I did an experiment in which I tried to show that the visibility of the interference fringes is related to the relative slit width in a double slit interferometer. In other words if one slit had a width that was twice the width of the other, would that change the visibility of the interference fringes. I thought it would because I assumed that by changing the width of the slits the intensity of the light passing through the slit would change, but now I am not so sure about that. All of the equations I have seen regarding intensity in a double slit interferometer do not indicate a linear relationship between intensity and slit width.
The fringe pattern is simply the fourier transform of the slit aperture, in 1 or two dimensions.
The fourier transform of a single slit ~ $sinc(x/a)$ where a is the width of the slit.
The fourier transform of a double (equal) slit ~ $cos(x/a)$ eqn.(1)
The fourier transform of two slits , one twice the width of the other is something like $a*cos(x/a) + b*sin(x/a)$
Which, using well known trigonometric identities is simply eqn. (1) displaced slightly along the x direction
there is no change in intensity