Why Does Wavelength Impact Passage Through a Hole?

[tobjackson]

Why does the wavelength of light have to be longer (or shorter?) than a hole diameter to pass through the hole?

Also, is this phenomenon the same reason AM radio signals cannot be picked up as well FM radio signals when driving under an overpass?

 

[Born2bwire]

It first depends on what materials are involved, the medium in the hole and the material that the hole is cut out in. The simplest case is to assume air and a perfect electrical conductor (PEC) respectively. Regardless of the material, it comes down to the boundary conditions, the properties of the fields that have to be satisfied along any interface. For a PEC, the tangential electric and normal magnetic field components must be zero on the surface and all field values are zero in the interior of the PEC. Taken into context the fact that the fields must satisfy the wave equations inside the hole, this requires that only specific field configurations can be supported when passing through the hole. If we have a square hole, then the longest wavelength possible is one-half the width of the hole. This allows the field to always be zero at both edges since the field must be sinusoidal in the cross-sectional plane of the hole. (the edges are a nodal point of the tangential electric field).

So when waves pass through a hole, the hole will only support specific modes. Waves that are not of a supported mode (or a combination of modes) will be attenuated as they pass through the hole. The attenuation is dependent upon the frequency and thickness of the hole. For most holes or tunnels, the lowest supported mode is on the order of a wavelength but it depends on the actual geometries and materials involved. All of this is discussed in the confines of waveguides. Your holes and tunnels and such can be approximated as waveguides and you look them up for further clarification.

But yes, this is why if you enter a tunnely you will receive FM radio and not AM radio. This is a common homework problem in engineering EM courses. Though nowadays some of the nicer tunnels will have repeaters in the tunnel to reradiate the nonpropagating radio signals of interest.

 

[tobjackson]

Thanks very much Born2bwire.

OK, so to make sure I understand: you are saying that the electric and magentic vectors of the wave must fit inside the confines of the hole, so that the peak-to-peak magnitude of the electric and magnetic vectors must be equal to or less than width of the hole. Is this correct?

If this is correct, then can you please tell me why we are assuming that there is a one-to-one correspondence between the wavelength and the peak-to-peak magnitude of the electric and magnetic vectors? I mean, what if the magnitude of the electric and magnetic vectors are much smaller than the wavelength (or much larger), then I would think the wave could pass (or not pass) through a hole that was smaller than 1/2 the wavelength.

Thanks very much,
tobjackson.

It first depends on what materials are involved, the medium in the hole and the material that the hole is cut out in. The simplest case is to assume air and a perfect electrical conductor (PEC) respectively. Regardless of the material, it comes down to the boundary conditions, the properties of the fields that have to be satisfied along any interface. For a PEC, the tangential electric and normal magnetic field components must be zero on the surface and all field values are zero in the interior of the PEC. Taken into context the fact that the fields must satisfy the wave equations inside the hole, this requires that only specific field configurations can be supported when passing through the hole. If we have a square hole, then the longest wavelength possible is one-half the width of the hole. This allows the field to always be zero at both edges since the field must be sinusoidal in the cross-sectional plane of the hole. (the edges are a nodal point of the tangential electric field).

So when waves pass through a hole, the hole will only support specific modes. Waves that are not of a supported mode (or a combination of modes) will be attenuated as they pass through the hole. The attenuation is dependent upon the frequency and thickness of the hole. For most holes or tunnels, the lowest supported mode is on the order of a wavelength but it depends on the actual geometries and materials involved. All of this is discussed in the confines of waveguides. Your holes and tunnels and such can be approximated as waveguides and you look them up for further clarification.

But yes, this is why if you enter a tunnely you will receive FM radio and not AM radio. This is a common homework problem in engineering EM courses. Though nowadays some of the nicer tunnels will have repeaters in the tunnel to reradiate the nonpropagating radio signals of interest.

 

[Bob S]

In microwave waveguides, when the wavelength of the signal frequency is below "cutoff" (when the wavelengh is longer that the transverse dimensions of the waveguide), the signal cannot propagate, and will be attenuated. See
http://www.k5rmg.org/calc/waveguide.html
and use the embedded Java calculator.

 

[Born2bwire]

The amplitude of the waves are immaterial, it is the wavelength that matters. The wave equation, derived from Maxwell's equations, and the boundary conditions caused by the waveguide will dictate that only certain field distributions can propagate. If we attempt to send a wave through that is not one of these propagating modes, then it will be attenuated as it travels through.

The magnitude of the electric and magnetic field vectors has no spatial correlation. They are not disturbances in space, they are simply the value of the fields at a point in space.

 

[tobjackson]

Thank you evryone for posting answers to my question. While I do not yet grasp the reason that wave propagation through a waveguide is restricted (among other restrictions) ) by the wavelength, I believe you have guided me so that I can pick it up from here and do some research to fully understand the reasons. Thanks very much.