1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Wavelength and Aperture Size

  1. Aug 7, 2008 #1
    This has bugged me for a while;

    In CD's for example the wavelegth of the laser is the limitng factor of the amount of data because of the spot size it makes. The spot size is diffraction limited, so the smaller the wavelength the smaller the allowed spot, OK so far.

    But when we are talking about something like a faraday cage to block out microwaves for example, we say the spacing of the cage 'bars' should be shorter than the wavelength, I have trouble understanding this, as the wavelength is along the normal to the cage surface, and the wavelength shoul not have a bearing on whether it 'gets in' or not.

    Why is it that the spacings (which are paralell to the polarisation vector) determine what kind of waves get through?
  2. jcsd
  3. Aug 7, 2008 #2
    Are you familiar with how a wire grid polariser works? (Particularly the direction of the polarisation relative to the grid, see wikipedia..)
  4. Aug 8, 2008 #3
    from Wiki:

    "For practical use, the separation distance between the wires must be less than the wavelength of the radiation, and the wire width should be a small fraction of this distance. This means that wire-grid polarizers are generally only used for microwaves and for far- and mid-infrared light. "

    This is what I have trouble with; why sould the longitudinal dimension (wavelength), be affected by objects that are transverse to it?
  5. Aug 8, 2008 #4
    Classically, it's not a matter of "blocking" the wave. Rather, eddy currents induced by the wave in the metal will produce their own wave to (at least approximately) cancel out the transmission. (Consider how active noise-cancelling headphones can work without requiring an air-tight seal.)

    Doing the math for Bragg diffraction is a simple example. In short, the ratio of wavelength to transverse dimensions becomes relevant because some of the rays involved have transverse components to their direction of propagation.
  6. Aug 8, 2008 #5
    say if we take one photon of wavelength 2X and want to pass it through a mesh of seperation X, I dont see whats to stop it. Surely if anything is to stop it it will be te amplitude of the wave which has dimensions in the transverse.

    So when we say it's wavelength dependent, we really mean it's energy dependant, right?
  7. Aug 8, 2008 #6
    Alas, completely incorrect.

    The amplitude does not have dimensions in the transverse, that is a common misconception (and part of my reason for drawing your attention to wire grid polarisers).

    The mesh size is indeed specifically dependent on wavelength (not energy either). Do you follow the trigonometry of Bragg diffraction (and of single slit diffraction)?
  8. Aug 8, 2008 #7

    Andy Resnick

    User Avatar
    Science Advisor
    Education Advisor

    How about if I turn it around- objects have a (electromagnetic) wavelength-dependent behavior due to certain physical length scales being present in the material? For wire-grid polarizers, the spacing of the wires is determined by the manufacturing process. For polarioid polarizing film, the spacing of the iodide-quinine crystals is much smaller due to the use of chemical fabrication. As it happens, the length scales in polaroid correspond to optical wavelengths, while the wire polarizers, having a larger spacing, is most effective in the mid-infrared and further out.

    Note that this is a different question than determining the extinction coefficient for the devices- and why the transmitted polarization is orthogonal to the physical orientation of the wires/crystals.

    Or am I missing the point?
  9. Aug 9, 2008 #8

    Ok, just wrote a long reply that got lost somewhere, but this is the jist of it;

    I get the trig, but I'm still having problems with it conceptually;

    If we want to make a Faraday cage for optical radiation, would we make a wire mesh with 100nm spacings?
  10. Aug 9, 2008 #9

    Andy Resnick

    User Avatar
    Science Advisor
    Education Advisor

    Pretty much, sure. I don't think it would be called a faraday cage- probably a mirror :)
  11. Aug 10, 2008 #10
    Would it?

    I have trouble with that, If we had a 100nm aperture you're telling me that if we fired optical photons at it, we would not be able to detect them on the other side?

    A mirror, usually Ag coated, has spacings (between xstal planes) much smaller than 100nm.

