Diffraction theory, power density and angle of incidence

[kagestodder]

TL;DR Summary

My question: How do i account for the angle of incidence so that there is a connection between the amount of power hitting the aperture is the same that also gets diffracted.

Hello everybody

I am currently looking at diffraction through a rectangular aperture. I am looking at an aperture which is large compared to wavelength, and am looking at diffraction at all angles behind the aperture in a distance which is approx equal to the size of the aperture. I am computing diffraction patterns computationally.

I have now hit somewhat of a wall wrt. understanding of the diffraction integrals. I am using kirchhoff diffraction integral and rayleigh-sommerfeldt diffraction integral, where the only difference between them is how the obliquity factor is defined(?).
The issue arises when i look at the problem wrt. power. Do the diffraction integral account for the angle of incidence of the plane wave hitting the aperture? Say a plane wave at an incidencde hits the aperture with some power density. The angle of incidence will affect the visible area of the aperture, and therefore the amount of energy hitting the aperture. When looking at kirchhoffs diffraction integral the two cosines of the integral will say that even when the incident wave hits the aperture at a large angle to the normal of the aperture, there will still be quite a powerfull diffracted field. But from a power perspective, almost nothing will hit the aperture. And the rayleigh-sommerfeldt diffraction integral has no parameter for the angle of incidence. Would it make sense to look at the incoming power desnity, convert this to a E-field value using the apparent aperture area and then use this as U0?

My question: How do i account for the angle of incidence so that there is a connection between the amount of power hitting the aperture is the same that also gets diffracted.

I hope this makes sense. Please correct me as i feel i am on deep water with this at the moment! :)

 

[tech99]

I think the way to look at this from a practical point of view is that the problem is the same as for a narrow mirror instead of a slit. In such a case it seems that the power hitting the projected area of the mirror should be used, and the reflected beam will be concentrated at the geometrical angle of reflection. The diffraction effects, however, will broaden the beam and will create sidelobes. I notice that you are very close to the slit, within the near radiation zone no doubt, so the incident power will vary across the aperture and in addition, the diffracted pattern will vary with distance.

 

[tech99]

Reading the question more closely it seems slightly simpler than I had supposed. Although the emergent rays are measured close to the slit, so the pattern will vary with distance, the incident power is uniform across the slit. The incident power will still be that striking the projected area of the slit.

 

[kagestodder]

Reading the question more closely it seems slightly simpler than I had supposed. Although the emergent rays are measured close to the slit, so the pattern will vary with distance, the incident power is uniform across the slit. The incident power will still be that striking the projected area of the slit.

Thanks

This was my assumption as Well. However it does not seem to hold such that the power hitting the slit (or mirror if we look at it like that) is not equal to integrating the power over a half sphere on the diffracted side of the aperture. This is the reason why I have become unsure.

So to me it seems like there is less power diffracted than hitting the aperture.

I compare
Pin = E0*E0*A
And
Pr = sum(Er*Er*2pi*r*r/ds), with ds being the sampling area of each point in the half sphere. And A the area of the aperture.

The values should be comparable and close to each other.

I may still be looking at this the wrong way.

 

[tech99]

Are you using the square of field strength when you calculate power?

 

[kagestodder]

Are you using the square of field strength when you calculate power?

Yes abs E squared!

I have been unable to find any sources or anything looking at these diffraction scenarios with regards to power in / power out. Any cases with results that I could copy the geometry from to verify my calculations could also help. For now it seems like the power hitting the aperture is more than gets diffracted on the other side of the aperture.