91影视

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.

$\mathbf{dsin}\left(胃\right)\mathbf{=}\mathbf{尘位}$

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

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

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

(a)

The wavelength of the sodium light is $\mathrm{位}=633\mathrm{nm}\left(\frac{{10}^{鈥�9}\mathrm{m}}{1\mathrm{nm}}\right)=6.33脳{10}^{鈥�7}\mathrm{m}$.

The slit width is$\mathrm{d}=20.0\mathrm{渭尘}\left(\frac{{10}^{鈥�6}\mathrm{m}}{1\mathrm{渭尘}}\right)=2.00脳{10}^{鈥�5}\mathrm{m}$

(b)

The angle for the third-order minimum is $85.0掳$

We apply the equation and solve it for $\mathrm{胃}$ where m=3 to find the angle formed at the third diffraction minimum.

$$\begin{gathered} \mathrm{胃}={\sin }^{鈥�1}\left(\frac{\mathrm{位}}{\mathrm{d}}\right) \\ ={\sin }^{鈥�1}\left(\frac{3脳6.33脳{10}^{鈥�7}\mathrm{m}}{2.00脳{10}^{鈥�5}\mathrm{m}}\right)=6.33脳{10}^{鈥�7}\mathrm{m} \\ =5.45掳 \end{gathered}$$

Therefore, $5.45掳$is the angle obtained by the third minimum.

We apply the equation $\mathrm{Dsin}\left(\mathrm{胃}\right)=\mathrm{尘位}$and solve it for D $\mathrm{m}=3$to determine the slit width that produces this minimum at $85.0掳$

$\mathrm{D}=\frac{\mathrm{尘位}}{\sin \left(\mathrm{胃}\right)}$

$$\begin{gathered} =\frac{3脳6.33脳{10}^{鈥�9}\mathrm{m}}{\sin \left(85.0掳\right)} \\ =1.91脳{10}^{鈥�6}\mathrm{m} \end{gathered}$$

Therefore, the width of the slit is localid="1654151821159" $1.91脳{10}^{鈥�6}\mathrm{m}$.