    Regardless, I'm still not satisfied, the trig of diffraction explains why you get a certain diffraction pattern for a given lambda etc, but wavelengths larger than the aperture can obviously get through in this case.

    Why then, does the cage block wavelengths of lambda larger than the aperture size?
  12. Aug 12, 2008 #11

    Andy Resnick

    User Avatar
    Science Advisor
    Education Advisor

    Whoa- hang on there. A wire grid polarizer consists of bizillions of regularly spaced wires. Well, hundreds, at least. That's a lot different from a single sub-wavelength aperture.

    In fact, subwavelength apertures are used extensively for near-field optics, in conjunction with scanning methods.

    Maybe I'm not understanding the original question?
  13. Aug 12, 2008 #12
    It might be useful to remember that the microwave radiation exists as a series of fields propagating through space. The electric field and magnetic field are at right angles to each other, and both of these fields are at right angles to the direction of propagation. While there is a wavelength along the direction of propagation, the wavelength along the direction of these two fields is what you are concerned with in your question. As long as the spacing of the bars is greater than one wavelength, and the diameter of the bars is very small, the fields pass through largely unimpeded. At a spacing of exactly one-half wavelength, theoretically the fields will be shorted out and at spacing of less than one-half wavelength they will be reflected. The mathematical treatment is somewhat complex, involving characteristic impedances, reflection coefficients, attenuation and phase shift constants, it gets a bit involved!
  14. Aug 12, 2008 #13
    ***the trig of diffraction explains why you get a certain diffraction pattern for a given lambda etc, but wavelengths larger than the aperture can obviously get through in this case. ***

    I assume that they can get through because the diffraction grating isnt conductive.
  15. Aug 12, 2008 #14
    I may be mistaken, but I believe the original poster was referring to the trade off between laser spot size and wavelength. The higher the frequency and shorter the wavelength, the more bandwidth is available to carry information, so more data can be carried at higher frequencies. However, because of the smaller spot size, the greater the data density in the spot, which has an upper limit. So there must be a trade off or an optimum point where the advantage of increasing the frequency and bandwidth is reduced by the disadvantage of the data density and greater difficulty in recovering the transmitted data. But I don’t think there is a similar analogy in microwave radiation and faraday screens. In this case the signal is blocked by the grids because of the physical spacing which constricts the allowable wavelength that can pass. So it may be a mistake to try and make an analogy between the limiting factor of spot size, which is data density, and the limiting factor of wavelength passing through a grid, which has to do with fields and conductors. But I would certainly be interested to know if such an analogy can be made.
  16. Aug 12, 2008 #15
    What I was really getting at was the relationship between the longitudinal wavelength and the size of the aperture through which it can pass.

    Granpa; usually difraction slits are made out of metal, well in my experience anyway.

    It seems a picture is emerging though, if we have a single slit, wavelengths below the slit slit spacing can pass through (diffracted), but if we have a regular array of slits it seems to block the radiation (polariser). If we stick two polarisers together, with the lines orthogonal, no light passes.

    Schroder; you say that the E and B fields also have a wavelength, but is this wave not propagating in the direction of the EM wave?
  17. Aug 12, 2008 #16
    Right. I do understand your question but it does take a bit of visual conceptualization to get the feel for this. I hope my explanation makes some sense: I usually refer to the E and H fields, but no matter, these fields are right angles to the direction of propagation and at right angles to each other. Of course, they do move along with the wave in the direction of propagation, but they move as a wave front. If you were directly in the path of the oncoming signal, and if you could see it coming, the E and H fields would appear as the wall of a steradian moving towards you. They extend out to the sides at a distance of one-half wavelength on each side, or one wavelength across. Depending on the type of polarization, if orthogonal the E field may be horizontal and the H field vertical or vice versa, or if circular polarization they may be rotating CW or CCW. They extend out to the sides, but they propagate forward, towards you. They cannot extend any further than one-half wavelength on either side, (in a perfectly directional wave) because the wave is moving forward at the velocity of light and these fields are changing in accordance with the amplitude of the wave. So by the time the forward wave reaches max amplitude, the field reaches max strength (one-half wavelength) then it must collapse in again. The amplitude of the wave does not determine the distance the field extends to the side, only the frequency and speed of light determine the wavelength. They don’t propagate to the side; they are dragged along in the direction of propagation of the wave front. The amplitude determines how dense the field is, or how many lines of force it contains. It may be difficult to visualize at first but if you give it a try you will see it.
  18. Aug 12, 2008 #17
    What are you talking about?
  19. Aug 12, 2008 #18
    The price of tea in China, what else?
  20. Aug 12, 2008 #19

