2 slits : diffraction always "enveloppe" of interference ?

[DoobleD]

In the double-slit diffraction experiment, two interferences are observed simultaneously : diffraction interference from each single-slit, and double-slit interference (where in the double-slit interference, diffraction is ignored because we considered theoretical slits way smaller than wavelength).

Intensity of the light reaching the screen behind the slits then looks like that :

My questions are :

1 - why is it the diffraction part the envelope of the double slit interference, and not the other way around ?
2 - Is it actually always in this way only ?

I've looked at 5 textbooks (undergrad level) and none of them justifies that. They all say something like : the diffraction act as an envelope, and the corresponding equation is the product of both equations for the single-slit diffraction and for the double-slit interference together.

 

[DoobleD]

Actually it kind of makes sense somehow because in the other way around, well, the light due to double-slit interference does not decrease in intensity on "left" and "right" of the screen. So I guess somehow, only the diffraction interference part, which has a decreasing intensity, can be the envelope. But that "somehow" is not really physical, I can't see the logical/physical justification of it.

 

[jtbell]

The width of the single-slit envelope varies inversely with the width of the slit(s): wider slits produce a narrower envelope and vice versa. Similarly, the spacing between the interference maxima varies inversely with the spacing between the slits.

In a multiple-slit setup, the spacing between the slits (center to center) has to be larger than the width of the slits, so the spacing between the interference maxima has to be smaller than the width of the diffraction envelope.

 

[DoobleD]

The width of the single-slit envelope varies inversely with the width of the slit(s): wider slits produce a narrower envelope and vice versa. Similarly, the spacing between the interference maxima varies inversely with the spacing between the slits.

In a multiple-slit setup, the spacing between the slits (center to center) has to be larger than the width of the slits, so the spacing between the interference maxima has to be smaller than the width of the diffraction envelope.

Very clear answer, as usual, thank you !

You wrote "the spacing between the slits (center to center) has to be larger". Why does it have to be that way? If we imagine a setup with space between slits shorter than width of each slit, then the envelope would be the slits interference instead of the diffraction right?

EDIT : I just realized that when measuring the spacing between the slits center to center, indeed the distance measured that way cannot be smaller than the width of the slits. Thus, the distance between each pair of wavelet source (one in each slit) is always greater than the width of one slit. You can ignore my last question, sorry.