Past exam of the physics 2 paper of the University of Oxford / 2013 resit / 8

In the Fraunhofer approximation, contributions from positions $\xi$ across the slit have relative phase $k\xi \sin \theta$. With uniform illumination, the field relative to its axial value isThus the single-slit diffraction intensity isOn the detector $\sin \theta =y/\sqrt{d^2+y^2}$; near the axis, $\beta \simeq \pi wy/(\lambda d)$. The central maximum takes the continuous limit $\sin \beta /\beta \to 1$, and the minima satisfy $w\sin \theta =m\lambda$ for nonzero integers $m$. The stated far-field condition justifies adding the aperture contributions with a linear phase.

The first single-slit minimum is at $\sin \theta =\lambda /w$, so the characteristic angular resolution of an aperture of width $D$ is of order $\lambda /D$. With $D=1$ m, $\lambda =550$ nm, and distance $\mathcal{D}=10^8$ m, this slit-based estimate isFor the actual circular lens, the Rayleigh criterion uses the first zero of the Airy disk, $\Delta \theta \simeq 1.22\lambda /D$. It gives $6.71\times 10^{-7}$ rad and a resolved separation of about $67$ m. The factor $1.22$ belongs to the circular aperture, not to the one-dimensional slit derivation.

For normal incidence, adjacent grating lines have optical path difference $s\sin \theta$. Constructive interference at order $m$ requires $s\sin {\theta }_m=m\lambda$. Differentiating at fixed $m$ yields the angular dispersionIt is zero in the zero order and increases in magnitude toward grazing diffraction. An order exists only if $|m\lambda |\le s$. The number of illuminated lines controls peak width and resolving power, but does not alter this angle-versus-wavelength relation.