Lec18 - these two curves correspond to the x values of the...

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Unformatted text preview: these two curves correspond to the x values of the solution to the transcendental equation. Another useful method to solve this equation is numerically, using, say, the Newton–Raphson method. The reason why we obtain this diffraction pattern is that as we go away from the central axis we have more and more destructive interference, which attenuates the envelope of the oscillating function. Problem : What is the ratio of intensities of the central peak maximum to the first of the secondary maxima? Solution : I β =0 I β =1 . 43 π = sinc 2 β | β =0 sinc 2 β | β =1 . 43 π = 1 sinc 2 β | β =1 . 43 π = β 2 sin 2 β β =1 . 43 π = 20 . 18 . 952 = 21 . 1 The intensity of the nearest secondary peak is only 4 . 7% that of the central intensity. Diffraction by a circular aperture : Many diffraction patterns arise from circular apertures, e.g., the pupil of the eye or the lens of a telescope. In the case of a rectangular aperture we found that the condition for dark fringes is a sin θ = mλ . For the first dark fringe, that marks the size of the central bright fringe, the condition for a rectangular aperture is...
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This note was uploaded on 07/19/2011 for the course PH 113 taught by Professor Liorburko during the Spring '09 term at University of Alabama - Huntsville.

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Lec18 - these two curves correspond to the x values of the...

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