 Solution

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.