Calculate Wavelength from Refractive Index

If I have a substance with a refractive index of 1.35, how do I calculate the wavelength of a 1GHz electromagnetic wave passing through it?

Mark Newman

Reply to
mnewman
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First of all, the definition of refractive index is n = sqrt(mu epsilon), where mu and epsilon are the relative permeability and relative permittivity (dielectric constant).

The (time) frequency of the light doesn't change when it enters the material, so the wavelength is proportional to the speed. (The wavelength is just the distance the wave travels in one period.)

The speed goes as 1/n, so the wavelength does as well.

However, the refractive index at 1 GHz is very unlikely to be the same as at optical wavelengths (which is where refractive index is commonly tabulated), so it's important to use the right values of mu and epsilon.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs 
Principal Consultant 
ElectroOptical Innovations LLC 
Optics, Electro-optics, Photonics, Analog Electronics 

160 North State Road #203 
Briarcliff Manor NY 10510 USA 
+1 845 480 2058 

hobbs at electrooptical dot net 
http://electrooptical.net
Reply to
Phil Hobbs

The permitivity would be 40. Conductivity is 1.

I am not sure of the mathematical relationship between the latter and permeability. Not having much luck on the net so far.

Would you be able to explain this and provide the final equation for my OP?

Mark Newman

Reply to
mnewman

** Hey - this is not Alfred E. Neuman is it ??

formatting link

... Phil

Reply to
Phil Allison

Conductivity is one what? Conductivity has units of siemens per square metre or equivalents. I'm assuming that that '1' is actually mu, and that the actual conductivity is small enough that you can usefully send RF through it.

If the material is ceramic or PVDF or something like that, where the magnetic response is weak, mu = 1 in an engineering approximation, so

n ~= sqrt(epsilon) ~=6.3.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs 
Principal Consultant 
ElectroOptical Innovations LLC 
Optics, Electro-optics, Photonics, Analog Electronics 

160 North State Road #203 
Briarcliff Manor NY 10510 

hobbs at electrooptical dot net 
http://electrooptical.net
Reply to
Phil Hobbs

And if it's not, you get imaginary numbers popping up in the permittivity, and therefore, refractive index. All that means is it's being attenuated sharply -- which is just another way of asking "can [you] usefully send RF"?

Isn't copper's |n| supposed to be something ridiculous at various frequencies, if you actually work it out? I must say, it sure takes a hell of long time for current to soak into a bar of the stuff.

Tim

--
Deep Friar: a very philosophical monk. 
Website: http://seventransistorlabs.com
Reply to
Tim Williams

Copper is a bit of a weird case, because it's a free-electron metal (the other two are gold and silver).

Above the plasma frequency, i.e. in the NIR and visible, free electron metals have an epsilon that is very nearly a negative real number, e.g. epsilon_r ~= -70+j7 at 1.24 microns, and declines very rapidly with frequency. Silver is even closer to a negative real, with epsilon_r ~= -81.5+j5 at the same wavelength.

This isn't that mysterious mathematically, of course--when there isn't much damping, there's almost 180 degree phase shift above resonance--but it gives rise to a lot of interesting physics, e.g. surface plasmons, light going through long skinny holes in metal films, huge resonances in silver nanoparticles, that sort of stuff.

To a leading-order approximation, a normal conductor has a complex epsilon given by

epsilon = epsilon_inf + j sigma/omega

(give or take a complex conjugation, depending on your sign convention), where epsilon_inf is a real-valued fit parameter, and sigma and omega are the conductivity and radian frequency as usual.

Copper's low-frequency conductance is around 60 MS/m, and epsilon0 is

8.8e-12, so at low frequency you'd have

epsilon_r ~ j 1e12/f(MHz).

If there weren't any resonances, this would drop to magnitude 1 at a wavelength of 28 nm in the extreme UV.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs 
Principal Consultant 
ElectroOptical Innovations LLC 
Optics, Electro-optics, Photonics, Analog Electronics 

160 North State Road #203 
Briarcliff Manor NY 10510 

hobbs at electrooptical dot net 
http://electrooptical.net
Reply to
Phil Hobbs

Isn't the definition of a metal that it has free electrons? How is copper different from, say, zinc or sodium? I've heard various rumors about this, but nothing analytical. And I say that as a physicist and ameteur chemist...

So |epsilon| is infinite at DC, which makes sense as there's no skin depth or anything, and still near infinite at most frequencies. At least, I think making an analogy with skin depth isn't too horrid. But it's just another way of saying, it still looks like a conductor until, oh, somewhere between THz and near IR, depending on use.

Tim

--
Deep Friar: a very philosophical monk. 
Website: http://seventransistorlabs.com
Reply to
Tim Williams

Metals have partly filled conduction bands, but the band shape doesn't have to be anything like the nice parabolas you get with free electrons. A near-parabolic band shape with weak interband coupling even at energies above the plasma frequency is what plasmon folks usually mean by "free-electron metal".

