Past exam of the physics A course of the University of Cambridge / 2018 / Paper 2 / Section B / B7 / e / Solution

For a rectangular slit centred at $x_c$, the one-dimensional Fraunhofer diffraction integral isThe two equal apertures have centres at $\pm D/2$; the glass subtracts their fields. ThereforeThis is phase-reversed double-slit diffraction: the narrow-slit interference factor is multiplied by the single-slit diffraction envelope. In the limit $a\to 0$, the normalized envelope tends to one, recovering part (d).

The interference minima and envelope minima obey, respectively,Near the axis these are ${\theta }_m\simeq 4m\mathrm{mrad}$ and envelope zeros ${\theta }_n\simeq 20n\mathrm{mrad}$. Here $D/a=5$, so every envelope zero already coincides with an interference minimum. The envelope strongly reduces the outer fringes; its central value is nonzero, but the total intensity still vanishes at $\theta =0$ because the two fields cancel. The sketch uses the small-angle coordinate; the equations above specify the exact angular zeros.