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Old April 5th 07, 04:35 PM posted to rec.radio.amateur.antenna
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Default Not understanding some parts of wave refraction

On Apr 5, 7:36 am, "MRW" wrote:
I am skimming thru the Propagation chapter of the ARRL handbook, and I
am having a difficult time understanding the shortening of wavelength
and the retainment of frequency. They have an equation showing that
wave velocity is: c = f*w (c = m/s, f = frequency, w = wavelength).
It also states that during refraction "the wavelength is
simultaneously shortened, but the wave frequency (number of crests
that pass a certain point in a given unit of time) remains constant."

I don't understand. If the wavelength is shortened, then shouldn't the
frequency increase instead of remaining constant?


Refraction occurs when an EM wave, having frequency f and wavelength w
enters a medium in which the speed of propagation (speed of light) is
different than vacuum. A medium with an index of refraction greater
than one produces a speed of light which is slower than in vacuum
(index of refraction is simply the ratio of vacuum speed to speed in
that medium). This changes the proportionality between frequency and
wavelength. Since w = c / f, the slower speed at a given frequency
will now have a correspondingly shorter wavelength. And, as f = c /
w, the slower speed at a given wavelength will now have a
correspondingly lower frequency.

I hope that makes sense.

73, Jim AC6XG

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