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Laser resonator exercise

  1. Jun 14, 2005 #1
    We've had a couple of exercises on lasers, and there's two I had some problems with:

    1. He-Ne-laser

    Given

    A He-Ne laser with a 'base wavelength' (might be a bad translation of terminology, for which I apologize) lambda= 632.8 nm;
    Level width delta(nu) = 1500 MHz ;
    Allowed deviation of laser frequency = 100 MHz ;

    This laser works in a resonator of length L, in a medium with n=1.

    Asked

    a) What is the length of the cavity so that the laser only has 1 longitudinal mode?

    b) What is the allowed deviation of length L, with the given deviation for the mode frequency?

    My problem

    a) is easy: We're working with a resonator, which is comparable to a Fabry-Perot resonator, and so the spread in frequency of longitudinal modes is:
    delta(nu) = c/(2*n*L) or delta(lambda) = lambda²/(2*n*L)

    1 longitudinal mode in the cavity, means that the spread in level width (which is also the spread of the gain curve of the laser) must be smaller than the spread-in-longitudinal-modes width:).

    So L < 10 cm .

    b) however I don't really get. My laser frequency can deviate 100 MHz... what does this mean exactly? Where does it deviate from? A deviation in length means that the spacing between modes changes.

    I made an odd derivation that seems to fit : but that doesn't do me any good if I can't understand it.. Derivation was:

    With L, spacing is : L = m*lambda1/(2*n), m =1,2,3...
    With L+dL, spacing is : L+dL = m*lambda2/(2*n), m=1,2,3...

    So dL = m/(2*n)* d(lambda)

    lambda = c/nu
    d(lambda) = -c/nu² *d(nu) =-lambda*d(nu)/nu

    so:

    dL/L = d(lambda)/lambda1= -d(nu)/nu1

    nu1 = c/base lambda = 3e8m/s / 632.8e-9m = 474 THz

    so dL = 21 nm, which is the solution that I should get.

    Still... I don't know.. I feel I'm not really understanding the physics behind this... I got this solution after 5 wrong solutions, and I still don't really see the reasoning behind it. Anyone care to clarify?

    2. Vertical Cavity Surface EMitting laser

    Given

    A gain curve in the shape of a parabola... It's easier to write down the mathematical characteristics... if x-values are lambda, y-values are gain values... then the parabola is centered around lambda =850 nm... and sections with the x-axis are at 865 nm and 835 nm (so delta(lambda) = 30nm).
    The top of the curve is a value g_max.

    The laser system is in a resonator of length L, and with mirrors on both sides, reflecting 99,9%.

    A cryptic point for me is the given : "Gain area: Quantum wells (L=2*lambda/n)";

    Asked

    a) What is g_max? (before threshold... ie. before the resonator reaches pumping threshold)
    b) What is delta(lambda) for points with gain that deviates less than 1% ?
    c) What is the number of modi in this area with cavity length =L?

    My answers

    To be honest...I don't really get it. a) I don't get at all.. wouldn't know how to start with it.

    b) I assumed they meant : delta(lambda) for points that deviate 1% from g_max. So I calculated the points on the parabola for which g = 0,99 * g_max (using the value g_max = 1940/m given as a solution), but I didn't get the right answer.

    c) This should be fairly easy to calculate..
     
  2. jcsd
  3. Jun 15, 2005 #2

    Claude Bile

    User Avatar
    Science Advisor

    With regard to question 1, part 2:

    Okay, the maths first. You got it right in a roundabout way, which may contribute to why you are confused with the physics. A more succinct way of working out the solution would be;

    [tex] \frac{dL}{L_0}=\frac{d\nu}{\nu_0} [/tex]

    Then rearranging for dL;

    [tex] dL=\frac{Ld\nu}{\nu_0} [/tex]

    Plugging in the numbers gives the correct result.

    All you are doing is working out the minimum shift in dL required to change the laser frequency by the maximum allowed amount from the central frequency. Central frequency is usually defined as the centre frequency of a 3 db gain bandwidth.

    Claude.
     
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