= Solution
The Fourier expression is the scalar <Fraunhofer diffraction> approximation for coherent, effectively monochromatic illumination with a uniform incident <plane wave> over the <aperture>. At a distant screen the aperture-dependent quadratic propagation phase must be negligible, roughly $D^2/(\lambda z)\ll1$ for transverse size $D$. Alternatively, a <thin lens> forms the angular <Fourier transform> in its back focal plane. That focal-plane construction is the appropriate one here; the aperture-to-lens distance need not itself be an enormous free-propagation distance. <aperture> features should be large compared with <wavelength> for this simple scalar transmission-mask model.
If complementary transmissions are $h$ and $1-h$, linearity gives $U_h+U_{1-h}=U_{\rm free}$. Away from the undiffracted forward beam, the uniform plane-wave contribution $U_{\rm free}$ vanishes in angular space, so $U_{1-h}=-U_h$ and their intensities agree. This is <Babinet's principle>. \b[It equates the off-forward diffracted intensities, not the transmitted power or the on-axis intensity including the direct beam.]
Let $X,Y$ be screen coordinates conjugate to <aperture> coordinates $x,y$, so $p\simeq kX/f$, $q\simeq kY/f$. With width $a=0.2\ {\rm mm}$ along $x$ and height $b=0.5\ {\rm mm}$ along $y$, define $\operatorname{sinc}u=\sin u/u$. The complementary rectangle has transform
$$
H(p,q)=ab\,\operatorname{sinc}(pa/2)\operatorname{sinc}(qb/2).
$$
Thus the obstruction's off-forward scattered component has
$$
I_{\rm sc}(X,Y)\propto
\operatorname{sinc}^2\!\left(\frac{\pi aX}{\lambda f}\right)
\operatorname{sinc}^2\!\left(\frac{\pi bY}{\lambda f}\right).
$$
Its dark lines occur at
$$
\boxed{X=m(0.500\ {\rm mm}),\qquad
Y=n(0.200\ {\rm mm}),\quad m,n=\pm1,\pm2,\ldots.}
$$
The central-lobe zero-to-zero size is $1.00\ {\rm mm}$ by $0.400\ {\rm mm}$; it is wider in the direction conjugate to the narrower obstruction. The central forward image of the illuminating beam must be distinguished from this scattered envelope in a real screen observation.
The source figure gives centre-to-centre separation $D=1.0\ {\rm mm}$. Choose the three rectangle centres at $x=-D,0,D$; an overall translation changes only the transform phase. Their scattered amplitude is proportional to
$$
H_3(p,q)=ab\,\operatorname{sinc}(pa/2)\operatorname{sinc}(qb/2)
[1+2\cos(pD)].
$$
Therefore
$$
\boxed{I_{\rm sc}(X,Y)\propto
\operatorname{sinc}^2\!\left(\frac{\pi aX}{\lambda f}\right)
\operatorname{sinc}^2\!\left(\frac{\pi bY}{\lambda f}\right)
\left[1+2\cos\!\left(\frac{2\pi DX}{\lambda f}\right)\right]^2.}
$$
Along the screen's $x$ direction, the three-element interference factor has principal maxima separated by \b[$\lambda f/D=0.100\ {\rm mm}$], under the single-rectangle envelope. Its zeros lie at $X=(m+1/3)(0.100\ {\rm mm})$ and $(m+2/3)(0.100\ {\rm mm})$. It also has weaker secondary maxima between principal orders. Every fifth principal order coincides with a single-rectangle zero and is missing; multiplication by the envelope slightly shifts maxima away from the nominal interference-factor positions. The figure plots the calculated scattered component and marks the separate direct-beam limitation.
\Image[/past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2016/paper-2-diffraction.png]
{title=Original calculated rectangular-obstruction and three-element Fraunhofer scattered intensities}
{description=Screen coordinates are physical conjugates of the obstruction coordinates. All plotted values describe the scattered component; the unresolved undiffracted illumination at the origin is excluded. Intensities are normalized, and no source figure is reproduced.}
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