Calculate the wavelength of light that produces its first minimum at an angle of $\mathbf{36}\mathbf{.}{\mathbf{9}}^{\mathbf{°}}$ when falling on a single slit of width \({\rm{1}}{\rm{.00 \mu m}}\).

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$\mathbf{380}\mathbf{}\mathbf{nm}$to.localid="1654146389787" $\mathbf{760}\mathbf{}\mathbf{nm}$As we all know, the wavelength of an electromagnetic wave in a medium is determined by

localid="1654148448638" ${\mathbf{\lambda }}_{\mathbf{n}}\mathbf{=}\frac{\mathbf{\lambda }}{\mathbf{n}}$

Where is the refraction index of the specified medium and is the wavelength of the electromagnetic wave in a vacuum. As we all know, any material's thickness is determined by

localid="1654148458726" $\mathbf{t}\mathbf{=}\frac{\mathbf{\lambda }}{\mathbf{n}}$

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 utilized 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.

$\mathbf{dsin}\mathbf{\left(}\mathbf{\theta }\mathbf{\right)}\mathbf{=}\left(m+\frac{1}{2}\right)\mathbf{\lambda }$

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.

The first order's minimum angle is $\mathbf{\theta }\mathbf{=}\mathbf{36}\mathbf{.}{\mathbf{9}}^{\mathbf{\theta }}$

The slit is a certain width$\mathrm{d}=1.00\mathrm{\mu m}\left(\frac{{10}^{‐6}\mathrm{m}}{1\mathrm{\mu m}}\right)=1.00\times {10}^{‐6}\mathrm{m}$

Solve the problem for the first time $\mathrm{m}=1$

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

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

Rearrange the equations to find the wavelength:

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

$$\begin{gathered} =\frac{(1.00\times {10}^{‐8}\mathrm{m})\times \sin (36.9^{°})}{1} \\ =6.01\times {10}^{‐7}\mathrm{m} \end{gathered}$$

Therefore, the required wavelength is $\mathrm{\lambda }=6.01\times {10}^{‐7}\mathrm{m}$