Consider a spectrometer based on a diffraction grating. Construct a problem in which you calculate the distance between two wavelengths of electromagnetic radiation in your spectrometer. Among the things to be considered are the wavelengths you wish to be able to distinguish, the number of lines per meter on the diffraction grating, and the distance from the grating to the screen or detector. Discuss the practicality of the device in terms of being able to discern between wavelengths of interest.

The electromagnetic spectrum includes visible light, as we all know. The electromagnetic wave's speed is determined by

$\mathbf{c}\mathbf{=}\mathbf{v\lambda }$

We also know that visible light has wavelengths ranging from $\mathbf{380}\mathbf{}\mathbf{nm}\mathbf{}\mathbf{to}\mathbf{}\mathbf{760}\mathbf{}\mathbf{nm}$. As we all know, the wavelength of an electromagnetic wave in a medium is determined by

${\mathbf{\lambda }}_{\mathbf{n}}\mathbf{=}\frac{\mathbf{\lambda }}{\mathbf{n}}$

Where $\mathbf{n}$is the refraction index of the specified medium and $\mathbf{\lambda }$is the wavelength of the electromagnetic wave in vacuum. As we all know, any material's thickness is determined by

$\mathbf{t}\mathbf{=}\frac{\mathbf{\lambda }}{\mathbf{n}}$

Young's double slit experiment, as we all know, had unexpected findings. As we all know, light interacts with little objects, such as the narrow slits utilised by Young. For a double slit to produce constructive interference, the path length difference must be an integral multiple of the wavelength.

$\mathbf{dsin}\left(\theta \right)\mathbf{=}\mathbf{m\lambda }$

For a double slit to produce destructive interference, the path length difference must be a half-integral multiple of the wavelength.

Where d is the distance between the slits, $\mathbf{\theta }$is the angle of the beam from its initial direction, and m is the interference order.

Discuss the device's usefulness in terms of distinguishing between wavelengths of interest.

Calculate the visible spectrum's shortest wavelength:

Calculate the maximal first-order angle as follows:

The path difference is supplied by, as we mentioned earlier in the concept session.

$\mathrm{dsin}\left(\mathrm{\theta }\right)=\mathrm{m\lambda }$

$\sin \left({\mathrm{\theta }}_1\right)=\frac{{\mathrm{\lambda m}}_{\mathrm{short}}}{\mathrm{d}}$

If we apply${\sin }^{‐1}$ to both sides, we get

$$\begin{gathered} \mathrm{\theta }={\sin }^{‐1}\left(\frac{{\mathrm{\lambda m}}_{\mathrm{short}}}{\mathrm{d}}\right) \\ ={\sin }^{‐1}\left(\frac{1\times 380\times {10}^{‐9}\mathrm{m}}{1\times {10}^{‐8}\mathrm{m}}\right) \\ =22.33° \end{gathered}$$

Calculate the visible spectrum's longest wavelength:

Calculate the maximal first-order angle as follows:

The path difference is supplied by, as we mentioned earlier in the concept session.

$$\begin{gathered} \mathrm{dsin}\left({\mathrm{\theta }}_1\right)=\mathrm{m\lambda } \\ \sin \left({\mathrm{\theta }}_2\right)=\frac{{\mathrm{m\lambda }}_{\mathrm{long}}}{\mathrm{d}} \end{gathered}$$

If we apply ${\sin }^{‐1}$ to both sides, we get

$$\begin{gathered} \mathrm{\theta }={\sin }^{‐1}\left(\frac{{\mathrm{\lambda m}}_{\mathrm{short}}}{\mathrm{d}}\right) \\ ={\sin }^{‐1}\left(\frac{1\times 760\times {10}^{‐9}\mathrm{m}}{1\times {10}^{‐6}\mathrm{m}}\right) \\ =49.46° \end{gathered}$$

The $\mathrm{\theta tan}$ is the ratio of the opposite side to the adjacent, as we know. Assume that the Adjacent is localid="1654143032911" $1\mathrm{m}$. Then there's

localid="1654143037143" $\mathrm{\theta tan}\frac{\mathrm{d}}{\mathrm{x}}$

We use the following relation to calculate the distance between the first greatest orders of the two wavelengths:

$\mathrm{D}=\mathrm{x}\left(\tan \left({\mathrm{\theta }}_2\right)-\tan \left({\mathrm{\theta }}_1\right)\right)$

$$\begin{gathered} =1\mathrm{m}\left(\tan \left(49346°\right)+\tan \left(22.33°\right)\right) \\ =0.758\mathrm{m} \end{gathered}$$

Therefore, the distance between the two wavelengths' initial maximum order is $0.758\mathrm{m}$