Question

(a) Find the maximum number of lines per centimeter a diffraction grating can have and produce a maximum for the smallest wavelength of visible light. (b) Would such a grating be useful for ultraviolet spectra? (c) For infrared spectra?

Short answer

(a) The maximum number of lines per centimeter is 26,300.

(b) Yes, in the ultraviolet spectrum, such a diffraction grating element would be useful. It would be able to generate additional diffraction pattern peaks as a result.

(c) No, such a diffraction grating element would be useless for the infrared spectrum because no peaks would be generated.

Step-by-step solution

Concept Introduction

The ultraviolet spectrum covers the wavelengths shorter than visible light.

The infrared spectrum includes wavelengths longer than visible light.

Determine The maximum number of lines per centimeter.(a)

The equation of the constructive interference can be expressed as

$\mathbf{\theta sin\lambda }$ …………………(1)

Where, d, $\mathrm{\lambda }$, $\mathrm{\theta }$and m is the separation between slits, the wavelength of the incident light, and the angle made with the maximum and order of maximum, respectively.

The number of lines per unit length in the diffraction grating element is given by the reciprocal of the distance , i.e.

$\mathrm{N}=\frac{1}{\mathrm{d}}$

As a result, equation (1) may be rewritten as follows:

$\mathrm{m\lambda }=\frac{\mathrm{sin\theta }}{\mathrm{N}}$…………………………(2)

Now, if we rewrite this equation in terms of the number of lines we have, we get $\mathrm{N}$.

$\mathrm{N}=\frac{\mathrm{sin\theta }}{\mathrm{m\lambda }}$ …………………………….(3)

The number of lines per unit length in a diffraction grating is greatest when the numerator is biggest and the denominator is smallest, i.e. when $\mathrm{sin\theta }=1$; and the smallest value that $\mathrm{N}$ may have is 1, hence the maximum number of lines per unit length is thus,

${\mathrm{N}}_{\max }=\frac{1}{\mathrm{\lambda }}$

And since it's asking for the maximum number of lines capable of producing diffraction maxima for the smallest value of wavelength in the visible spectrum, and since the smallest value of wavelength in the visible spectrum is $380\mathrm{nm}$, we replace it with $380\mathrm{nm}$in the above equation, giving us,

$$\begin{gathered} {\mathrm{N}}_{\max }=\left(\frac{1}{380\mathrm{nm}}\right)\left(\frac{1\mathrm{nm}}{{10}^{‐9}\mathrm{m}}\right) \\ =2.63\times {10}^{‐1} \end{gathered}$$

This means that the maximum number of lines per meter of this diffraction grating element is role="math" localid="1654072425428" $2.63\times {10}^6$ lines, and as there are $100\mathrm{cm}\mathrm{in}1\mathrm{m}$, we simply divide by 100 to get the number of lines per cm, and therefore the number of lines per $\mathrm{cm}$is thus,

${\mathrm{N}}_{\max }=2.63\times {10}^3{\mathrm{cm}}^{‐1}$

Addendum: It's worth noting that equation (2) can be rewritten as follows:

$\mathrm{sin\theta }=\mathrm{mN\lambda }$ ………………..(3)

If we round the number of lines to keep the number of significant figures the same, we must be careful not to round to a larger number, since this would indicate that$\mathrm{sin\theta }>1$which is not a legitimate answer; as a result, we always round to a lesser number in this case as an exception.

Therefore, 26,300 lines per cm.

Determine whether the grating be useful for ultraviolet spectra(b)

Therefore, we use the value of ${\mathrm{N}}_{\max }$ from part (a) and substitute it in equation (3), and using a value of $100\mathrm{nm}$ for UV wave as an example, we get:

$$\begin{gathered} \mathrm{sin\theta }=\mathrm{m}\times 2.63\times {10}^6{\mathrm{m}}^{‐1}\times 100\mathrm{nm}\left(\frac{{10}^{‐9}\mathrm{m}}{1\mathrm{nm}}\right) \\ =0.263\times \mathrm{m} \end{gathered}$$

And since the sine function's maximum value is 1, this means that the value of $\mathrm{m}$will be,

$$\begin{gathered} \mathrm{m}=\frac{1}{0.263} \\ =3.8 \end{gathered}$$

Because$\mathrm{m}$is an integer, it can take up to four values, or in other words, there will be a maximum of four orders in addition to the center maximum. As a result, a diffraction grating of this type might be beneficial in the UV spectrum.

Therefore, yes, in the ultraviolet spectrum, such a diffraction grating element would be useful. It would be able to generate additional diffraction pattern peaks as a result.

Determine whether the grating be useful for infrared spectra(c)

So, using a value of $1000\mathrm{nm}$as an infrared wave value in equation (3), however, we find that the maximum value of $\mathrm{n}$is,

$$\begin{gathered} \mathrm{sin\theta }=\mathrm{m}\times 2.63\times {10}^6{\mathrm{m}}^{‐1}\times 1000\mathrm{nm}\left(\frac{{10}^{‐9}\mathrm{m}}{1\mathrm{nm}}\right) \\ =2.63\times \mathrm{m} \end{gathered}$$

And since the sine function's maximum value is 1, this means that the value of $\mathrm{m}$will be,

$$\begin{gathered} \mathrm{m}=\frac{1}{2.63} \\ =0.38 \end{gathered}$$

Because $\mathrm{m}$can only take integer values, there is no first-order maximum, hence such diffraction grating elements would be useless in the infrared spectrum.

Therefore, no, such a diffraction grating element would be useless because no peaks would be generated.