Question

Find the wavelength of light that has its third minimum at an angle of $\mathbf{48}\mathbf{.}\mathbf{6}\mathbf{°}$when it falls on a single slit of width localid="1654252496415" $\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{\mu m}$.

Short answer

The required wavelength is $\mathrm{\lambda }=107.5^{‐7}\mathrm{m}$.

Step-by-step solution

Concept Introduction

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 a 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 utilized by Young. For a double slit to produce constructive interference, the path length difference must be an integral multiple of the wavelength.

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

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

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

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

Given the required data

The third order's minimum angle is $\mathrm{\theta }=48.6°$.

The slit width is$\mathrm{d}=3\mathrm{\mu m}\left(\frac{{10}^{‐6}\mathrm{m}}{1\mathrm{\mu m}}\right)=3\times {10}^{‐6}\mathrm{m}$

Find the wavelength of the falling light.

Calculate for the third order, i.e., $\mathrm{m}=3$.

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:

$$\begin{gathered} \mathrm{\lambda }=\frac{\mathrm{dsin}\left(\mathrm{\theta }\right)}{\mathrm{m}} \\ =\frac{\left(3\times {10}^{‐6}\mathrm{m}\right)\times \sin \left(48.6°\right)}{3} \\ =7.5\times {10}^{‐7}\mathrm{m} \end{gathered}$$

Therefore, the wavelength of falling light is $\mathrm{\lambda }=7.5\times {10}^{‐7}\mathrm{m}$.