Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 338 / 2 / c

Choose photodetector $x$ along increasing main grating dispersion, and $y$ along increasing wavelength for the cross-disperser. The echellogram shows two order traces and their intersections with the $y$ axis, which passes through the optical camera optical axis.

For fixed incidence, $d\beta /d\lambda =m/(d\cos \beta )>0$, so increasing wavelength goes to the right along each order. The cross-disperser also sends increasing wavelength upward in the chosen orientation; the traces therefore slope upward. At $x=0$, ${\lambda }_m=K/m>{\lambda }_{m+1}=K/(m+1)$, so the trace labeled $m$ lies above the trace labeled $m+1$. Reflecting either plotting axis would reverse its arrow but leave the optical content unchanged. The cross-disperser separates orders that would overlap in the main dispersion direction.

Let ${\beta }_0$ be the direction of the optical camera optical axis. The grating equation on that axis isThe central wavelengths in adjacent diffraction orders are therefore ${\lambda }_m=K/m$ and ${\lambda }_{m+1}=K/(m+1)$, with separation$K$ is the fixed order-times-wavelength product directed onto the optical camera axis, or the optical path difference between adjacent grooves in that direction. In a Littrow configuration, $K=2d\sin \alpha$. This adjacent-central-wavelength spacing is a conventional free spectral range of an echelle grating; at high order it is approximately ${\lambda }_m/m$.

There is a qualification in the wording of the printed definition. The exact interval for which order $m$ lies closest to the $y$ axis is not, in general, identical to the spacing of two adjacent order centres. For a concrete counterexample take ${\beta }_0=0$, $0<\alpha <\pi /2$, and a photodetector mapping $x=f_{\mathrm{cam}}\tan \beta$. Equality of the distances from the axis for orders $m,m+1$ means ${\beta }_m=-{\beta }_{m+1}$, so their common boundary is $\lambda =K/(m+1/2)$. The other boundary with order $m-1$ is $K/(m-1/2)$. Consequently the literal closest-order interval has widthfor $m\ge 2$ and accessible non-grazing orders. This differs from the printed $K/[m(m+1)]$. For a nonzero ${\beta }_0$, the exact boundary is determined by ${\beta }_m+{\beta }_{m+1}=2{\beta }_0$ and need not even have the simple half-order form. The two definitions agree to leading order $K/m^2$ when $m$ is large. Thus the printed expression is correct as an adjacent-centre convention or a high-order coverage estimate, rather than an exact consequence of the literal nearest-axis definition. Gap-free coverage should be checked using the actual photodetector mapping and neighbouring order endpoints.