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Laser intensity as a function of distance

  1. Apr 25, 2006 #1
    i built a laser in my lab...
    it consists of one Ne-He tube with a concave mirror at one end and a window at the other end.
    i just put another mirror infront of tube's window to get the laser.

    i got one question i cant find an answer to, the lab's guide said the intensity of the laser depends on the distance of the two mirrors.
    and i cant see why, i mean, the only reason i can find is that for some positions more light is being reflected back and forth between the mirrors so the laser is more powerful (but thats position, not distance.. and i always re-tune the mirror so it will produce the most powerful laser for that distance), and for some distances theres not enough light reflecting between them so no lasering effect is present.

    but apart from that... is there anything im missing?
    i mean im supposed to have a fit to the graph, and no function apart from heaviside comes to mind.
     
    Last edited: Apr 25, 2006
  2. jcsd
  3. Apr 25, 2006 #2

    Claude Bile

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    Unfortunatly the term 'intensity' carries a double meaning these days. The proper, radiometric unit for intensity is Watts per steradian, however it is commonly used with units of Watts per square metre as well (Irradiance is the proper name for the quantity with units W/m^2). To get a satisfactory answer to your question, you need to clarify exactly what you mean by Intensity, whether you mean W/sr or W/m^2.

    Claude.
     
  4. Apr 25, 2006 #3
    i meant irradiance, and its always measured at the same distance from the back mirror, the only thing that i change is the position of the front mirror...
     
  5. Apr 26, 2006 #4

    Claude Bile

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    When you increase the separation of your mirrors, you are doing two things which could contribute to the decrease in irradiance. Firstly it may increase the far-field divergence of the beam, which will be evident as a change in spot size as you vary the mirror separation. Whether this effect appears depends on other cavity parameters. Secondly, the beam inside the cavity spreads out spatially, which increases losses within the cavity (assuming that you filled the gain medium to begin with). This leads to an increased lasing threshold and hence a lower output power and thus irradiance.

    So, first thing to do, check that the spot size does not change with mirror separation, if not, you can be fairly confident that the drop in irradiance is due to an increase in intracavity losses.

    Claude.
     
  6. Apr 27, 2006 #5
    hmm.. im not sure i fully understand it.
    correct me if im wrong:
    the first reason you mentioned was the widening of the beam between the mirrors due to their finite size, am i right?
    and the second reason was the shape of the front of the beam?
    between the two mirrors is a tube of certain length with the gain medium, i dont add ne-he between them when i move them apart.

    sorry if i got you wrong... guess my english is not as good as i thought it to be.

    and finally, how do i calculate the theoretical irradiance as a function of the mirror distance?
     
  7. Apr 27, 2006 #6

    pervect

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    Try playing with, for instance

    http://www.iap.uni-bonn.de/lehre/ss01_laserphysics/GaussianBeam.html

    Your laser system can be described as a Gaussian beam (note the curved mirror).

    You should be able to see by playing with the java applet that for a certain specified curvature of the mirror, there is an optimum cavity length which minimizes the beam divergence. This occurs when the black dot is in the center of the attached "stability diagram".

    For details on the physics, read up on "Gaussian beams". I'm not sure what level you are at - google, for instance, finds

    http://www.mellesgriot.com/products/optics/gb_1.htm

    as one source to start reading about the theory.
     
  8. Apr 27, 2006 #7
    thank you very much.
     
  9. Apr 27, 2006 #8

    Claude Bile

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    The widening of the beam is due to diffraction.
    The second reason is due to intrinsic losses within the cavity, absorption and scattering are two such sources of loss.
    Not sure off the top of my head. If you could express the far field divergence as a function of mirror separation, that would be a step in the right direction, though such analysis typically neglects variations in intracavity losses.

    Claude.
     
  10. Apr 28, 2006 #9
    ok, so i got you right with the first reason :biggrin: you get a diffraction because only a finite source area is used (as always), if you got an infinite concave mirror (whice is nearly flat, so you dont get a sphere) there would be no diffraction.

    anyway, thank you for explaining, i'll try calculating it right after i'll refresh my memory, and see if it fits.
     
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