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What determines whether in a double slit experiment the interference pattern will be horizontal or vertical for example? Is it mostly seen horizontal because the slits are small enough horizontally for interference pattern to be seen? Also, I sometimes see a "rectangular" interference pattern when using a lens to focus the light. How does that happen? Shouldn't we expect a circular pattern?

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TL;DR

  1. If the slits are sufficiently long the diffraction pattern will be spaced in the same direction as the slits.
  2. If the length of the slits is of the order of the wavelength then the diffraction pattern will appear "rectangular".
  3. One should only expect a circular interference pattern if the aperture has a circular symmetry for example a circular hole.

Explanation

The interference pattern observed from the double-slit experiment produced by particles is the same as that produced by light due to wave-particle duality. However, the scales of the interference patterns will differ as the scale depends on the wavelength.

The interference pattern can be calculated using the Fresnel-Kirchhoff diffraction integral. There are two commonly used limits of this integral, the Fresnel limit (screen or point source close to aperture) and the Fraunhofer limit (screen and source far from aperture).

For this question, we will consider the Fraunhofer limit as the results are easier to understand intuitively and also easier to calculate as in the Fraunhofer limit the amplitude of the interference pattern on a parallel screen is proportional to the Fourier transform of the aperture function (a function that is $1$ where light (or particles) can pass through and $0$ where they cannot). The intensity is then just the square magnitude of the amplitude.

Thus, for a single slit (a rectangle) then the aperture function $h\left(x,y\right)=\Pi\left(x/w\right)\Pi\left(x/h\right)$ the intensity of the interference pattern is:

$$I\propto\operatorname{sinc}^2\left(\frac{1}{2}wk\sin\left(\theta\right)\right)\operatorname{sinc}^2\left(\frac{1}{2}hk\sin\left(\phi\right)\right)$$

where $\Pi\left(x\right)$ are Rectangular functions ("top hat functions"), $w$ and $h$ are the width and height of the slit, while $k\equiv\frac{2\pi}{\lambda}$ where $\lambda$ is the wavelength and finally $\theta$ and $\phi$ are the angles in the $x$ and $y$ directions to the normal. We can approximate $\sin\left(\theta\right)=\frac{x}{L}$ and $\sin\left(\phi\right)=\frac{y}{L}$ when close to the centre of the interference pattern.

For a double-slit we get an extra factor of $\cos^2\left(\frac{1}{2}Dk\sin\left(\theta\right)\right)$ where $D$ is the slit separation so:

$$I\propto\operatorname{sinc}^2\left(\frac{1}{2}wk\sin\left(\theta\right)\right)\operatorname{sinc}^2\left(\frac{1}{2}hk\sin\left(\phi\right)\right)\cos^2\left(\frac{1}{2}Dk\sin\left(\theta\right)\right)$$

I recommend plotting these equations (replacing $\sin\left(\theta\right)=\frac{x}{L}$ and $\sin\left(\phi\right)=\frac{y}{L}$) using plotting software (for example Desmos) and tweaking the parameters to see how this affects the interference pattern. You should find as the slits get longer there appears to be no variation vertically, as the slits get narrower then in the double-slit experiment the intensity drops off less slowly.

Finally, for a circular hole, we will get Airy disks (the Wikipedia page linked has some nice images of diffraction patterns).

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  • $\begingroup$ Thank you so much! Regarding the lens question, I did not mean (and maybe I should have clarified that) the pattern due to diffraction from a circular aperture, but just interference from two plane waves using an interferometer for example. Why doesn't a lens focusing the light make this "Airy pattern" also? $\endgroup$
    – Darkenin
    Jun 28 at 9:21
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    $\begingroup$ Any circular aperture (a lense can also be viewed as a magnifying component and a circular aperture) will in the Fraunhofer regime produce Airy disks. However, the ratio of the wavelength to the aperture size sets the scale of the diffraction pattern and so for large apertures only the central maxium will be visible so the interference pattern apears to vanish but as the apeture gets smaller more and more fringes can be seen. $\endgroup$
    – Chris Long
    Jun 28 at 9:37
  • $\begingroup$ In regards to the shape pattern observed in an interferometer see this section on Wikipedia: en.wikipedia.org/wiki/Michelson_interferometer#Configuration specifically figure 3 which shows how tilting the mirrors will change the interference pattern and the text describes how the fringes are conic sections. $\endgroup$
    – Chris Long
    Jun 28 at 9:42
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If the slits are vertical, the light will diffract left and right, creating a horizontal pattern.

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