Refractive medium: Calculating change of wavelength

Mark L. Fergerson

The article referenced states that even though the wavelength changes, the frequency does not. How is this possible?

Can anyone provide an example using the figures in my OP?

Martin Crawford

The dielectric properties of water are complex at radio frequencies. See for example:

John

I _think_ that the refractive index is just the ratio of the speed of light in a vacuum and the speed of light in the medium. If that's true, then the wavelength will reduce directly as a factor of the refractive index.

But it's been a loooooooong time since I've needed to know this -- Wikipedia, or Google, should be your friend here.

Keep in mind that refraction only tells the whole story if the material is lossless and not frequency selective compared to the spectrum of the signal you're sending through it.

F (cycles/second) * Lambda (meters/cycle) = Speed(meters/second)

There's complications, of course, if the material has more than one kind of internal wave propogation (Karo corn syrup has two indexes of refraction for visible light).

The speed of light is less than in a vacuum. Art

The very simplest definition of refractive index is

n = /

Basically you only have to consider continuity of the wave E field at the surface boundary to see how the only possible solution when its speed slows down is for the wavelength to shorten proportionately.

So wavelength in refractive media = lambda_vacuum/n

It is left as an exercise for the reader to prove this.

There is actually a tiny refractive index of air vs vacuum, but I cannot be bothered to look up what it is at 900MHz.

So your concrete number example is 15.7/1.3 = 12.077cm

There is a subtle distinction between phase velocity of the wave crests and group velocity of a signal travelling in a dispersive medium like a waveguide or edge of a sharp spectral line which is usually the fundamental misunderstanding of physics at the root of any claims for FTL signalling by electronics engineers.

Interesting fact:

Minor corrections for imperfect vacuum in historic experiments used to measure the speed of light in a vacuum have been responsible for systematic errors that at one point far exceeded the error bars.

Plotting contemporaneous speed of light with error bars against time is very informative - basically every subsequent experimenter that refined this method reproduced the original (very famous) experimenters error exactly. It was only when a new even more accurate technique came along that the discrepancy was observed and the systematic error corrected.

One of the Relativity texts of the 1970's had this plot in as a salutary lesson about assuming that famous experimentalists were infallible and measurement errors free from systematic effects. (actually if anyone knows of an online retailer with the original book I would love to obtain a copy - sadly I can't recall the title)

Empirical fact. See Martin Brown's post for expansion.

Plug in the numbers and do the math. Easy peasy.

Mark L. Fergerson

On Fri, 05 Apr 2013 07:57:27 +0100, Martin Brown wrote

Thanks to everyone for their helpful answers.

Martin Crawford

El 04-04-13 4:35, snipped-for-privacy@teledyne.com escribió:

Hello Martin,

Water at 900 MHz doesn't behave as a lossless dielectric like PE, PTFE or PP, especially when it is conducting due to contamination.

The phase velocity in a medium:

Vph = 1/Re{(sqrt(u*e) }

Re{..} = Real part of ... u = complex permeability (4*pi*10e-7 H/m for water), e = complex permittivity for the medium (er*8.854e-12 F/m)

If you convert u*e to polar notation, taking square root from a complex number is just halving the argument and taking the root from the modulus.

Re(Z) = |Z|*cos(arg(Z))

Though the real part of relative permittivity (er) is around 78 over a realtive wide range of frequencies, above 1 GHz, er drops with increasing frequency. The imaginary part (the loss part) changes with frequency, and depends on the conductivity (due to contaminations).

For 900 MHz and pure water, er = 78-j8 for 900 MHz and seawater, er = 78-j60

When discussing pure water you may use er = 78 + j0 as a first approximation, this saves you from complex calculus. This will result in Vph = 0.34e8 m/s (that is 0.11*c0)

Wavelength follows from lambda = c/f = vph/f c = propagation velocity (m/s), f = frequency (Hz)

lambda (900 MHz) = 0.34e8/900M = 38 mm.

Best regards,

El 05-04-13 13:24, snipped-for-privacy@teledyne.com escribió:

Hello Martin,

If a good answer is important for you, I would spend some time on this topic. You can't use the refractive index for optical frequencies for the radio frequency range.

Is your challenge related to pure water, or for example seawater?

Best regards,

Context?

It may be useful to ask on sci.optics.

Joe Gwinn

El 04-04-13 11:25, snipped-for-privacy@teledyne.com escribió:

Hello Martin,

If you are familiar with complex calculus it is not that difficult to calculate the wavelength and attenuation. However you need to know the complex permittivity of the material at the operating frequency. When using magnetic material, you also need the complex permeability. You generally can't use tables that are valid for optical frequencies.

Complex permittivity is mostly specified as relative epsilon in the form of eps.r' and eps.r''

Complex permittivity = eps = 8.854e-12(eps.r' - j*eps.r'')

Eps.r''/eps.r' = tan(delta).

The complex material properties fit into the formula for the complex propagation constant (gamma).

Gamma = Alpha + j*Beta = j*omega*sqrt(u*eps)

For non magnetic materials u = 4*pi*e-7 H/m, eps = complex permittivity.

You may know that: Beta = 2*pi/lambda and

After some calculus, you will arrive at:

Lambda = 1/( f*Re{sqrt(u*eps)} )

Only when the material is low loss (so Eps.r''/eps.r'