E Field Propagation & Metal Bars

[likephysics]

I was watching this E & M lecture -http://www.youtube.com/watch?v=XtHsVSW2W3E&feature=SeriesPlayList&p=C2CEECFD938FD494&index=28
There is a demo at ~45:00 into the lecture, where a structure with metal bars is introduced between 2 waveguides. I am trying to figure out what exactly happens between the E field and the metal.

When the E filed and metal bars are oriented in the same direction, does the E field from the waveguide setup a current in the metal bars, which cancel out the original E field?

 

[Born2bwire]

Without viewing the video, I bet that the metal bars acts as a polarized filter. Anytime an electromagnetic wave is incident with a conductor, it will induce currents. These currents will induce secondary waves that cancel out the incident field. In the case with the bars, the currents will be restricted to more or less flowing along the bars. Thus, when the incident wave is polarized such that its electric field is aligned with the bars, currents will be excited that will cancel out the propagation of the electric field through the bars. If the field is polarized normally to these bars, the air gap prevents the currents from being excited properly and thus the wave passes through more or less unimpeded. The spacing of the bars is important since you want them to be close enough to put up an effective screen when aligned with the polarized wave but at the same time it must be large enough not to greatly impact incident waves of cross-polarization.

EDIT: Yep, that's it. One way to think about is that in the far-field, the incident radiation will look like a plane wave. A plane wave is excited ideally by an infinite plane of current that flows in the same direction as the electric field of the excited wave. So, presenting a fine screen with conductors that run in the direction as the incident electric field of the incident "plane" wave will allow an approximation of the currents that would be excited should that plane wave strike a true solid conductor. This is why currents in the off-direction contribute negligibly to the reflected field.