Question

Visible light is incident perpendicularly on a diffraction grating of 200 rulings/mm. What are the (a) longest, (b) second longest, and (c) third longest wavelengths that can be associated with an intensity maximum at θ = 30.0°?

Short answer

(a) The longest wavelength is 625 nm for the 4^th order of maxima.

(b) The 2^nd longest wavelength is 500 nm for the 5^th order of maxima.

(c) The 3^rd longest wavelength is 417 nm for the 6^th order of maxima.

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

role="math" localid="1663059226081" $\mathit{d}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathbf{=}\mathit{m}\mathit{\lambda }\mathbf{}\mathbf{}\mathbf{}\mathbf{for}\mathbf{}\mathit{m}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{}\mathbf{1}\mathbf{,}\mathbf{}\mathbf{2}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}$(maxima)

Where d is the width of the grating element, which is equal to role="math" localid="1663059212479" $\frac{\mathbf{w}\mathbf{i}\mathbf{d}\mathbf{t}\mathbf{h}\mathbf{}\mathbf{o}\mathbf{f}\mathbf{}\mathbf{g}\mathbf{r}\mathbf{a}\mathbf{t}\mathbf{i}\mathbf{n}\mathbf{g}}{\mathbf{N}\mathbf{u}\mathbf{m}\mathbf{b}\mathbf{e}\mathbf{r}\mathbf{}\mathbf{o}\mathbf{f}\mathbf{}\mathbf{s}\mathbf{l}\mathbf{i}\mathbf{t}\mathbf{s}}$.

Here the number of slits per unit length is 200 rulings/mm. Therefore, the width of the grating element will be

role="math" localid="1663059196577" $$\begin{gathered} \mathit{d}\mathbf{=}\frac{\mathbf{w}}{\mathbf{N}} \\ \mathbf{=}\frac{\mathbf{1}}{\mathbf{200}\mathbf{}\mathbf{r}\mathbf{u}\mathbf{l}\mathbf{i}\mathbf{n}\mathbf{g}\mathbf{s}\mathbf{/}\mathbf{m}\mathbf{m}} \\ \mathbf{=}\mathbf{0}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}}\mathbf{}\mathit{m} \end{gathered}$$

Determine the wavelengths in each case.

For $\text{m}=1$, the wavelength of the intensity maxima at$\theta =30°$ is

$$\begin{gathered} {\lambda }_1=d\sin \theta \\ =\left(0.5\times {10}^{-5}m\right)\sin 30° \\ =0.25\times {10}^{-5}m \\ =2500nm \end{gathered}$$

For $\text{m}=3$, the wavelength of the intensity maxima at$\theta =30°$ is

$$\begin{gathered} {\lambda }_3=\frac{d\sin \theta }{m} \\ =\frac{\left(0.5\times {10}^{-5}m\right)\sin 30°}{3} \\ =0.0833\times {10}^{-5}m \\ =833nm \end{gathered}$$

For $\text{m}=4$, the wavelength of the intensity maxima at$\theta =30°$ is

$$\begin{gathered} {\lambda }_4=\frac{d\sin \theta }{m} \\ =\frac{\left(0.5\times {10}^{-5}m\right)\sin 30°}{4} \\ =0.0625\times {10}^{-5}m \\ =625nm \end{gathered}$$

For $\text{m}=5$, the wavelength of the intensity maxima at$\theta =30°$ is

$$\begin{gathered} {\lambda }_5=\frac{d\sin \theta }{m} \\ =\frac{\left(0.5\times {10}^{-5}m\right)\sin 30°}{5} \\ =0.05\times {10}^{-5}m \\ =500nm \end{gathered}$$

For $\text{m}=6$, the wavelength of the intensity maxima at$\theta =30°$ is

$$\begin{gathered} {\lambda }_6=\frac{d\sin \theta }{m} \\ =\frac{\left(0.5\times {10}^{-5}m\right)\sin 30°}{6} \\ =0.0417\times {10}^{-5}m \\ =417nm \end{gathered}$$

As the visible range is 400 nm to 700 nm, the order of maxima from 4 to 6 is considered. Hence the order of wavelengths of diffraction maxima is

${\lambda }_4>{\lambda }_5>{\lambda }_6$