Explain the physical and historical significance of Young\'s interference experi

Question

Explain the physical and historical significance of Young's interference experiment. Besides the historical meaning of this discovery, include the following in your writing: A description of the diffraction of light by a narrow slit and the effect of narrowing the slit. With sketches, describe the production of the interference pattern in a double-slit interference experiment using monochromatic light. Sketch the double- slit interference pattern, identifying what lies at the center and what the various bright and dark fringes are called (such as "first side maximum" and "third order")

Explanation / Answer

Diffraction pattern from a single slit A laser illuminates a single slit and the resultant patten is projected on a distant screen. The sketch shows the view from above a single slit. Let's assume that the slit is constant width and very tall compared with that width, so that we can consider the system as two-dimensional. With light at normal incidence, the pattern is symmetrical about the axis of the slit. On a distant screen, the light arriving on the axis from all points in the slit has travelled an equal distance from the slit, so the centre of the pattern is a maximum. The next question is what determines its width. A larger view of the diffraction pattern. Notice the broad central maximum, and the equally spaced, successively weaker maxima on either side. First order minima This animated sketch shows the angle of the first order minima: the first minimum on either side of the central maximum. We call the slit widtha, and we imagine it divided into two equal halves. 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. a point at the very top of the lower half of the slit. Consider parallel rays from both points, at angle to the axis of symmetry. (Why parallel? Because the screen is distant. Typically in diffraction experiments, the slit is ~ 10 µm wide, while the distance to the screen might be ~ 1 m.) The ray from the distance a/2 below has to travel an extra distance (a sin /2). If this distance is half a wavelength, i.e. if a sin = then they are /2 out of phase and they interfere destructively. 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 = , they all interfere destructively. So the first mimimum has sin = /a. On the other side of the axis of symmetry, sin = –/a is also a minimum. These two minima limit the broad central maximum. The angle of the first order minima Higher order minima An argument like the one applies if, in our imagination, we divide the slit into any even number of equal slices. THe diagram shows a division into four. Each half is divided into quarters, and light from a source in the first quarter cancels that from one in the second quarter. Similarly, sources in the third quarter are nullified by those in the fourth quarter. So this diagram represents the second order minima, where sin = /(a/2), or sin = 2/a. For the nth order minima, we have a sin = n , where n is an integer, but not zero. Remember that, on the axis where = 0, there is a minimum, so the minima are equally spaced in sin , except either side of the central maximum. 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 = is usually good. Next we calculate how the intensity varies with sin . Geometry for the second order minima Intensity I() To calculate I(), we use Huygens' construction and phasor addition. The geometry for the phasor sum, The arc made of very many phasors has the same length as the central amplitude A0. The slit of width a is divided into N slits, each of width a. The path difference between rays from successive slices of the slit are equal, and so too are the angles between successive phasors. The slices of the slit all have equal width and length, so the lengths of all the phasors are equal. For a large number of slices, the phasors approximate the arc of a circle, as shown in red. We'll call the angle that this subtends 2. Now 2 is also the phase difference between the first and the last phasor, which is 2a sin /. = a sin /. The phasor sum has magnitude A, which we can write as R.sin , where R is the radius of the arc formed by the phasor sum. On the axis, where the phasors are all in phase, the phasor sum is the straight line shown in red at right. This is the amplitude of the diffraction patter at /= 0, which we call A0. By the definition of angle, 2 = A0/R, which gives us R = A0/2. So the magnitude of the phasor sum is A = 2R.sin = A0sin /. As we've seen in several previous chapters, the intensity I is proportional to square of the amplitude A. So, using I0 for the intensity at the centre of the pattern, we have I = I0(sin /)2 where = a sin /. Graphing I() with the phasor sum The animated phasor diagram (left); the plot of I() and, below it on the same scale, the resulting diffraction pattern. The vertical black line scanning across the graph is in step with the phasor diagram. As we remarked when looking at the intenstiy of Young's experiement in Interference, the eye does not respond linearly to intensity. To my eyes, at least, the difference in brightness seems much less than the difference in the intensity graph. Varying the slit width A slit with variable width. As the sit is narrowed, the pattern expands. DIffraction effects are most noticeable when the slit with a is not very much larger than the wavelength . This apparatus allows us to vary the slit width, but the pattern is still projected on a distant screen to make the diffraction effects