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Questions about lasers and diffraction

  1. Sep 9, 2009 #1
    Okay, so I've been thinking about diffraction experiments and light, but not having a laboratory, I would like to know some specifics about what happens in a diffraction experiment. I know that laser light diffracts when it passed through a narrow aperture, but does it diffract as it passes out the opening at the end of a laser? Does laser light diffract from a point in midair, if so can you detect it by looking at the laser beam at a small angle. Third question: Does the edge of the laser have to actually hit the sides of the slit to diffract or can it diffract when passing through a narrow aperture comparable to the wavelength of the incident light but still larger than the laser light diameter? Thanks for any help.
  2. jcsd
  3. Sep 9, 2009 #2


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    Welcome to PF. :smile:

    Typically, the opening is designed to be considerably larger than the diameter of the laser beam, so that it does not create any additional diffraction.

    I don't really understand this question. Laser beams do diffract, even if you don't place an aperture in its path.

    Since laser beam diameters are much larger than the wavelength, the conditions you indicate are not possible.
  4. Sep 10, 2009 #3
    These are some very advanced questions! But you don't really need a lab to experiment. A simple laser pointer can demonstrate lots of nifty effects. You can use razor blades to make variable slits and rectangles.

    First note that the beam from a laser is self-confining. The beam propagates though space based on equations that confine the beam although it does tend to expand as you go far away from the source. The simplest beam like this is called "first order mode" and has a gaussian intensity profile. The modes are the result of the two mirrors that make the laser cavity. The mathematical solution of that system determines the propagation of the beam through all space (unless it's blocked by an aperture etc.) I told you, you were getting into complex stuff here.

    Second, as to how diffraction works, it basically works like this. If I have a plane EM wave in space (can be light or radio etc.) and it falls upon a hole that blocks all but part of it. On the other side of the hole will be seen a pattern of radiation. At some distance (known as the far field) certain patterns can be observed. For a standard rectangular hole the patterns will be the classic sinx/x forms. The relationship between the hole geometry and the final observed "diffraction pattern" is that there is a Fourier transform relationship between the geometry of the hole (Intensity pattern on it) and the angle of propagation away from the hole. For large distances from the hole and narrow patterns the angles are directly proportional to horizontal distances on your screen (sin X approximately equals x) But the true transform relationship is between geometry and angle. (not displacement on the screen)

    Hence a rectangular hole is like a single square electrical pulse which has the typical sinx/x transform pattern which is what is observed on your screen. If the pulse is narrow the spread in frequency is wide (pattern spreads) and if pulse is wide most energy is at DC (light mostly goes straight through the hole (angle = 0)). Note that the hole is 2D so that the transform is also 2D. Hence there is one transform in X and a second one in y. Whichever dimension is smallest has the widest pattern. If the hole is circular you have to use a transform with circular symmetry. I think you may have bitten off more than you can chew here!

    As for hitting the sides, remember that the beam has a Gaussian cross-section. Hence it really doesn't "end" anywhere. There are always "wings" that kit the hole. Now those wings may be so weak you see very little diffraction, but remember that is it the intensity pattern over a region of space that is the source and the far field of that source is the transform of that pattern. It really doesn't matter HOW the original source geometry was generated. OK?
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