![]() ![]() By the definition of angle, 2 α = A 0/ R, which gives us R = A 0/2 α. This is the amplitude of the diffraction patter at θ /= 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 α, where R is the radius of the arc formed by the phasor sum. Now 2 α is also the phase difference between the first and the last phasor, which is 2π a sin θ/ λ. We'll call the angle that this subtends 2 α. 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 δa. 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 θ. 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 θ = θ is usually good. Remember that, on the axis where θ = 0, there is a minimum, so the minima are equally spaced in sin θ, except either side of the central maximum. So this diagram represents the second order minima, where sin θ = λ/ (a/2), or sin θ = 2λ/ 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.Īn 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 θ = –λ/ 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 a sin θ = λ, If this distance is half a wavelength, i.e. The ray from the distance a/2 below has to travel an extra distance ( a sin θ/2). Typically in diffraction experiments, the slit is ~ 10 µm wide, while the distance to the screen might be ~ 1 m.) (Why parallel? Because the screen is distant. Consider parallel rays from both points, at angle θ 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. In fact, each ray from the slit will have another to interfere destructively, and a minimum in intensity will occur at this angle.This animated sketch shows the angle of the first order minima: the first minimum on either side of the central maximum. A ray from slightly above the center and one from slightly above the bottom will also cancel one another. Thus a ray from the center travels a distance \(\lambda / 2\) farther than the one on the left, arrives out of phase, and interferes destructively. In Figure 2b, the ray from the bottom travels a distance of one wavelength \(\lambda\) farther than the ray from the top. However, when rays travel at an angle \(\theta\) relative to the original direction of the beam, each travels a different distance to a common location, and they can arrive in or out of phase. When they travel straight ahead, as in Figure 2a, they remain in phase, and a central maximum is obtained. (Each ray is perpendicular to the wavefront of a wavelet.) Assuming the screen is very far away compared with the size of the slit, rays heading toward a common destination are nearly parallel. These are like rays that start out in phase and head in all directions. According to Huygens’s principle, every part of the wavefront in the slit emits wavelets. Here we consider light coming from different parts of the same slit. The analysis of single slit diffraction is illustrated in Figure 2. (b) The drawing shows the bright central maximum and dimmer and thinner maxima on either side. ![]() The central maximum is six times higher than shown. Monochromatic light passing through a single slit has a central maximum and many smaller and dimmer maxima on either side. ![]()
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