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from different elements of the wave front in the form of propagating spherical wave, e^{(2i r/)}/r, where r is the distance from the point in question to the element of surface on the wave front, and is the wavelength. This spread of wave is somewhat similar to the pattern (in the wall behind) generated by firing a machine gun at an iron plate with two slits on it. Most of the bullets through the slits would align with each, some may be scattered sideways by hitting the edges although the chance becomes rarer as deviation getting further. One big difference is that bullets don't interfere with each other (story ascribed to Feynman). Anyway, assuming that the slit is very narrow, the two waves from slit 1 and 2 can be represented as:_{1} = Ae^{(2i r1/)}/r_{1} ---------- (66a)_{2} = Ae^{(2i r2/)}/r_{2} ---------- (66b) | ||

## Figure 09 Double-slit Exp. |
## Figure 10 Probabilities [view large image] |
where A is the normalization constant, r_{1}, and r_{2} are the distance from the slits to the screen as shown in Figure 09, which lays out the x-y plane only, while the z direction is perpendicular to the |

The total probability is : P

The normalization constant A can be evaluated by integrating Eq.(68) numerically over the angle from -/2 to +/2 :

1 = (8A

For the values of d and quoted below, the integral is equal to about 0.390, which gives A = 0.566D. For rapid oscillation of the interference term, i.e., when d >> , the integral approaches a value of /8, giving A

It is impossible to produce a graph with realistic data for light wave using home computer and Basic programming. Figure 10 is produced for some sort of matter wave with = 0.001 cm, d = 0.005 cm, and D = 10 cm. The "separate terms" together is plotted in green, the "interference terms" in blue, and the total probability is in red. The only running variable in the numerical computation is sin, which varies from -0.8 to +0.8 in steps of 0.001. Since cos

d sin = n , where n = 0, 1, 2, 3, ...

and minima occur at :

d sin = (n/2) , where n = 1, 2, 3, ...

The avid readers may have already noticed that the probability goes negative in the interference terms every half oscillating cycle (see Eq.(67b) or the blue curve in Figure 10). One explanation assures us that only the total probability P

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