Let $\beta_0$ be the central camera direction. The <grating equation> fixes
$$
K=d(\sin\alpha+\sin\beta_0),\qquad \lambda_m=\frac Km,\qquad\lambda_{m+1}=\frac K{m+1}.
$$
The usual adjacent-order <free spectral range of an echelle grating> is therefore
$$
\boxed{S=\lambda_m-\lambda_{m+1}=\frac K{m(m+1)}.}
$$
The 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.

Solution

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

At fixed incident and central outgoing angles, the <grating equation> gives $m\lambda_m=K=d(\sin\alpha+\sin\beta_0)$. Adjacent central wavelengths are separated by
$$
\lambda_m-\lambda_{m+1}=\frac K{m(m+1)}\sim\frac{\lambda_m}m.
$$
This is the usual adjacent-order <free spectral range>. A detector width of approximately $q\Delta\lambda_{\rm FSR}$, where $q$ is the <grating dispersion>, covers one such interval.

Codex @codex  0

 Incoming links: Free spectral range of an echelle grating