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Single-slit diffraction and wavelength

  1. Oct 22, 2006 #1
    I keep hearing that single-slit diffraction only occurs when the magnitude of the wavelength of the wave is larger than the slit. But according to the diffraction equation asin(theta) = m(lamda), minima can never occur when the wavelength is greater than the slit. I've tried reading my textbook many times, looking at online articles on diffraction, but none seem to answer my question. Could someone please help me on this?
  2. jcsd
  3. Oct 22, 2006 #2


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    That is simply not correct. I hope you're not seeing this in a real physics textbook. Sloppily written popular presentations and Web sites, on the other hand... :yuck:

    That is correct. When the wavelength equals the width of the slit, the first minimum occurs at 90 degrees, that is, the entire "field of view" in front of the slit is filled by the central peak of the diffraction pattern. When the wavelength is longer, the "sides" of the central peak get chopped off at the 90-degree point on both sides, but the intensity still has a maximum at the center and decreases as theta increases in either direction.
  4. Oct 22, 2006 #3
    this is from wikipedia: "The most conceptually simple example of diffraction is single-slit diffraction in which the slit is narrow, that is, significantly smaller than a wavelength of the wave."
  5. Oct 24, 2006 #4
    can someone please help?
  6. Oct 24, 2006 #5


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    The diffraction from a slit that is less than a wavelength does not exhibit the interference pattern you see from a wider slit. Look at the picture at Wikipedia. There are no minima, just circular waves leaving the slit. Diffraction is still ocurring, i.e., the barrier around the slit does not create a shadow.

    An ideal double slit experiment would use slits that are less than a wavelength to get semicircular waves from the slits. Then the observed interference pattern would be purely double slit interference.
    Last edited: Oct 24, 2006
  7. Oct 24, 2006 #6
    You're interested in diffraction, rather than interference? Diffraction will occur regardless of the relation between slit width and wavelength, it is the effect whereby light doesn't all go in quite a straight line. A narrow slit experiment is an extreme example where, though you would expect the light to shine only in one precise direction, it is actually visible from almost every direction. You can see another example in your shadow if you slowly move two fingers very close to each other. Your equation seems to be for interference, the effect whereby light originating from different positions (ie. along a not-quite-so-narrow slit) overlaps to enhance or cancel out in different directions.
  8. Oct 24, 2006 #7


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    In fact, if you remove one side of the slit completely, leaving only one edge and producing a "slit" that is (semi-)infinitely wide, you still get diffraction around the remaining edge. This is often called "knife edge diffraction".
  9. Oct 24, 2006 #8


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    Which is also a problem with any telescope where a secondary mirror is held by "spider vanes" crossing the field of incoming light to the main objective. Even a secondary without vanes, as in an SCT or Maksutov, causes diffraction from the edges of the secondary simply because it (the secondary) is there.

    With vanes, you get the famous "difraction spikes" seen in so many photos. With just a secondary you get diffraction "smear" where all wavelengths are scattered across the whole image which lowers both contrast and resolution. The best view, and photos, you see come from refractors where all the objects, especially stars, are sharp and pinpoint.

    Off subject, I know, but I couldn't resist.....:biggrin:
  10. Oct 24, 2006 #9
    So if I'm understanding you correctly, the asin(theta) = m(lamda) for minima in diffraction where 'a' is the width of the slit is used only whenever interference ALSO occurs in the diffraction of light from the sinlge-slit? If so, then I've been wasting so much time reading the textbook over and over just for some slight misinterpretation of the text!

    I greatly thank everyone for their help!
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