91影视

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

$\mathbf{c}\mathbf{=}\mathbf{惫位}$

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

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

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

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

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.

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

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.

The diffraction grating's number of lines per centimeter is $\mathrm{N}=125$lines per cm. This helps us to determine the distance

between two slits

$\mathrm{d}=\frac{1\mathrm{cm}}{\mathrm{N}}=\left(\frac{1\mathrm{cm}}{125}\right)\left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right)=8脳{10}^{鈥�5}\mathrm{m}$

The incident light wave has a wavelength of $\mathrm{位}=600\mathrm{nm}\left(\frac{{10}^9\mathrm{m}}{1\mathrm{nm}}\right)=6.00脳{10}^{鈥�7}\mathrm{m}$

The distance between the screen and the double slits is $\mathrm{x}=1.50\mathrm{m}$..

The following rule determines the distance between neighboring fringes:

\(\begin{aligned}{}{\rm{\Delta y}} & = \frac{{{\rm{x\lambda }}}}{{\rm{d}}}\\ &= \frac{{{\rm{1}}{\rm{.50\;m}} \times {\rm{6}}{\rm{.00}} \times {\rm{1}}{{\rm{0}}^{{\rm{ - 7}}}}\;{\rm{m}}}}{{{\rm{8}} \times {\rm{1}}{{\rm{0}}^{{\rm{ - 5}}}}\;{\rm{m}}}}\\ &= {\rm{1}}{\rm{.125}} \times {\rm{1}}{{\rm{0}}^{{\rm{ - 2}}}}{\rm{\;m}}\left( {\frac{{100\;cm}}{{1\;m}}} \right)\\ &= 1.125\;cm\end{aligned}\)

Therefore, the distance between the fringes is \({\rm{\Delta y}} = 1.125\;{\rm{cm}}\).