Question

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?

Short answer

Using the diffraction grating, approximately 4838 rulings were required to resolve the two wavelengths, 590.159 nm, and 590.220 nm.

Step-by-step solution

Diffraction grating

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.

Determine the average wavelength and 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.