Assume that the limits of the visible spectrum are arbitrarily chosen as $\mathbf{430}$and$\mathbf{680}\mathit{n}\mathit{m}$. Calculate the number of rulings per millimeter of a grating that will spread the first-order spectrum through an angle of $\mathbf{20}\mathbf{.}\mathbf{0}\mathbf{°}$$20.0°$

Diffraction is a phenomena in which the interference or the bending of waves occurs around the corners of the opening or obstacle through or on which it strikes.

The diffraction formula is following,

$$\begin{gathered} \mathrm{n\lambda }=2\mathrm{dsin\theta } \\ \mathrm{\lambda }=\text{wavelength} \\ \mathrm{d}=\text{spacing} \\ \mathrm{\theta }=\text{bragg angle} \\ \mathrm{n}=\text{order of diffraction} \end{gathered}$$

The shortest Wavelength${\lambda }_1=430nm$

The longest Wavelength${\lambda }_2=680nm$

The diffraction formula,

$$\begin{gathered} {\lambda }_1=d\sin \theta \\ {\lambda }_2=d\sin \left(\theta +\Delta \theta \right) \\ \sin \left(\theta +\Delta \theta \right)=\sin \theta \cos \Delta \theta +\cos \theta \sin \Delta \theta \end{gathered}$$

From above equation,

$sin\theta =\frac{{\lambda }_1}{d}$

And,

$\cos \theta =\sqrt{1-{\left(\frac{{\lambda }_1}{d}\right)}^2}$

After replacing the$\sin \theta$ and$\cos \theta$ in ${\lambda }_2$equation,

${\lambda }_1\cos \Delta \theta +\sqrt{d^2-{{\lambda }_1}^2}\sin \Delta \theta ={\lambda }_2$

The value of d will be,

$d=\sqrt{\frac{{\left({\lambda }_2-{\lambda }_1\cos \Delta \theta \right)}^2+{\left({\lambda }_1\sin \Delta \theta \right)}^2}{{\sin }^2\Delta \theta }}$

Substitute all the value in the above equation.

$$\begin{gathered} d=\sqrt{\frac{{\left(680 \text{nm}-430 \text{nm}\times \cos 20°\right)}^2+{\left(430 \text{nm}\sin 20°\right)}^2}{{\sin }^{2}20°}} \\ d=914 \text{nm} \\ d=9.14\times {10}^{-4} \text{mm} \end{gathered}$$

$$\begin{gathered} \frac{1}{d}=\frac{1}{9.14\times {10}^{-4} \text{mm}} \\ \frac{1}{d}=1.09\times {10}^3 \text{ruling per mm} \end{gathered}$$

Hence the number of ruling is, $1.09\times {10}^3 \text{ruling per mm}$.