Single Slit Diffraction A ray passing through the center of the slit will have a path length exactly λ/2 greater than that from the source at A, and hence these two waves will interfere destructively. Now consider a point adjacent to A; the light emitted with have a path difference exactly λ/2 different to light from a point just above the central point, and will cancel it out. Similarly, for every point between Aand the center, there will be a corresponding point λ/2 away between the center and B that will destructively interfere with it. Hence there is effectively no light emitted it the direction θmand it corresponds to a minimum. A similar situation arises when the path difference between Aand Bis any whole number of wavelengths--such a situation is shown in , iv). On the other hand, when the path difference between Aand Bis a half-integer multiple of the wavelength, such as in iii), there will only be partial cancellation. All the emitters between Aand a point one-third of the way to B
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