Single Slit Fraunhofer Diffraction

[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?

 

[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.

 

[Daniel Gallimore]

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

 

[sophiecentaur]

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.

 

[Daniel Gallimore]

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.