 Solution

Let $f_t$ be the optical telescope focal length, $x=f_t\theta$ the physical optical slit width, $f_{\mathrm{coll}}$ the collimator focal length, and $A_{\mathrm{coll}}=D{f}_{\mathrm{coll}}/f_t$ the incident collimated beam diameter. For the grating equation $m\lambda =d(\sin \alpha +\sin \beta )$, differentiation at fixed wavelength gives $|d\beta /d\alpha |=\cos \alpha /\cos \beta$. Thus the monochromatic optical slit image width isThe incident and emergent optical pupil widths are $A_{\mathrm{coll}}=W\cos \alpha$ and $A_{\mathrm{cam}}=W\cos \beta$. Multiplying cancels the anamorphic magnification of a grating:This is conservation of the Lagrange optical invariant in the dispersion plane. It uses paraxial optical slit angles, an unvignetted optical pupil and matching full-width conventions for $\theta ,p,D,A_{\mathrm{cam}}$.