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examine the effect the secondary disturbances from AB will have at a point 0. Draw CP perpendicular to AB. P is called the pole of the wave with respect to C. On PA and PB mark off points W, X, Y, Z such that WC is equal to PC plus one-half the wave length of the light, XC equal to PC plus two halves the wave length, etc. Then when the crest of a wave from P reaches C, the crest of the wave leaving W at the same time will be half a wave length behind, or there will be the trough of a wave from W there. In like manner there will be a crest from X, a trough from Y, etc. Intermediate points will have intermediate phase relations.
Where we have crest and trough superimposed this way we have interference, cancellation of the waves. Thus all the waves coming from AB are not effective in producing illu
mination at C.
A
F/qurs. >5
If we consider the whole wave instead of having only two points such that WC equals PC plus half the wave length of the light, there are an infinite number of points in a circle around P that satisfy this condition. In the same way there ai*e circles through XX, YY, etc. These circles divide the plane wave into a circle and concentric rings that are known as half period elements. The waves from each of these elements interfere
with waves from adjacent elements.
To get the effective illumination at C we must sum up the effect of all the elements. There is a change in area of elements as one goes outward from P. There is also a change in intensity of illumination with increasing obliquity of the rays. The result of these two is that practically all of the effective illumination comes from comparatively few zones surrounding the pole. For elements farther out there is almost complete cancellation.
By mathematical and practical demonstration it has been found that the illumination at C is equal to onehalf the illumination from the first element plus one-half the illumination from the last element. In cases where the last element is far away the second term is negligible. However, if we have P at quite a distance from C, and use a camera aiaphrac m to cut out the more distant elements, the effect of this second term is marked. Suppose we stop down until we have an even number of half period elements. Then the first and last elements will be out of phase when they reach C and there will be little light. If we have an uneven number of elements, the first and last will be in phase. They will reinforce and there will be more light at C than if there were no diaphragm there. Because of this rather strange action of interference it is true that we would get more light at C from the first element alone than from the whole wave.
How does this apply to the rectilinear propagation of light? Rectilinear propagation is only an approximation due to the short wave length of light. As in Figure 1 the waves do go around the corner, but they interfere and produce darkness.
At point Q, Figure 1, there will be illumination, but not the full illumination of the wave, for half the first elements of the wave have been cut off. Below Q the light will grow gradually dimmer as more and more of the effective elements are cut off by YP. Finally, say at J, all the effective or only partly interfering
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