# Young's double slit experiment, slit separation

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

• That depends on the detail of your calculation. Do you account for the width of the slits in your calculation? Or do you treat them as infinitesimally thin? I'm guessing the latter, in which case that should answer your question... – lemon Nov 14 '17 at 15:26
• Well...it looks a bit complex because there are some terms i'm not really sure to understand...but the actual question was to find the fringe seperation...and from what you said I guess we were just meant to use the distance between the centres to solve the question Am i right? – Lee Nov 14 '17 at 16:16
• Yes the main fringe spacing is determined by the center to center spacing of the slits. – Floris Nov 14 '17 at 17:12

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 $w$).