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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.

In the simple slit-image approximation, the projected slit width is $w=x{f}_{\mathrm{cam}}/f_{\mathrm{coll}}$. Equating this width to the separation $q\Delta \lambda$ of barely resolved features, using the grating dispersion, givesThis recovers the stated spectral resolving power under the assumption that the grating has unit anamorphic magnification.

For arbitrary distinct $\alpha$ and $\beta$, the anamorphic magnification of a grating must be included. At fixed wavelength, the grating equation gives $|d\beta /d\alpha |=\cos \alpha /\cos \beta$, so the slit image instead has width $w=x{f}_{\mathrm{cam}}\cos \alpha /(f_{\mathrm{coll}}\cos \beta )$. Thus the general slit-limited resolving power of a grating isThe two expressions agree in the Littrow configuration, $\alpha =\beta$. Without that condition or the unit-magnification approximation, the quoted $\cos \beta$ expression is not the general slit-limited result. Finite grating size, detector sampling, and optical aberrations can lower the actual resolution further.

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.

A slit of width $x$ at the focus of a collimator subtends $x/f_{\mathrm{coll}}$. Its projected width is $x{f}_{\mathrm{cam}}\cos \alpha /(f_{\mathrm{coll}}\cos \beta )$. Dividing this width by the grating dispersion gives $\Delta \lambda =xd\cos \alpha /(m{f}_{\mathrm{coll}})$. Thus the spectral resolving power is $R=m\lambda f_{\mathrm{coll}}/(xd\cos \alpha )$. Omitting the anamorphic factor gives a denominator $\cos \beta$ instead; these agree in the Littrow configuration.