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OPhO 2021, Експеримент — Задача 5. Problem 2.1

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2021 Online Physics Olympiad: Invitational Contest — Experimental Exam · 8 т.

Условие

Recall that the intensity of the interference pattern from two thin slits behaves like $\cos^2(x)$ . Is it possible to have a series of thin slits such that the amplitude pattern behaves exactly like the intensity pattern from the two thin slits case? Specifically, the amplitude $A_1(x)$ from one pattern of slits behaves like the intensity $I_2(x)$ from another pattern of slits if:

\frac{I_2(x)}{I_{2,\max}} = \frac{A_1(x)}{A_{1,\max}} \qquad (3)

What sort of aperture would create such an $A(x)$ ? Verify this experimentally. You will notice that the pattern will deviate towards the edge. At what angle does this deviation become significant? Note that “significant” is subjective, so you will need to provide justification for how you define significant.

Solve this problem using Fourier Optics (There are at least 2 separate ways to do so. As long as you borrow concepts from Fourier Optics, you will receive full points):

Here are four potentially useful references from Wikipedia. Any result you use that is not in these references must be derived. This holds for the following two problems as well.
- Fraunhofer Diffraction Equation
- Fourier Optics
- Convolution
- Fourier Transform
Remember that asking for help on public forums or seeking help from other students, other teams, professors, i.e. is strictly prohibited.

Решение

Any aperture pattern can be represented by an aperture function, where a slit centered at $x_0$ that reduces the amplitude to $f$ of the original can be represented by $f \cdot \delta(x - x_0)$ where $\delta(x)$ is the delta function.

It turns out that taking the Fourier transform of the aperture function can give the amplitude function. It may seem strange that we are transforming a variable with dimension of length $x_1$ (Which represents the location of the slits) to another dimension of length $x_2$ (which represents the location at the screen), when Fourier transforms typically take variables to their inverses (i.e. time to frequency). However, this is not the case. We are actually taking the transform from $x_1$ to $\dfrac{x_2}{\lambda D}$ .

While we will not go into the details, it turns out taking the Fourier transform of the amplitude function will also get us the aperture function (i.e. transforming $x_2$ into $x_1$ ).

For a double slit, the intensity function is $\cos^2\left(\pi d\dfrac{x}{\lambda D}\right) = \cos^2(\pi d x_2)$ . We can use the identity $\cos^2(x) = \dfrac{1}{2}(1 + \cos(2x))$ to write:

\cos^2(\pi d \cdot x_2) = \frac{1}{2} + \frac{1}{2}\cos(2\pi d \cdot x_2). \qquad (10)

Taking the Fourier Transform, we get:

\begin{aligned} \hat{f}\left[\frac{1}{2} + \frac{1}{2}\cos(2\pi d \cdot x_2)\right] &= \hat{f}\left[\frac{1}{2}\right] + \hat{f}\left[\frac{1}{2}\cos(2\pi d \cdot x_2)\right] \\ &= \frac{1}{2}\hat{f}[1] + \frac{1}{2}\hat{f}[\cos(2\pi d \cdot x_2)] \\ &= \frac{1}{2}\delta(x_1) + \frac{1}{2}\left(\frac{\delta\left(x_1 - \frac{2\pi d}{2\pi}\right) + \delta\left(x_1 + \frac{2\pi d}{2\pi}\right)}{2}\right) \\ &= \frac{1}{2}\delta(x_1) + \frac{1}{4}\delta\left(x_1 - d\right) + \frac{1}{4}\delta\left(x_1 + d\right) \end{aligned}

and to normalize it (since only the relative amplitudes matter), we’re left with:

\delta(x_1) + \frac{1}{2}\delta(x_1 - d) + \frac{1}{2}\delta(x_1 + d). \qquad (15)

We can verify this experimentally. Setting

slits = [
  [0*wavelength, 1*wavelength, wavelength, wavelength, 1],
  [-2*wavelength, 1*wavelength, wavelength, wavelength, 0.5],
  [2*wavelength, 1*wavelength, wavelength, wavelength, 0.5],
]

will give us the following plot, where we have plotted the intensity pattern of the previous aperture pattern for reference:

Simulated plot titled 'Intensity as a Function of Distance' with legend 'Amplitude of New' (blue) and 'Intensity of Old' (orange); vertical axis 0.0 to 1.0, horizontal axis -200 to 200. Both curves peak at 1.0 at 0 and fall to near zero around ±125 to ±135; towards the edges the orange curve rises to about 0.65 and the blue one to about 0.48.

Notice that this is not perfect. This is because the Fourier Transform uses the far-field approximation, but the location of the first minimum isn’t necessarily small enough.

Оригинал в Архива: experiment.pdf