![]() The maxima lies between the minima and the width of the central maximum. Similarly, for the nth fringe, division of the slit can take place into 2n parts and this condition can be used as: As such, one can obtain a dark fringe.įor the next fringe, division of the slit can take place into 4 equal parts of a/4 and the same logic can be applied. Therefore, at θ = sin − 1λa, there would be destructive interference because any ray emanating from a point has a counterpart that produces destructive interference. ![]() Furthermore, the path difference must be out of phase by λ2, with λ being the wavelength.įor a ray emanating from any point in the slit, there exists another ray at a distance a/2 from which destructive interference can take place. Moreover, one can consider any arbitrary pair of rays at a distance a/2.įor a dark fringe, the path difference must produce destructive interference. Furthermore, it is possible to consider any number of ray pairings that start from a distance a/2 from one another. The path difference exhibited by the top two rays is:Īn important point to remember is that this calculation is valid only if D is very large. Furthermore, consider a pair of rays whose emanation takes place from distances a/2 from each other. Also, the division of the slit can take place into zones of equal widths a/2. In order to describe the pattern, one must first look at the condition for dark fringes. Now, one can identify the angular position of any point on the screen by ϑ whose measurement takes place from the slit centre which divides the slit by a/2 lengths. ![]() x`D is the separation between slit and source. Single Slit Diffraction Formula of Single Slit DiffractionĬonsider that the slit width a << D.
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