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B Why do EM waves of longer wavelengths spread out more?

  1. Mar 25, 2017 #1
    Why do longer wavelengths spread out more than shorter wavelengths?
    What is the physics principle/law which explains why radio waves spread out more than optic waves in free space?
     
  2. jcsd
  3. Mar 25, 2017 #2

    sophiecentaur

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    The way that all waves basically behave (EM / sound / ocean etc) when they encounter an obstacle or a gap in a wall, is exactly the same and it is related to wavelength. The process is called Diffraction. Huygens (born 1629) first proposed that you could determine how a wave would progress, thought in terms of 'secondary wavelets', which are constantly forming on the 'wave front' and then add together to form the next step of the wave. Search "Huygens Construction" to see loads of links. In that wiki link, you can see the basics of the process, which later developed into the modern theory of diffraction. The effect scales with wavelength so a wide aperture of object will have the same effect of 'spreading' a wave with long wavelength as a narrow obstruction of similar shape on a proportionally shorter wavelength. Radio waves are just the same as light waves but the wavelength is many millions of times longer.
     
  4. Mar 25, 2017 #3

    jfizzix

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    The reason why a beam of visible light spreads out less than a beam of radio waves of the same area can be understood from the uncertainty principle.

    The uncertainty principle tells us that waves cannot be both well-confined in position and well-confined in momentum.
    The amount that a beam of light spreads depends on its momentum distribution which gives you the distribution of directions of propagation.
    With a really short wavelength (i.e., a higher momentum), the same uncertainty in momentum corresponds to a smaller uncertainty in propagation direction.
    Therefore, a tightly confined beam will have a larger spread of directions if it is of a longer wavelength.

    In this way, one can see that a narrow beam of long wavelength radiation must spread out faster than a short wavelength beam.
     
  5. Mar 25, 2017 #4

    sophiecentaur

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    Dear old Huygens didn't know about QM when he came up with his Mechanical explanation. Both models have their place, though. HUP is basically an inequality so that makes it difficult to be as precise as Huygens. :smile:
     
  6. Mar 25, 2017 #5

    ZapperZ

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    I'm going to tee off what sophiecentaur has asked, which is have you see examples of wave behavior in a ripple tank? What you are asking is NOT unique to EM wave. It is a property of diffraction in ALL waves.

    We can painfully go over Huygens principles, etc., but this particular aspect needs to be clarified so that you know the underlying principle involved here, which is wave mechanics, not just EM wave.

    Zz.
     
  7. Mar 25, 2017 #6

    sophiecentaur

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    It's almost more a matter of Maths than Physics, actually because it is just Patterns following a basic mathematical axiom.
    It is strange that the same Maths can apply to so many different phenomena. - right down to two sheep plus two sheep makes four sheep and the same for lollipops and pounds.
    All that's necessary for the wave thing to be common is that the propagation medium needs to be uniform and isotropic. When it's not, you get other effects as well.
     
  8. Mar 25, 2017 #7
    If you're familiar with the equation for Fraunhofer diffraction from a single slit, the first minimum occurs at an angle θ given by sinθ = λ/w where λ is the wavelength and w is the slit width. The angular width of the central diffraction peak would be 2θ. For a given slit width, the larger the value of λ the wider the central peak will be. But it's best to think in terms of the ratio λ/w.
     
  9. Mar 25, 2017 #8
    The thing that confused me was the fact that this Phenomena is usually expressed from the point of view of "Obstacles" and "Slits", while I'm speaking of "Free Space" and "No Obstacles". But, if the Antenna itself that radiates the beam would be considered like that slit, where radiation is produced from it, then I understand.

    Thank you all.
    Please let me know if my above understanding is not correct.
    I will need to go through the related formulas now.
     
  10. Mar 26, 2017 #9

    sophiecentaur

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    Yes. It's like having a hole in space, letting the EM energy through. That analogy is a bit over-simple because the effective size of that hole is much wider than just the thin metal wire but it holds.
    You can think of the diffraction pattern of an infinitely wide slit (i.e. free space) as having a max just in the direction of propagation.
     
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