Question

You've found an unlabeled diffraction grating. Before you can use it, you need to know how many lines per it has. To find out, you illuminate the grating with light of several different wavelengths and then measure the distance between the two first-order bright fringes on a viewing screen $150cm$behind the grating. Your data are as follows:

Use the best-fit line of an appropriate graph to determine the number of lines per $mm$.

Short answer

The grating light at multiple wavelength is$800lines/mm.$

Step-by-step solution

Step: 1 Diffraction grating:

A light beam is created by scratching a flat piece of transparent material with multiple parallel scratches. The material may be scratched with a great number of scratches per centimetre.

Step: 2 Equating:

Grating equation as

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

Angle by

$$\begin{gathered} Y_m=L\tan \left({\theta }_m\right) \\ \tan \left({\theta }_m\right)=\frac{Y_m}{L} \\ {\theta }_m={\tan }^{-1}\left(\frac{Y_m}{L}\right) \\ \frac{1}{d}=\frac{\sin \left({\theta }_m\right)}{m\lambda }. \end{gathered}$$

We cannot ignore localid="1649153588019" $m$since it will be unity because we are provided the distance between the two initial fringes. We may also compute the angles of each wavelength and scatter the wavelength and the sine of the angle in a plot for each of these wavelengths. The graphic below, created in Excel, shows what we would end up with.

Step: 3 Obtaining value:

The Table and graph as follows,

λ	2Y	Y	Y/H	Θ(rad)	Θ (deg)	Sin(Θ)
4.30E-07	109.6	54.8	0.36533	0.350269	20.06893	0.34315
4.80E-07	125.4	62.7	0.418	0.395927	22.68493	0.385663
5.30E-07	139.8	69.9	0.466	0.43608	24.98552	0.422389
5.80E-07	157.2	78.6	0.524	0.482663	27.65454	0.464139
6.30E-07	174.4	87.2	0.581333	0.526581	30.17086	0.50258
6.80E-07	194.8	97.4	0.649333	0.575906	32.99701	0.544595

The combination of the sine of the angles over the wavelength is what we're searching for, as previously stated. This is only the slope of the scattered plots' line. We can see that this slope is $800,000lines/unitlength$or $800lines/mm$.