Hi I am wondering why the relative permittivity given by a source for water is 80 and the refractive index is 1.333 is this due to differences in wavelengths. If so can anyone point me to a source which has the relative permittivity at visible wavelengths. Also would i be right in saying that at ~20degreesC the relative permeability of water would be about 1 because it isnt magnetic. Thanks Alex
Permittivity is generally measured at a frequency of a few kHz, whereas refractive index is measured at optical frequencies, so to answer your question, yes, it is to do with the different wavelengths these quantities are measured at. To get the relative permittivity at visible frequencies, just work it out from the refractive index - I assume you know how the two are related. For any dielectric, the relative permeability can be assumed to be 1 to a very good approximation. Claude.
I am doing a project on calculating the refractive index of substances. I have already done one experiment using optical methods snells law etc but wanted to do my second experiment in a different manner by looking at the electromagnetic properties of the substance and working out the refractive index that way. I was going to measure permittivity by calculating the capacitance between two plates accross a substance and then using that to work out the permitivity using E=AC/d where A = area E = permitivity of substance and d = distance between plates. I just wanted a source that i could check my results against to see what should turn out but i spose i could just calulate what the relative permitivity should be equal to 1.333^2 for water(shouldnt it). = 1.777F/m As a further experiment i was going to use different substances and see how their electromagnetic properties affected the refractive index. Alex
Yes You may also want to investigate other methods of measuring the refractive index, such as elliptometry. You may also wish to investigate anisotropic materials (materials whose refractive index depends on the direction of E). Claude.
I have been trying to do this experiment for a while now and now i am using a multimeter with a capacitance setting to calculate the capacitance across a container full of water. Then i am working out the permittivity using E=C*d/A I keep ending up getting a relative permittivity of 80. I have seen this quoted as the permittivity for water from online sources. The question is now how do i convert that value for that frequency to a value that will be realistic for visible wavelengths. 1.33^2 for visible frequencies is what i was expecting. Does the permittivity change for different frequencies? If so is there a way/formula that can change between permittivity at different wavelengths. Thanks Alex
80 is the correct value at low frequencies, up into low microwaves (S-band). In the far infrared (IR), vibrational modes of the H2O molecule are excited which dramatically change the complex dielectric constant, introducing as well large absorption. This was discovered during WW2 at the Radiation Lab; absorption by water vapor in the air put an end to using ever shorter wavelengths for radar to improve accuracy. These modes involve the behavior of chemical bonds, so they cannot be predicted from low frequency behavior. The dielectric constant has even more dramatic structure at optical frequencies. The absorption suddenly drops to near zero in a narrow band of frequencies where, not surprisingly, nature took advantage of having a window--it's what we call the visible band. Here's a web page that's pretty technical but it shows the frequency dependence you're asking about at least into the IR http://www.lsbu.ac.uk/water/microwave.html Jackson has an in depth discussion in his book Classical Electrodynamics (sect. 7.5 in the 2nd edition) that covers frequencies up to X-ray.
looks very interesting So is the only method of working out refractive index of water using snells law as that is all i can find at the moment. http://www.lsbu.ac.uk/water/images/microw22.gif Could this be used to work out n but wouldnt the absorbtion need to be a complex value to reduce the refractive index to visible levels. (lambda*absorbtion/4piR)^2 < 0 in order to decrease the refractive index due to the relatvie permittivity been so high. Alex
I'm afraid I don't follow what you are asking in the last paragraph. You can think of the real and imaginary parts of the dielectric constant as independent. That's not strictly true (especially around a resonance like the vibrational modes I mentioned earlier, where real and imaginary parts are related by what's called the Kramers-Kronig relations), but please realize that absorption is both high and low at various IR and optical frequencies.
Think im going out of my depth here Im trying to say for this equation http://www.lsbu.ac.uk/water/images/microw22.gif to hold true at visible wavelengths n^2 = 1.33^2 therefore the RHS must equal that therefore if Er is 80 then the rest would have to be negative to make the equation hold true. Alex
Ok, I see the confusion. At optical frequencies where you would use Snell's law to measure index of refraction, the permittivity is 1.3^2, not 80. 80 is the value at low frequencies (up to a few GHz in the microwave).
Right so the value for the relative permittivity changes with wavelength but is there a way to work out the relative permittivity of a substance at visible wavelengths without knowing the refractive index. Alex
Not without advanced E&M, quantum mechanics, condensed matter physics, at the least, and big computer codes.
Calculating refractive index difference is possible - this is done for stoichiometric glasses (glass where the refractive index varies with the level of dopants ions in the glass) for example. Essentially, given some dopant concentrations you can work out the refractive index of your glass - though the methods for doing this generally rely on empirical data anyway, and is rarely done via first principles. Claude.
The subject of the experiment was liquid water and i was going to try and find the different methods for calculating the refractive index. I thought it would be more iteresting to calculate it from the electromagnetic properties of the water rather than simply measuring angles and using snells law. It would be really helpful if there was a formula that i could use to connect the relative permmitivity of 80 at z frequency to the relative permittivity of x at y frequency. If you havnt already guessed im not too clued up on these kind of things after all this is only a high school project. Alex