Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 338 / 2 / a / v / Solution

Substitution of $m\lambda =d(\sin \alpha +\sin \beta )$ givesThe incident and outgoing optical path length contributions between the two ends of the illuminated diffraction grating add. Their total is $W\sin \alpha +W\sin \beta$, so $L$ is the edge-to-edge optical path difference in the selected diffracted direction. This is a length, not a dimensionless resolving power. With $N_g=W/d$ illuminated grooves, $L=m{N}_g\lambda$ and the ideal diffraction grating diffraction-limited spectral resolving power is $R_g=m{N}_g=L/\lambda$. A wide optical slit generally gives a smaller slit-limited $R$; this distinction limits how far the slit-width scaling may be extrapolated. All of these formulae assume the stated reflection-grating sign convention.