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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.