On Thu, 05 Apr 2007 09:23:13 -0700, Jim Lux wrote:

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?

frequency stays the same, but since it's moving slower, c is smaller, so
lambda (wavelength) is shorter.

Same thing goes on in coaxial cable.. the wave propagates in a
dielectric with a propagation speed, say, 66% of the free space speed.
In such a case, a one wavelength long piece of coax for 30 MHz is 6.6
meters, not 10 meters (the free space wavelength)

The challenge, of course, would be in getting the opposite phenomenon to
occur (propagation faster than free space)...but that's a topic for a
different day.

jim

Speedy Gozales did it, but that's also a topic for a different day.

Walt, W2DU