$Diffraction angle$

$diffraction angle$

In black: the calculated diffraction pattern due to a single slit of width a. I 0 for the intensity at the centre of the pattern, we have So the magnitude of the phasor sum isДЄs we've seen in several previous chapters, the intensity I is proportional to square of the amplitude A. By the definition of angle, 2 О±В = A 0/ R, which gives us RВ = A 0/2 О±. This is the amplitude of the diffraction patter at Оё /= 0, which we call A 0. On the axis, where the phasors are all in phase, the phasor sum is the straight line shown in red at right. The phasor sum has magnitude A, which we can write as R.sin О±, where R is the radius of the arc formed by the phasor sum. Now 2 О± is also the phase difference between the first and the last phasor, which is 2ПЂ aВ sin Оё/ О». We'll call the angle that this subtends 2 О±. For a large number of slices, the phasors approximate the arc of a circle, as shown in red. The slices of the slit all have equal width and length, so the lengths of all the phasors are equal. The path difference between rays from successive slices of the slit are equal, and so too are the angles between successive phasors. The slit of width a is divided into N slits, each of width Оґa. The arc made of very many phasors has the same length as the central amplitude A 0. Next we calculate how the intensity varies with sin Оё. We can note too that, for light diffracting the throught slits, the slit is usually much wider than a wavelength, so the pattern is usually very small, so the approximation that sin Оё = Оё is usually good. Remember that, on the axis where Оё = 0, there is a minimum, so the minima are equally spaced in sin Оё, except either side of the central maximum. So this diagram represents the second order minima, where sin Оё = О»/ (a/2), or sin Оё = 2О»/ a. Similarly, sources in the third quarter are nullified by those in the fourth quarter. Each half is divided into quarters, and light from a source in the first quarter cancels that from one in the second quarter. These two minima limit the broad central maximum.ДЄn argument like the one applies if, in our imagination, we divide the slit into any even number of equal slices. On the other side of the axis of symmetry, sin Оё = вЂ“О»/ a is also a minimum. Now, for every point in the top half of the slit, there is one in the bottom half a distance a/2 below and, at the angle that satisfies aВ sin Оё = О», If this distance is half a wavelength, i.e. The ray from the distance a/2 below has to travel an extra distance ( aВ sin Оё/2). Typically in diffraction experiments, the slit is ~ 10В Вµm wide, while the distance to the screen might be ~ 1В m.) (Why parallel? Because the screen is distant. Consider parallel rays from both points, at angle Оё to the axis of symmetry. a point at the very top of the lower half of the slit. Using the Huygens' construction, we consider a point at the very top of the slit, and another point a distance a/2 below it, i.e. We call the slit width a, and we imagine it divided into two equal halves. This animated sketch shows the angle of the first order minima: the first minimum on either side of the central maximum.

$Diffraction angle$