I Single Slit Fraunhofer Diffraction

1. Jul 12, 2017

Daniel Gallimore

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

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

Any recommendations?

2. Jul 13, 2017

sophiecentaur

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. Jul 13, 2017

Daniel Gallimore

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 $\sin(\beta/2)/(\beta/2)$ function. And if the integral is incorrect, I need some pointers to correct it.

4. Jul 13, 2017

sophiecentaur

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. Jul 13, 2017

Daniel Gallimore

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.