 Solution

The convolution of two aperture functions in one dimension isWith Fourier transform convention $\hat{f}(q)=\int f(x)e^{-iqx}dx$, the convolution theorem statesThe first identity follows by substituting the convolution integral, putting $v=x-u$, and separating the resulting double integral into $\int f(u)e^{-iqu}du$ times $\int g(v)e^{-iqv}dv$. For an absolutely integrable pair the interchange is justified by absolute convergence; finite optical apertures satisfy suitable integrability assumptions.

In Fraunhofer diffraction, the observed field is the Fourier transform of the transmitted complex aperture field. For two identical slits, let $g(x)$ be one slit and use a pair of Dirac delta distributions to locate the centres:Consequently $\hat{t}(q)=2\hat{g}(q)\cos (qD/2)$. For a rectangular slit, $\hat{g}=a\mathrm{sinc}(qa/2)$, giving a single-slit diffraction envelope times the double-slit interference factor in intensity. A $\pi$ phase reversal replaces the plus between the deltas by a minus, giving $\sin (qD/2)$ instead, as in B7.

More generally, an $N$-slit array is a single-aperture function convolved with a finite comb. Its field contains the factorThis separates the narrow diffraction grating maxima from the individual-aperture envelope. The same reasoning holds in two transverse dimensions. Multiplying an existing aperture by an additional finite mask instead convolves their field transforms and can broaden features. These are operations on complex fields: intensity is the squared modulus of the final field and cannot generally be substituted into the field-level convolution theorem.