A beam of light consists of two wavelengths, 590.159 nm, and 590.220 nm, that are to be resolved with a diffraction grating. If the grating has lines across a width of 3.80 cm, what is the minimum number of lines required for the two wavelengths to be resolved in the second order?

A set of a large number of slits is called a diffraction grating to resolve the incident light into its component wavelengths. The diffractions at angles $\mathit{\theta }$ for N slits are given by

$\mathit{d}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathbf{=}\mathit{m}\mathit{\lambda }\mathbf{}\mathbf{}\mathbf{}\mathbf{for}\mathbf{}\mathit{m}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{}\mathbf{1}\mathbf{,}\mathbf{}\mathbf{2}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}$(maxima)

Where d is the width of the grating element, which is equal to $\frac{\mathbf{w}\mathbf{i}\mathbf{d}\mathbf{t}\mathbf{h}\mathbf{}\mathbf{o}\mathbf{f}\mathbf{}\mathbf{g}\mathbf{r}\mathbf{a}\mathbf{t}\mathbf{i}\mathbf{n}\mathbf{g}}{\mathbf{N}\mathbf{u}\mathbf{m}\mathbf{b}\mathbf{e}\mathbf{r}\mathbf{}\mathbf{o}\mathbf{f}\mathbf{}\mathbf{s}\mathbf{l}\mathbf{i}\mathbf{t}\mathbf{s}}$.

And the resolving power R of two observed wavelengths is

$\mathit{R}\mathbf{=}\frac{{\mathbf{\lambda }}_{\mathbf{a}\mathbf{v}\mathbf{g}}}{\mathbf{\Delta }\mathbf{\lambda }}\mathbf{=}\mathit{N}\mathit{m}$

Where${\mathit{\lambda }}_{\mathbf{a}\mathbf{v}\mathbf{g}}$ is the average of the two wavelengths, and$\mathit{\Delta }\mathit{\lambda }$is wavelength width.

Two wavelengths that are resolved with diffraction grating are 590.159 nm and 590.220 nm. The average of these is

$$\begin{gathered} {\lambda }_{avg}=\frac{{\lambda }_1+{\lambda }_2}{2} \\ =\frac{590.159nm+590.220nm}{2} \\ =590.190nm \end{gathered}$$

And the wavelength width will be

$$\begin{gathered} \Delta \lambda ={\lambda }_2-{\lambda }_1 \\ =590.220nm-590.159nm \\ =0.061nm \end{gathered}$$

Inserting these two terms in the resolving power equation to determine the number of slits

$$\begin{gathered} \frac{{\lambda }_{avg}}{\Delta \lambda }=Nm \\ N=\frac{{\lambda }_{avg}}{m\Delta \lambda } \\ N=\frac{590.190nm}{2\left(0.061nm\right)} \\ N\approx 4838 \end{gathered}$$

Hence the minimum number of rulings required is 4838.