Question

For your science fair project you need to design a diffraction grating that will disperse the visible spectrum $\left(400-700nm\right)$over$30.0^{\circ }$ in first order.
$\left(a\right)$ How many lines per millimeter does your grating need?
$\left(b\right)$What is the first-order diffraction angle of light from a sodium lamp $\left(\lambda =589nm\right)$?

Short answer

Part $\left(a\right)$

$\left(a\right)$The lines per mm needed for grating is$n=1231.$

Part $\left(b\right)$

$\left(b\right)$The angle of light from sodium lamp is${\theta }_1=46.5.$

Step-by-step solution

Step: 1 Equating equation: (part a)

In diffraction grating,has

$d\sin \left({\theta }_1\right)=\lambda$

The small angle at each wavelength as

localid="1649135282502" $\begin{array}{l}d\sin \left(\theta \right)=400\mathrm{nm}\\ d\sin \left(\theta +{30}^{\circ }\right)=700\mathrm{nm}\\ \frac{\sin \left(\theta \right)}{\sin \left(\theta +{30}^{\circ }\right)}=\frac{4}{7}\\ \sin \left(\theta \right)=\frac{4}{7}\sin \left(\theta +{30}^{\circ }\right)\end{array}$

Step: 2 Finding angle: (part a)

By trignomentry,

$\begin{array}{l}\sin (a+b)=\sin a\cos b+\cos a\sin b\\ \frac{7}{4}\sin \theta =\left(\sin \theta \cos {30}^{\circ }\right)+\left(\cos \theta \sin {30}^{\circ }\right)\\ \frac{7}{4}\sin \theta =\frac{\sqrt{3}}{2}\sin \theta +\frac{\cos \theta }{2}\\ \left(\frac{7}{4}-\frac{\sqrt{3}}{2}\right)\sin \theta =\frac{\cos \theta }{2}\\ 0.884\sin \theta =\frac{\cos \theta }{2}\\ \frac{\sin \theta }{\cos \theta }=\tan \theta =\frac{1}{2\times 0.884}\\ \theta =\tan {}^{-1}\left(\frac{1}{2\times 0.884}\right)\\ \theta ={29.5}^{\circ }.\end{array}$

Step: 3 Calculating numbers: (part a)

Calculating distance and numbers by

$$\begin{gathered} d=\frac{400\mathrm{nm}}{\sin \left({29.5}^{\circ }\right)} \\ d=812.3\mathrm{nm} \\ d=812.3\times {10}^{-6}\mathrm{mm}. \\ n=\frac{1}{812.3\times {10}^{-6}\mathrm{mm}} \\ n=1231. \end{gathered}$$

Step: 4 Angle of light in sodium lamp: (part b) 

The angle of fringe for any wavelength

$\begin{array}{l}\sin \left({\theta }_1\right)=\frac{\lambda }{d}\\ {\theta }_1={\sin }^{-1}\left(\frac{\lambda }{d}\right)\\ {\theta }_1={\sin }^{-1}\left(\frac{589\mathrm{nm}}{812.3\mathrm{nm}}\right)\\ {\theta }_1={46.5}^{\circ }.\end{array}$