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At 21:26 9/24/99 +0000, you wrote:
Could someone give me an explanation that could be understood by high
school students as to what the wavelength has to do with the amount of
diffraction.
We know that the most diffraction occurs when the wavelength is large
compared to the opening, but why?
Any help would be appreciated.
Gerry White
A graphical construction is drawn with a slit shown as a break in
a horizontal line. The light rays emanating from each side of the
opening are compared.
For the two rays issuing straight ahead, we suppose the waves
are in phase, and so we say that this direction is favored.
We then consider other directions to one side of straight ahead,
say to the left.
At some angle, the left ray is one wavelength shorter than the
right ray, so that this angle is also reinforced.
A similar argument shows that intermediate directions cancel the
wave.
A triangle drawn at the slit with its adjacent side normal to the rays
diffracting to the left and its hypotenuse forming the width of
the slit and its opposite side being one wavelength long leads
immediately to a rule connecting wavelength, slit width and
diffraction angle.
This rule depends on a wavelet theory of retransmission, or a
probability wave propagating thoughout - much the same concept
I suppose at this level.
But to answer your question, if the wavelength is very small
compared to the slit, the angle is small and different orders
merge in a general penumbra.
Is this the discussion you wanted?
brian whatcott <inet@intellisys.net>
Altus OK