# Distance between two slits in the double slit experiment

I was looking at some problem set in MIT QM course, and noticed the following problem:

Electrons of momentum $$p$$ fall normally on a pair of slits separated by a distance $$d$$. What is the distance, $$w$$, between adjacent maxima of the interference fringe pattern formed on a screen a distance $$D$$ beyond the slits? note: You may assume that the width of the slits is much less than the electron de Broglie wavelength.

Now the deveriation of the answer itself isn't too difficult, but I'm confused about the note (italicized).

If I assume the width of the slits is much less than the wavelength, namely $$d << \lambda$$, then in $$\sin{\theta}= \dfrac{m \lambda}{d}$$, since $$\dfrac{\lambda}{d}$$ is a large number, there would be no angle that satisfies the equation. I feel like $$d$$ just cannot be smaller than $$\lambda$$. Also, if I think physically, it'd be very hard to create two slits that are only separated by such a short length.

What am I missing? I just feel like I'm misreading or misunderstanding something basic here. Can anyone enlighten me?

$$d$$ is the distance between the slits, not the width of the slits themselves. The note at the end of the problem concerns the width of the slits. As the slits are narrower than the electron de Broglie wavelength, only a single electron (or rather a column of electrons) can pass through the slit at a time. The distance between the slits, $$d$$, is not constrained by the problem.