But I looked around and could not find any information relating the effect of the thickness of the slits when they are particularly small.
When the slits have small width (size in direction perpendicular to propagation), then the depth of the slit (size in direction of propagation) can have a pronounced effect on the intensity of output light. Namely, suppose the diaphragm with the slits is made of metal. Let incident light wave be polarized so that $\vec E$ vector is parallel to the slit, and let the wavelength be larger than twice the slit width. Then the wave wouldn't be able to be a standing wave between the slit walls, it would have to be "compressed" to have smaller wavelength to pass through still satisfying boundary conditions. This means that wave vector component in the direction perpendicular to propagation will be higher. The wave still has to maintain its frequency, so total wave vector magnitude must remain the same (because $|\vec k|=\omega/c$). Since (in two dimensions) we have
then to have $|k_x|>|\vec k|$, we must have $k_y^2<0$. This leads to an effect known as evanescent wave, where the wave, instead of oscillating like a sinusiod, falls off exponentially. So, as the depth of the thin slits increases, intensity of outgoing light will decrease exponentially.
If the wave is polarized in the direction parallel to slit width, there'll be no such effect, and the wave will pass virtually without change in amplitude, since boundary conditions on the metallic slit walls are "easier" in this case.
The shape of outgoing wave, unlike amplitude, won't change much: since the source is smaller than half wavelength, it's effectively a point source.