Single slit, optics, special case

[fluidistic]

I've thought about the case when the wavelength of the incident EM waves on a single slit is much greater than the width of the slit. When we have $$\lambda >> D$$. It could be the case of a radiowave incidating over a single slit whose width is day a nanometer large.
According to Hecht, the irradiance is worth $$I(\theta)=I(0)\frac{\sin \beta}{\beta}$$ where $$\beta=\frac{kD \sin \theta}{2}$$. Thus $$\beta=\frac{\pi D \sin (\theta)}{\lambda}$$.
In my case this reduces to $$\beta \approx 1$$ and therefore $$I(\theta)\approx I(0)$$ for all $$\theta$$... This means that the intensity over the screen does not almost change and is therefore almost constant whatever theta is. Now I realize that I(0) is likely a very small value, in other words it's like if I spread the incident wave over a gigantic screen.
So basically the wave can pass through the slit but the intensity on a far screen (Fraunhaufer conditions) is very small and spread almost equally everywhere on the screen. Is my interpretation right?

 

[Andy Resnick]

I'd have to go back over the derivation, but IIRC that formula for irradiance is valid only in the far-field. The near field distribution (the Fresnel region) is much more complex than the Frauhofer distribution.

http://en.wikipedia.org/wiki/Near-field_diffraction_pattern

 

[fluidistic]

I'd have to go back over the derivation, but IIRC that formula for irradiance is valid only in the far-field. The near field distribution (the Fresnel region) is much more complex than the Frauhofer distribution.

http://en.wikipedia.org/wiki/Near-field_diffraction_pattern

Thanks for the info Andy. I was considering the far field thus (?) Fraunhaufer diffraction.

I mean the screen where I observe the irradiance is very far from the slit. Is my reasoning valid?

 

[jtbell]

If you have a flat viewing-screen, as you get further away from the center, the light has to travel a longer distance from the slit, so you have a 1/r^2 decrease in intensity, multiplied onto your $I(\theta)$.

If you have a semicircular viewing-screen, with the slit at the center of curvature, so all points on the screen are the same distance from the slit, than I think your conclusion is valid. (I'm going from my memory of the assumptions that go into the derivation of Fraunhofer diffraction.)

 

[fluidistic]

If you have a flat viewing-screen, as you get further away from the center, the light has to travel a longer distance from the slit, so you have a 1/r^2 decrease in intensity, multiplied onto your $I(\theta)$.

If you have a semicircular viewing-screen, with the slit at the center of curvature, so all points on the screen are the same distance from the slit, than I think your conclusion is valid. (I'm going from my memory of the assumptions that go into the derivation of Fraunhofer diffraction.)

I think I understand what you mean.
In fact I(0) is definied as being the maximum of irradiance (or intensity, I don't remember) viewed on the viewing-screen. So I'm totally safe not using a 1/r^2 relation. I'm sorry that I haven't clarified it.

 

[Redbelly98]

There are two complications to this problem.

1. One must take the obliquity factor into account, if you are looking at angles that are far off-axis. This multiplicative factor has a maximum for θ=0. So the the intensity will diminish away from the central maximum, even for a semicircular screen.

2. My recollection about large wavelengths is: for a wavelength comparable to, or larger, than the slit width, we need to know about the boundary conditions at the slit border. The electric field doesn't abruptly go to zero at the slit boundary, though that is a reasonable approximation when the wavelength is a lot smaller than the slit size.

 

[Andy Resnick]

Thanks for the info Andy. I was considering the far field thus (?) Fraunhaufer diffraction.

I mean the screen where I observe the irradiance is very far from the slit. Is my reasoning valid?

Well, my point is that if you have a sub-wavelength aperture, the far-field pattern will not be given by the Fraunhofer distribution. In fact, it's likely that the irradiance will simply be zero.