Refractive medium: Calculating change of wavelength

[[email protected]]

Does anyone know the procedure (formula) for calculating wavelength of
radio waves, expressed in MHz, according the refractive index of the
transmission medium?

For example, water has a refractive idex of about 1.3. The normal
wavelength of a 900MHz signal is 15.7cm. What does the latter change
to when propagating through a sealed container of water?

Martin Crawford

 

[[email protected]]

Does anyone know the procedure (formula) for calculating wavelength of
radio waves, expressed in MHz, according the refractive index of the
transmission medium?

For example, water has a refractive idex of about 1.3. The normal
wavelength of a 900MHz signal is 15.7cm. What does the latter change
to when propagating through a sealed container of water?

Martin Crawford

http://www.rpi.edu/dept/phys/Dept2/APPhys1/optics/optics/node7.html


Mark L. Fergerson

 

[[email protected]]

http://www.rpi.edu/dept/phys/Dept2/APPhys1/optics/optics/node7.html


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

 

[John Walliker]

Does anyone know the procedure (formula) for calculating wavelength of
radio waves, expressed in MHz, according the refractive index of the
transmission medium?

For example, water has a refractive idex of about 1.3. The normal
wavelength of a 900MHz signal is 15.7cm. What does the latter change
to when propagating through a sealed container of water?

Martin Crawford

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

http://www.lsbu.ac.uk/water/microwave.html

John

 

[whit3rd]

Does anyone know the procedure (formula) for calculating wavelength of

radio waves, expressed in MHz, according the refractive index of the

transmission medium?

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).

 

[Artemus]

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 speed of light is less than in a vacuum.
Art

 

[Martin Brown]

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?

The very simplest definition of refractive index is

n = <speed of light in vacuum>/<speed of light in material>

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)

 

[[email protected]]

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

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

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

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


Mark L. Fergerson

 

[[email protected]]

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

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

Thanks to everyone for their helpful answers.

Martin Crawford

 

[Wimpie]

El 04-04-13 4:35, [email protected] escribió:

Does anyone know the procedure (formula) for calculating wavelength of
radio waves, expressed in MHz, according the refractive index of the
transmission medium?

For example, water has a refractive idex of about 1.3. The normal
wavelength of a 900MHz signal is 15.7cm. What does the latter change
to when propagating through a sealed container of water?

Martin Crawford

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,

 

[Wimpie]

El 05-04-13 13:24, [email protected] escribió:

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


Thanks to everyone for their helpful answers.

Martin Crawford

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,

 

[[email protected]]

On Thu, 04 Apr 2013 13:35:54 +1100, [email protected] wrote:

Can someone please explain the difference in methodolgies suggested by
Martin Brown and Wimpie?

The result for the same frequency of 900MHz in water appears to differ
substantially, eg. 38mm and 120mm.

Which one applies to my OP?

Martin Crawford

 

[Wimpie]

El 04-04-13 11:25, [email protected] escribió:

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

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

36% Penetration depth (skin depth) = 1/alpha [m]

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' <<1), the
formula will convert to:

Lambda = (lambda.freespace)/sqrt(eps.r), where eps.r is a just a real
number because we assumed "low loss" material.

If you do some google search you may find graphs that show wavelength
versus frequency for water, but knowing the formulas may give you a
means to check third party data.

Best regards,