Question

(a) Sodium vapor light averaging 589 nm in wavelength falls on a single slit of width 7.50 µm. At what angle does it produce its second minimum?

(b) What is the highest-order minimum produced?

Short answer

(a) At the second minimum, the angle created is $9.04°$.

(b) The largest minimum order quantity generated is 12 units.

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.

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

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.

Given data

The wavelength of the sodium light is $\mathrm{\lambda }=589\mathrm{nm}\left(\frac{{10}^{‐9}\mathrm{m}}{1\mathrm{nm}}\right)=5.89\times {10}^{‐7}\mathrm{m}$

The separation of the slits is$\mathrm{d}=7.50\mathrm{\mu m}\left(\frac{{10}^{‐6}\mathrm{m}}{1\mathrm{\mu m}}\right)=7.50\times {10}^{‐6}\mathrm{m}$

Find the angle that produces its second minimum(a)

To get the angle produced at the second minimum, we use the equation $\mathrm{d}\mathrm{s}\mathrm{i}\mathrm{n}(\theta )=\mathrm{m}\lambda$$\mathrm{a}\mathrm{n}\mathrm{d}\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{v}\mathrm{e}\mathrm{i}\mathrm{t}\mathrm{f}\mathrm{o}\mathrm{r}\theta \mathrm{w}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{e}\mathrm{m}=2$

$$\begin{gathered} \mathrm{\theta }={\sin }^{‐1}\left(\frac{\mathrm{\lambda m}}{\mathrm{d}}\right) \\ ={\sin }^{‐1}\left(\frac{2\times 5.89\times {10}^{‐7}\mathrm{m}}{7.50\times {10}^{‐6}\mathrm{m}}\right) \\ =9.04° \end{gathered}$$

Therefore, at the second minimum, the angle created is $9.04°$.

Find the highest-order minimum produced(b)

When $\sin \left(\mathrm{\theta }\right)$is at its maximum, i.e. the angle is $90°$, the highest order minimum is created.

We solve the equation$\mathrm{dsin}\left(\mathrm{\theta }\right)=\mathrm{m\lambda }$to reach the highest order and solve it for m.

$$\begin{gathered} =\frac{\left(7.50\times {10}^{‐6}\mathrm{m}\right)\times \sin \left(90°\right)}{5.89\times {10}^{‐7}\mathrm{m}} \\ =12.7 \\ =12\mathrm{order} \end{gathered}$$

Therefore, the largest minimum order quantity generated is 12 units.