Single Slit Fraunhofer Diffraction

In summary, this is not homework. An Optics course a few years ago gave me the standard formula for irradiance, but I don't remember how to get from the electric field to the irradiance. I found a working of the problem in a book on Optics and was able to replicate the result in notation that was more familiar to me.
  • #1
Daniel Gallimore
48
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Just to clarify, this isn't homework. I took an Optics course a few years ago, but in the time between then and now, I've lost all my notes.

Suppose we have an infinitely tall slit of width [itex]a[/itex] and a parallel screen a distance [itex]r_0[/itex] away. Let the [itex]x[/itex] axis be parallel to the width of the slit, with the origin at the center of the slit (such that the slit extends from [itex]x=-a/2[/itex] to [itex]x=a/2[/itex]). The [itex]z[/itex] axis extends toward the screen from the origin. An angle [itex]\theta[/itex] locates a point [itex]P[/itex] on the screen (with respect to the [itex]z[/itex] axis) a distance [itex]x=X[/itex] along the screen. A ray drawn from the origin to [itex]P[/itex] is length [itex]r[/itex].

Given a plane wave incident on the slit, what is the formula for the irradiance of the diffracted wave as it encounters the screen?

I know the answer is similar (if not equal to) [tex]I=I_0\left(\frac{\sin(\beta/2)}{\beta/2}\right)^2[/tex] where [tex]\beta\equiv ka\sin\theta[/tex] I also know that [tex]E=-\frac{ikE_0}{2\pi r_0}\int_{-a/2}^{a/2} e^{ikr} \, dx[/tex] To get from the electric field to the irradiance is not a problem.

Any recommendations?
 
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  • #2
This is the standard formula, I think. The problem is with the limits of the integral. For a real case and a narrow slit, the illumination across the slit is not uniform so the profile should be included in the formula.
 
  • #3
sophiecentaur said:
This is the standard formula, I think. The problem is with the limits of the integral. For a real case and a narrow slit, the illumination across the slit is not uniform so the profile should be included in the formula.
Does saying the incident wave is a (coherent) plane wave not specify the profile? If not, please assume the illumination is uniform across the width of the slit. Also, I know this is just a toy example, but I'd like to get it down before tackling anything more realistic, like a nonuniform incident wave front.

What I want to know is how to get from the integral to to the [itex]\sin(\beta/2)/(\beta/2)[/itex] function. And if the integral is incorrect, I need some pointers to correct it.
 
  • #4
Daniel Gallimore said:
If not, please assume the illumination is uniform across the width of the slit
If you assume that, then no problem. But a slit in a screen of finite, practical thickness could be several wavelengths deep and, depending on the material (say metal) it will definitely affect the distribution of luminosity at the way out of a narrow slit. It just depends on how far you want to go, I think. Thinking in terms of a microwave slot through a thick sheet of metal, there will be induced currents around the edges of the slit which will alter the effective width.
Imo, the problem of a narrow slit would probably be best approached not as the limiting case of a wide slit but rather as a microwave slot antenna.
 
  • #5
sophiecentaur said:
If you assume that, then no problem. But a slit in a screen of finite, practical thickness could be several wavelengths deep and, depending on the material (say metal) it will definitely affect the distribution of luminosity at the way out of a narrow slit. It just depends on how far you want to go, I think. Thinking in terms of a microwave slot through a thick sheet of metal, there will be induced currents around the edges of the slit which will alter the effective width.
Imo, the problem of a narrow slit would probably be best approached not as the limiting case of a wide slit but rather as a microwave slot antenna.
I found a working of the problem in a book on Optics and was able to replicate the result in notation that was more familiar to me.

Thank you for your comments.
 
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Related to Single Slit Fraunhofer Diffraction

1. What is Single Slit Fraunhofer Diffraction?

Single Slit Fraunhofer Diffraction is a phenomenon that occurs when a wave, such as light or sound, passes through a single slit. It causes the wave to spread out and create a diffraction pattern on a screen or surface beyond the slit.

2. How does the width of the slit affect the diffraction pattern?

The width of the slit has a direct effect on the diffraction pattern. As the width of the slit decreases, the diffraction pattern becomes wider and more spread out. Conversely, as the width of the slit increases, the diffraction pattern becomes narrower and more concentrated.

3. What is the difference between Single Slit Diffraction and Double Slit Diffraction?

The main difference between Single Slit Diffraction and Double Slit Diffraction is the number of slits involved. Single Slit Diffraction occurs when a wave passes through a single slit, while Double Slit Diffraction occurs when a wave passes through two closely spaced slits. This results in different diffraction patterns.

4. How does the wavelength of the wave affect the diffraction pattern in Single Slit Fraunhofer Diffraction?

The wavelength of the wave also plays a role in the diffraction pattern. As the wavelength increases, the diffraction pattern becomes wider and more spread out. This is because longer wavelengths are able to diffract more around the edges of the slit. Conversely, shorter wavelengths result in a narrower diffraction pattern.

5. What real-world applications use Single Slit Fraunhofer Diffraction?

Single Slit Fraunhofer Diffraction has several practical applications, such as in optics and spectroscopy. It is used in the design of optical instruments, such as telescopes and microscopes, to improve image resolution. It is also used in spectroscopy to analyze the properties of materials based on the diffraction pattern produced by passing light through them.

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