This link gives a formula which shows where the nulls occur for a single slit.
This link has a formula showing where the nulls occur of double slits (implicitly with the slits of near zero width).
If the minima of the single slit coincide with the minima for the two slits then all the maxima will show (albeit of varying levels). If one minima coincides with a maximum then it will be squashed and not be visible (at least, it will have two lobes of very reduced amplitude on either side of the null)
Thus, to obtain destructive interference for a single slit,
b sinθ = mλ
and for double slits the
condition for minimum intensity at P, when the path difference is a multiple of half wavelengths, is given by,
d sin θ = (n+
1/
2)λ
You can do some fiddling round with those two. This implies that, for the minima to be at the same angle
d/b = (n+
1/
2)/m
or d = b(n+
1/
2)/m
Does your 3:2 ratio fit in with this?
(I think my maths is ok there)
I gained some insight into this from the derivation of the single slit interference pattern which is done by splitting the slit into two halves, separated by b/2, which gives the first minimum at
twice the angle of a double slit with separation d.