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Let the telescope and collimator have focal lengths $f_t,f_c$, and let $w=f_t\theta$ be the physical slit width. The collimated beam diameter in the dispersion direction is $A_c=f_{c}D/f_t$, so $w{A}_c/f_c=\theta D$.

The diffraction grating changes both the angular width and the beam diameter. At fixed wavelength, differentiating the grating equation gives $|d\beta /d\alpha |=\cos \alpha /\cos \beta$. Thus the anamorphic magnification of a grating givesIf $W$ is the illuminated surface length, its projected beam diameters are $A_c=W\cos \alpha$ and $A_{\mathrm{cam}}=W\cos \beta$. Multiplication cancels the anamorphic factors:ThereforeThis is a one-dimensional optical-invariant relation: a grating cannot independently magnify the slit and shrink the corresponding beam without compensating angular changes. The calculation uses local paraxial imaging about each instrument’s chief ray and an unclipped beam.

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.