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One for the physics guru's

Discussion in 'Off Topic [BG]' started by i_got_a_mohawk, Mar 22, 2009.

  1. Trying to calculate the dielectric constants for two substances. One with refractive index 1.51, one with refractive index of 0.47.

    Using in a system where 670nm light is being used.

    670nm is 1.85eV (I think).

    I'm looking over the equations and I'm just lost and confused. Anyone got any advice or a simple walkthrough of how to approach the problem?

  2. machine gewehr

    machine gewehr

    Sep 17, 2005
    I called you nerd a few days back and you denied it.
    Now,I'm studying computers and I tell you its not easy at all but what you posted,man...You're one big nerd.:p

  3. I know nothing about this, but......

    someone must have done a spreadsheet for this so you can just plug in the numbers.........

    Try university sites.
  4. Ok, ok.

    I know I'm a nerd :( , heck, yesterday with the missus, was down to my backup rubber (blueberry!) and made a Dr Manhattan comment :oops: .
  5. I've tried looking and can't find anything other than the equations I need to use. I just can't figure out how to do it tho. Damned complex numbers.

    Heck, even if I could just figure out what to do with this extinction coefficient part I'd be able to plug away at it :bawl: , wish I knew how to use things like matlab now :(
  6. machine gewehr

    machine gewehr

    Sep 17, 2005
    Just keep the thread on top so some one will surely help,lots of sharp minds here.

    Last week,I learned I had to submit a java work just 2 hours before the deadline.First thing I did was come here and start writing for help.Luckily my friend was able to help me before I hit the post thread button.:p
  7. Lucky :p

    I've had this for a while, deadline is still a while away. It's part of a greater report, but I'm just stuck on this part :(
  8. CDRhom

    CDRhom William Murderface's Bass Tech Gold Supporting Member

    Feb 27, 2009
    Fort Worth, TX
  9. fdeck

    fdeck Supporting Member Commercial User

    Mar 20, 2004
    Madison WI
    HPF Technology LLC
    I'm a physicist, and have taught electrodynamics, but I don't want to just give away the answer. So I will offer just a faint hint instead. We throw around a whole bunch of different parameters... permeability, permittivity (both "absolute" and "relative"), refractive index, dielectric constant, and so forth. It would be worth finding all of the definitions and relationships between those quantities that you can find in your textbook or a trustworthy online reference. Stare at those relationships and see which variables you can eliminate.
  10. I'm not looking for just an answer, I'm more looking for help on how to crack it, so your help is very much appreciated.

    The main thing I really don't understand is the ik (imaginary number and extinction coefficient?).

    I've had a look about and just can't seem to find a solid way in which to work out the imaginary part of the equations. On saying that, I have a hard enough time working with regular numbers, let alone complex ones (well, this is the first time I've ever done anything with complex numbers, really wish I had taken a masters more in the vein of my undergrad)!
  11. I know that if done simply, the dielectric constant is the root of the refractive index, which is the way I tried to go backwards in order to get the answer. But it seemed to mess with the results quite a lot by missing out the imaginary component.

    Cheers for the links :)

    If it's of any interest the experiment is to do with surface plasmons.

    And on that note, something has just clicked and I may have this . . .
  12. Ok, maybe not. Bugger.

    Looking for an angle around 80-90 degrees. Got my answer to (theta) = Sin(-1) 1.08.

    I realised I had made a daft mistake earlier on in the maths. But sadly, looks like I still need the imaginary part to knock it down a bit to get under 1 :(
  13. hbarcat

    hbarcat Supporting Member

    Aug 24, 2006
    Rochelle, Illinois
    Ahem . . .

    Other terms for the relative static permittivity are the dielectric constant, or relative dielectric constant, or static dielectric constant. These terms, while they remain very common, are ambiguous and have been deprecated by some standards organizations. The reason for the potential ambiguity is twofold. First, some older authors used "dielectric constant" or "absolute dielectric constant" for the absolute permittivity E rather than the relative permittivity. Second, while in most modern usage "dielectric constant" refers to a relative permittivity, it may be either the static or the frequency-dependent relative permittivity depending on context.
    - Wikipedia

  14. Most of that is over my head. It might be because I've been working away on one thing or another for the best part of the day and it's almost 1:30 am and my brain is shutting down :p.

    However, I'm pretty sure it's frequency dependant, if not, I don't understand why we're to make use of the wavelength of the laser.

    I dare say, if I only fail this one thing . . . it wouldn't be toooo terrible . . .

  15. My brain hurts.
  16. Dertygen


    Dec 21, 2008
    A-Town, Colorado
  17. fdeck

    fdeck Supporting Member Commercial User

    Mar 20, 2004
    Madison WI
    HPF Technology LLC
    This is just slightly involved. The solution to the wave equation for the electric field has something like exp(i(k x - w t)) where k is the so called propagation vector and w is the angular frequency. I don't know where you are at with this stuff, but is this familiar?

    If k is real, then the entire exponent is imaginary, and it is just the plane wave that propagates forever.

    Now let k be a complex number a + i*b. Now if we put this into the wave solution, we get:

    exp(-b x)*exp(i(a x - w t))

    This is the plane wave again, but with a simple exponential decay term in front of it. Thus, you have an expression for the electric field in a medium where the wave is being extinguished as it passes through the medium. An example is trying to shine a light through a piece of metal. Many common metals have tabulated values for complex refractive index.

    Is this useful? Yes. The complex refractive index is used in computing the detailed behavior of reflection from metal surfaces, or transmission through thin metal films such as mirror sunglasses.

    My suggestion is: Work through the derivations a few times until you are comfortable with the complex exponential notation. It's your friend. Note that the "hyperphysics" site is pretty cool, and most of the Wikipedia entries on physics are decent.
  18. fdeck

    fdeck Supporting Member Commercial User

    Mar 20, 2004
    Madison WI
    HPF Technology LLC
    It's good to spend a bit of time and see if you can crack this nut, but if you really get up against a wall, let me know.
  19. It looks kinda familiar. I see a similarity with the complex number k and the dielectric constant equation where:

    E = E' + i*E''

    (where E is epsilon)

    I need to wake up a bit and come back to this, thanks again for all the help so far ! :D

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