(a) Show that a 30,000-line-per-centimeter grating will not produce a maximum for visible light. (b) What is the longest wavelength for which it does produce a first-order maximum? (c) What is the greatest number of lines per centimeter a diffraction grating can have and produce a complete second-order spectrum for visible light?

Visible light refers to the visible region of the electromagnetic spectrum that is, the range of wavelengths that trigger brightness and color perception in humans. It lies between 380 nm to 760 nm.

Grating lines on the diffraction grating are 30000 liner per cm, which means that the

distance between the two slits is$\mathrm{d}=\left(\frac{1\mathrm{cm}}{30000}\right)\left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right)=3.34\times {10}^{‐7}\mathrm{m}$

The equation for the constructive interference can be given as

$\mathrm{dsin\theta }=\mathrm{m\lambda }$…………………..(1)

Solving equation (1) for the minimum ${\mathrm{\lambda }}_1$when and $\mathrm{\theta }=90.0°$.

$$\begin{gathered} \mathrm{\lambda }=\frac{\mathrm{dsin\theta }}{\mathrm{m}} \\ =\frac{\left(3.34\times {10}^{‐7}\right)\times \sin \left(90.0°\right)}{1} \\ =3.34\times {10}^{‐7}\mathrm{m}\left(\frac{1\mathrm{nm}}{{10}^9\mathrm{m}}\right) \\ =334\mathrm{nm} \end{gathered}$$

Therefore, this grating cannot create a visible light maximum because the shortest wavelength of visible light is $380\mathrm{nm}$.

So, Compare the potential values of $\mathrm{\lambda }$, the visible light wavelength range using the result of section (a).

The maximum wavelength that this grating can produce the first-order maximum is not within the range of visible light wavelengths.

Therefore, the maximum wavelength of light falling on the grating is shorter than the visible spectrum's lower bound

Solving equation (1) for the maximum$\mathrm{\lambda },\mathrm{and}\mathrm{m}=2,\mathrm{and}\mathrm{\theta }=90.0°$. Here the value of maximum wavelength for the visible light spectrum is 760 nm. Therefore,

$$\begin{gathered} \mathrm{d}=\frac{{\mathrm{m\lambda }}_{\max }}{\mathrm{sin\theta }} \\ =\frac{2\times 760\mathrm{nm}}{\sin \left(90.0°\right)}\times \left(\frac{{10}^{‐9}\mathrm{m}}{1\mathrm{nm}}\right) \\ =1.52\times {10}^{‐6}\mathrm{m} \end{gathered}$$

Then the number of lines per cm will be

$$\begin{gathered} \mathrm{n}=\frac{1\mathrm{cm}}{1.52\times {10}^{‐6}\mathrm{m}}\left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right) \\ =6579 \end{gathered}$$

Therefore, The number of lines per centimeter $6579$.