Pretty much. The method of moments sorta kinda almost works (if you squint at it) down to about 10 microns, and then falls apart. The conductivity isn't really infinite at DC, of course--the pole isn't really at DC.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs 
Principal Consultant 
ElectroOptical Innovations LLC 
Optics, Electro-optics, Photonics, Analog Electronics 

160 North State Road #203 
Briarcliff Manor NY 10510 

hobbs at electrooptical dot net 
http://electrooptical.net
Reply to
Phil Hobbs

Yes, conductivity is 1 S/m and permittivity is 40.

How is the wavelength of 1GHz within this medium to be determined?

Mark Newman

Reply to
mnewman

The conductivity contributes an additional term epsilon'' = j sigma/omega.

Thus at 1 GHz, epsilon_r = 40 + j (sigma/omega)/epsilon_0

which is

epsilon_r = 40+ j 1.59E-10/8.8E-12

epsilon_r = 40 + j 18

so n = sqrt(epsilon_r) ~= 6.5 + j 1.5,

again choosing the sign of the imaginary part so that it's a loss term.

That's pretty lossy if you're planning to use anything more than a few percent of a wavelength's worth.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs 
Principal Consultant 
ElectroOptical Innovations LLC 
Optics, Electro-optics, Photonics, Analog Electronics 

160 North State Road #203 
Briarcliff Manor NY 10510 USA 
+1 845 480 2058 

hobbs at electrooptical dot net 
http://electrooptical.net
Reply to
Phil Hobbs

Mmm, k-vectors and energy bands... one thing I didn't see much of (or forgot a lot of?).

Physicists do like to approximate local minima with parabolas (e.g., molecular QHO), but the point is, it breaks down (into further nonlinearity, i.e., the real function has a different form, which at least requires more Taylor terms in the approximation, or a better approach), and the inconvenience means it simply doesn't work as well (linearity; well... quadraticity) when you're pushing those limits (how the frequency response, plasmon physics, whatever, works out)?

Since, after all, you started with an approximation; op-amps aren't ideal integrators either, they just do a good enough job of it under most approximations.

Tim

--
Deep Friar: a very philosophical monk. 
Website: http://seventransistorlabs.com
Reply to
Tim Williams

Thank you for clarifying this.

Mark Newman

Reply to
mnewman

"Brendon"

** Go away and develop half a brain - f****it.

Trolls asking damn silly questions like the OP are NOT to be fed, courted or treated like gods.

This NG has become completely ruined because of them.

And you are just another one.

... Phil

Reply to
Phil Allison

I wouldn't be so harsh about it but I do wonder the value in performing suc h a calculation in the first place. I mean are we talking measuring frequen cy by a waveguide after a signal has passed through some sort of lens or so mething ? Sounds kinda like measuring a brick building by splashing water o n it......

Seriously, if there is a reason I would really be interested to know exactl y what it is. If it is just a futile exercise in mathematics, there are pro bably groups that could give better answers and be much more receptive to t he question.

Reply to
jurb6006

Alternatively we could just move on and quit agonizing over something so minor. The OP seems happy, and AFAICT hasn't corrupted the local youth. Built anything interesting lately?

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs 
Principal Consultant 
ElectroOptical Innovations LLC 
Optics, Electro-optics, Photonics, Analog Electronics 

160 North State Road #203 
Briarcliff Manor NY 10510 USA 
+1 845 480 2058 

hobbs at electrooptical dot net 
http://electrooptical.net
Reply to
Phil Hobbs

I wouldn't be so harsh about it but I do wonder the value in performing such a calculation in the first place. I mean are we talking measuring frequency by a waveguide after a signal has passed through some sort of lens or something ? Sounds kinda like measuring a brick building by splashing water on it......

Seriously, if there is a reason I would really be interested to know exactly what it is. If it is just a futile exercise in mathematics, there are probably groups that could give better answers and be much more receptive to the question.

** The post has all the hall marks of a home work question.

The alleged "substance" is fictional.

The only thing that gets near ( RI = 1.33 permittivity 60, conductivity 1 S/m ) is salt water.

Students sometimes do lab experiments passing microwaves through pure water and then adding salt to see the attenuation.

FYI the attenuation of salt water at 1GHz is about 560 dB/m !!!!!!

.... Phil

Reply to
Phil Allison

Maybe..take the case of having a radar set INSIDE a quonset hot and tracking birds at a few miles or so.. Wouldn't delving into mathematical models be interesting in attempts to explain that? Wouldn't that get one into resistivity, permittivity, etc - the area being discussed?

Reply to
Robert Baer

Actaully, you can make a lot of interesting microwave filters by inserting thoughtfully shaped pieces of plastic in a waveguide segment. A bit of arcana that i picked up helping a coworker test just such devices.

?-)

Reply to
josephkk

Comes to mind,

formatting link
of course, many designs are out there.

Lumps of dielectric in a waveguide is analogous to step changes in the width or height of the waveguide (probably not quite equivalent, due to dispersion and crap, but narrowband it should be very similar), which is again similar to changing the width of microstrip or stripline in a monolithic (PCB or otherwise) filter.

And you can get really interesting behavior by putting lumps of magnetized ferrite inside said waveguide, particularly inside a cross polarized structure (magic tee, or etc.).

Tim

--
Deep Friar: a very philosophical monk. 
Website: http://seventransistorlabs.com
Reply to
Tim Williams

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