= Solution
Let $f_t$ be the <optical telescope> <focal length>, $x=f_t\theta$ the physical <optical slit> width, $f_{\rm coll}$ the <collimator> <focal length>, and $A_{\rm coll}=Df_{\rm coll}/f_t$ 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 is
$$
p=x\frac{f_{\rm cam}}{f_{\rm coll}}\frac{\cos\alpha}{\cos\beta}.
$$
The incident and emergent <optical pupil> widths are $A_{\rm coll}=W\cos\alpha$ and $A_{\rm cam}=W\cos\beta$. Multiplying cancels the <anamorphic magnification of a grating>:
$$
\frac{pA_{\rm cam}}{f_{\rm cam}}=\frac{xA_{\rm coll}}{f_{\rm coll}}=\theta D,
\qquad
\boxed{L=R\theta D=\frac{RpA_{\rm cam}}{f_{\rm cam}}.}
$$
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 $\theta,p,D,A_{\rm cam}$.
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