^ "Fraunhofer, Joseph von (1787-1826) - from Eric Weisstein's World of Scientific Biography".As the spread of wavelengths is increased, the number of "fringes" which can be observed is reduced. If the spread of wavelengths is significantly smaller than the mean wavelength, the individual patterns will vary very little in size, and so the basic diffraction will still appear with slightly reduced contrast. it consists of a range of different wavelengths, each wavelength is diffracted into a pattern of a slightly different size to its neighbours. In all of the above examples of Fraunhofer diffraction, the effect of increasing the wavelength of the illuminating light is to reduce the size of the diffraction structure, and conversely, when the wavelength is reduced, the size of the pattern increases. Since you now understand the geometrical shift of the minimum along with a, you can make sense of this condition optically: If a phase shift of $\lambda/2$ results in a destructive interference, then shifting that phase by a whole $\lambda$ should again result in a destructive interference.I ( x, y ) ∝ sinc 2 ( π W x λ R ) sinc 2 ( π H y λ R ) ∝ sinc 2 ( k W x 2 R ) sinc 2 ( k H y 2 R ) Non-monochromatic illumination places with complete destructive interference. This is the condition for all minima in the intensity pattern, i.e. $xdist=a*sin(\theta)=m*\lambda$ (where m is a non-zero positive integer) In fact, you will notice that this works with all multiples of 2, so let's make the algebraic: Turns out this is the condition for the second minimum. If we use the diagram from the video you posted, the a/4 condition would mean that we have the distance of $xdist=\lambda/2$ higher up and therefore the angle of the resulting beam would be angled higher. What is the difference between a/2 and a/4: Think about it geometrically first. That leaves us with 1 entire section that does not destructively interfere, no minimum. Same thing with quadrant 2 and 4, hence we also have a minimum here.įor a/5 it does not work: We split the beam up into 5 sections, with 4 of them we can make the same argumentation as before. quadrant 1 and 3 destructively interfere. Beam 1 from quadrant 1 would destructively interfere with beam 1 from quadrant 3, etc. Every beam in the top half would find a beam corresponding in the bottom half that would lead to destructively interference.įor a/4 we have the same case: We can split the beam up into 4 quadrants each containing an arbitrary number of beams. Think about it this way: The a/2 condition essentially says that if you were to divide the beam up into two halves with an equally arbitrary resolution of beams in each half, then the first beam of the top half (at 0 measuring from the top of the slit) would destructively interfere with the first beam of the second half (at a/2 measuring from the top of the slit). This how is the variation in intensity is explained. When $xdist$ does not equal $a/2$, then you cannot do this 'pairing up' and some portion of the beam will not destructively interfere. That must be the case for some angle of the diffraction, and when that is the case you can 'pair' up a portion of the beam with another portion of the beam and you will get complete diffraction. The fact that $xdist = a/2$ is a condition not an assumption.
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