question is...In making calculations involving slit seperation,say "a"...will "a" just be the seperation between their centres or is it necessary to subtract the value of the width from the sepration between the centresto get the value of a...I checked my textbook but there isn't anything about that
Usually when you compute the diffraction pattern of two slits of finite width, you consider the aperture to be the convolution of two infinitesimal slits, and a finite aperture. Now as you may know, the Fraunhofer diffraction pattern is the Fourier transform of the aperture function. The convolution theorem tells us that the Fourier transform of a convolution of two functions is the product of the Fourier transforms of the individual functions.
What that means is that if you consider the centers of the slits to be a distance $a$ apart, and they each have a width $w$, then the diffraction pattern will be the product of the pattern you expect from two infinitesimal slits that are distance $a$ apart, and the diffraction pattern of a single aperture of width $w$.
This will look like a cosine function (spacing determined by $a$) whose amplitude is modulated by a sinc function (shape determined by $w$).
See for example this question.
