= Solution
In <Fresnel diffraction>, the propagation <wave phase> varies quadratically across the aperture and that curvature must be retained. For a transverse scale $w$ at propagation distance $L$, its size is of order $\pi w^2/(\lambda L)$. In <Fraunhofer diffraction>, that quadratic term is negligible, approximately $w^2/(\lambda L)\ll1$, while the linear phase encoding the observation angle remains. A lens can also form the far-field angular pattern in its <focal plane>.
For a normally incident <plane wave> and aperture transmission $h(x,y)$, the far-field <complex amplitude> is its <Fourier transform>:
$$
\psi(q_x,q_y)\propto\iint h(x,y)e^{-i(q_xx+q_yy)}\,dx\,dy,
\qquad q_x=\frac{2\pi}{\lambda}\sin\theta_x.
$$
The detected <light intensity> is proportional to $|\psi|^2$. For identical parallel slits, write the transmission as a sum of shifted copies of one slit. The <Fourier shift theorem> gives one single-slit transform multiplied by the sum of positional <phase factors>:
$$
\psi(q)\propto a\operatorname{sinc}(qa/2)\sum_j e^{-iqx_j},\qquad
\operatorname{sinc}u=\frac{\sin u}{u}.
$$
The <single-slit diffraction> envelope sets the broad scale, while coherent interference between slit centres produces the finer array pattern. One sums <complex amplitudes> before squaring, rather than adding independent slit intensities.
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