Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 338 / 1 / b

The echellogram contains nearly parallel traces, one for each diffraction order. Choose the main echelle grating dispersion to increase wavelength toward the right and the cross-disperser to increase wavelength upward. At a fixed horizontal coordinate, order $m+1$ has a shorter wavelength than order $m$, so it lies below order $m$ in this convention.

Let ${\beta }_0$ be the central camera direction. The grating equation fixesThe usual adjacent-order free spectral range of an echelle grating is thereforeThe constant $K$ has dimensions of length and is the path difference between adjacent grooves for light directed along the camera axis; it is also the fixed product $m{\lambda }_m$. Near the Littrow configuration, $K=2d\sin \alpha$. The detector width corresponding to this interval is approximately $qS$, with grating dispersion $q$.

There is a convention issue in reading the question literally. Adjacent order-center separation is exactly the expression above. A partition assigning each wavelength to whichever order lands closest to the vertical axis has boundaries halfway in detector displacement, not at adjacent order centers. In the small-angle detector approximation its boundaries for order $m$ are $2K/(2m+1)$ and $2K/(2m-1)$, giving width $4K/(4m^2-1)$. Both conventions give $S\sim K/m^2$ for high orders, but their exact finite-$m$ widths differ. The quoted expression is the conventional adjacent-order spacing.