    Thank you for this excellent explaination, I think I have it now, if we have a higher frequency, the E and B fields don't have enough time to propagate any further out than half a wavelength of the light from the centre of the circle (if you're looking at it head on).

    So passing through an aperture larger than the wavelength, does not effect the E and B field, only the photons that are very close (1/2 lambda) to the edge of the aperture are diffracted.

    "The amplitude determines how dense the field is, or how many lines of force it contains."

    So the amplitude of the EM wave determines the density of the field. If we have a high power laser for example, does this mean that the E and B fields are the same spatially (1 lambda in total) but the collective amplitude of the EM wave is huge. People tend to talk about large E fields in laser physics, but a large field has the same spatial dimensions (1 lambda)?

    Is this correct?
    Last edited: Aug 12, 2008
  21. Aug 12, 2008 #20
    You got it, mate! That was the key thing, not enough Time to extend any more before they have to collapse again. G'day:smile:
  22. Aug 12, 2008 #21
    Did you just make this nonsense up yourself?

    Cite your reference.
  23. Aug 13, 2008 #22
    While there is no reason for me to engage in a discussion with you “cesiumfrog”, I will say that nothing in the “nonsense” I have written contradicts the established electromagnetic theory first formulated by James Clerk Maxwell circa 1870, which is my reference. In fact, Maxwell’s Equations are only true if at every point in space E is proportional to H, and at right angles to the direction of propagation and also if the velocity of electromagnetism has a fixed value c which is all that I have said. Perhaps if you tried understanding what was written you would not be so fast to make derisive comments. End of discussion!
  24. Aug 13, 2008 #23

    Andy Resnick

    User Avatar
    Science Advisor
    Education Advisor

    Maxwell's equations hold for much more general conditions that you claim. One restriction you place (E proportional to H) doesn't make any sense.
  25. Aug 13, 2008 #24
    I followed the link in your post to Maxwell's equations, and the following is cut and paste from there:

    The differential form states that the curl of the magnetic field strength is the sum of the free current density and the time differential of the electric displacement field.

    The integral form states that the integral of the magnetic field strength around a closed loop is equal to the sum of the net free current passing through any surface enclosed by the loop and the time derivative of the flux of the electric displacement field through that same surface.

    The differential form states that the curl of the electric field is equal to the negative [partial] derivative of the magnetic field with respect to time.

    The integral form states that the integral of the electric field around a closed loop is equal to the negative [partial] time derivative of the magnetic flux through any surface bounded by that loop.

    In other words: the circulation of the electric field is proportional to the rate of change of the magnetic field and the circulation of the magnetic field is proportional to the rate of change of the electric field.

    Does that make any sense?
  26. Aug 13, 2008 #25

    Andy Resnick

    User Avatar
    Science Advisor
    Education Advisor

    It indeed does make sense to me- that's not the issue. Does it make sense to you?

    I claim it does not, because you equated the proportionality of two quantities (E, H) in post #22 with the proportionality of spatial and temporal derivatives of those same quantities in post #24.

    Note to moderators: I did not intentionally create a link to Maxwell's equations in my post- was that automated?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook