Two spectral lines at wavelengths $580,\text{nm}$ and $590,\text{nm}$ are to be resolved in the fourth order ($m=4$) spectrum of a diffraction grating whose line spacing is $1.20,\mu \text{m}$. Using Rayleigh’s criterion, what is the minimum total number of illuminated slits $N$ required?

Step-by-step solution

1. Rayleigh’s resolving power for a diffraction grating is $R=mN$ where $m$ is the order and $N$ is the total number of illuminated lines.
2. The resolving power also equals $\frac{lambda}{Deltalambda}$, where $Deltalambda$ is the difference between the two wavelengths and $lambda$ is taken as the average wavelength.
3. Compute the average wavelength: ${\lambda }_{\text{avg}}=\frac{580+590}{2}=585,\text{nm}=5.85\times 10^{-7},\text{m}$.
4. Compute $Deltalambda=590-580=10,\text{nm}=1.0\times 10^{-8},\text{m}$.
5. Set $R=\frac{{\lambda }_{\text{avg}}}{Deltalambda}=mN$. Thus $N=\frac{{\lambda }_{\text{avg}}}{m,Deltalambda}=\frac{5.85\times 10^{-7}}{4\times 1.0\times 10^{-8}}\approx 14.6$.
6. Since $N$ must be an integer, take $N_{min}=15$.

Concept tested: Resolving power of diffraction grating

Published 21 September 2026 · How this solution was prepared and checked