Past exam of the physics A course of the University of Cambridge / 2016 / Paper 2 / Section B / B8

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 $D^2/(\lambda z)\ll 1$ 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 $U_h+U_{1-h}=U_{\mathrm{free}}$. Away from the undiffracted forward beam, the uniform plane-wave contribution $U_{\mathrm{free}}$ vanishes in angular space, so $U_{1-h}=-U_h$ and their intensities agree. This is Babinet's principle. 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 $a=0.2\mathrm{mm}$ along $x$ and height $b=0.5\mathrm{mm}$ along $y$, define $\mathrm{sinc}u=\sin u/u$. The complementary rectangle has transformThus the obstruction's off-forward scattered component hasIts dark lines occur atThe central-lobe zero-to-zero size is $1.00\mathrm{mm}$ by $0.400\mathrm{mm}$; 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 $D=1.0\mathrm{mm}$. Choose the three rectangle centres at $x=-D,0,D$; an overall translation changes only the transform phase. Their scattered amplitude is proportional toThereforeAlong the screen's $x$ direction, the three-element interference factor has principal maxima separated by $\lambda f/D=0.100\mathrm{mm}$, under the single-rectangle envelope. Its zeros lie at $X=(m+1/3)(0.100\mathrm{mm})$ and $(m+2/3)(0.100\mathrm{mm})$. 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.