The source is large, spatially incoherent. Each point $P$ of the source creates its own fringe system. For a non-centered point of the source, identified by ${{x}_{P}}$ the additional path is ${{\delta }_{P}}=\frac{{{x}_{P}}D}{S}$
The fringes remain visible if ${{\delta }_{P}}\ll \frac{\lambda }{2}$
At worst, for points at the edge of the source ${{x}_{P}}=\frac{s}{2}$ , which gives $\frac{s}{2}\frac{D}{S}\ll \frac{\lambda }{2}$ or $\frac{s}{S}\ll \frac{\lambda }{D}$
This criterion is simple $\frac{s}{S}$ because is the angle under which is seen the source if one is placed at the level of the two holes.
Sorry for my poor english !
If the slits are on top of each other, then the light travelling through each slit goes the same distance and therefore has the same phase.
In this case, the distance between the fringes is infinite.
On the other hand, if the slits are very far apart, then even a small angle incurs a large path difference, so the fringes are very close together.
Thus we have reasoned that the distance between the fringes goes down as $d$ goes up.
Consider the formula written by OP:
$$d = \frac{m \lambda}{\sin \theta} \, .$$
The first intensity maximum occurs when $m=1$ giving
$$ \sin \theta_1 = \frac{\lambda}{d} \, .$$
Expanding the $\sin$ to lowest order we get
$$\theta_1 = \frac{\lambda}{d}$$
which says that increasing $d$ makes the angle of the first maximum smaller, as we predicted above.
From this reasoning, we see that OP's formula is probably correct and that putting the $\sin$ in the numerator would give the wrong behavior.
Note that I didn't actually derive the correct answer, I just showed that moving the $\sin$ function from denominator to numerator would probably be incorrect.
That said, given the definitions in the question, the correct formula for the maxima of the two slit interference is in fact
$$d = \frac{m \lambda}{\sin \theta}$$
as written by OP.
Best Answer
If d>D, practically none of the light from the two slits will overlap on the screen. Without overlap there is no interference.
If the slits are extremely thin, light will spread over a wider angle from each slit, so D can be smaller and still allow overlap -- and interference -- between the light from the two slits.
To get a good sense of how the slit width affects the beam spread from a slit, see Fraunhofer diffraction.