cl Young's experiment with finite slit width. Top: a Young's experiment with slit separation d and slit width a. Bottom: a single slit diffraction patter with the same slit width a. In the Young's experiement in Interference, we didn't mention the effect of finite slit width. From the equations above, we can see that, if d is an integral number of times the slit width (d = na), then the nth interference fringe is absent: neither slit raidates power at this angle so there are no rays for constructive interference. Some assumptions must be made for this description of the single slit diffraction pattern: The slit size is small, relative to the wavelength of light. The screen is far away. Cylindrical waves can be represented in 2D diagrams as cicular waves. The intensity at any point on the screen is independent of the angle made between the ray to the screen and the normal line between the slit and the screen (this angle is called T below). This is possible because the slit is narrow. point1 Consider a slit of width a, light of wavelength l, and a smaller than l. When the light encounters the slit, the pattern of the resulting wave can be calculated by treating each point in the aperature as a point source from which new waves spread out. pointb Let L represent the distance between the slit and the screen. Let T represent the angle between the wave ray to a point on the screen and the normal line between the slit and the screen. point2 The top part of the figure to the left is an imitation of a single slit diffraction pattern which may be observed on the screen (there would really be more blending between the bright and dark bands, see a real diffraction pattern at the top of this page). Below the pattern is an intensity bar graph showing the intensity of the light in the diffraction pattern as a function of sin T. Most of the light is concentrated in the broad CENTRAL DIFFRACTION MAXIMUM. There are minor seconday bands on either side of the central maximum. The first DIFFRACTION MINIMUM occurs at the angles given by sin T = l / a I will mention now that the intensity of light is proportional to the square of its amplitude. This will come into play later on. With the equation: sin T = l / a (*) note that the width of the central diffraction maximum is inversely proportional to the width of the slit. If we increase the width size, a, the angle T at which the intensity first becomes zero decreases, resulting in a narrower central band. And if we make the slit width smaller, the angle T increases, giving a wider central band. But why are there these bands of light? And how can we derive the equation (*) for the location of the central diffraction minimum? The equation (*) is the result of analysis of the PATH DIFFERENCE between light rays coming from the top and the bottom of the slit, and how this path difference relates to our discussion on INTERFERENCE. Recall that we are considering points within the aperature as point sources from which new waves spread out. In the diagrams below the waves have been drawn from a side view, rather than a top view of wavefronts. This is to help us compare the phase of the waves. path1 The quantity a sin T is called the path difference between the two light rays. We can see that along the parallel wave rays, the bottom wave has already completed about two-thirds of its cycle when the top wave begins its cycle. This means that in this diagram the two light rays have a path difference of about 2/3 x 2p or 4p / 3. Now, remember that the slit width, a, is only a few hundred nanometers in size. And so even the light waves from the very top and very bottom of the slit are essentially right on top of each other, as well as all the waves inbetween. This means that they interfere, and the resultant wave's amplitude equals the sum of the individual wave amplitudes, by the superposition of waves. This also means that for the top and bottom light waves, their phase difference is equal to their path difference, which in this example is about 4p / 3. path2 At T = 0, when the wave rays follow the normal line directly to the screen, a sin T = 0. This means that the path difference and the phase difference of all the waves is zero. Hence the waves are all in phase, and constructive interference has the resultant wave's amplitude equal to the sum of all the individual wave's amplitudes. This explains the very bright central band around sin T = 0. With all the waves in phase, we have the largest resultant wave amplitude possible. And since the intensity of light is proportional to the square of its amplitude, the pattern on the screen has a very intense central band at this angle, that is when T = 0. As T varies slightly from zero, a sin T also varies slightly from zero, as does the phase difference of the waves. This results in the interference of all the waves being not totally constructive, and so the intesity of the central band decreases while moving slightly away from sin T = 0, as we saw in the intesity bar graph above. path3path4 When a sin T = l, something special happens. Here the path difference between the top and bottom light rays equals one wavelength (l). That is, they are in phase. (Left figure) This means that the path difference between the top ray and the ray just below the midpoint of the slit (a/2), is half a wavelength (l/2). (Right figure) And a path difference of half a wavelength corresponds to a phase difference of p. That is, the top wave, and the wave just below the midpoint of the slit, are out of phase, and therefore cancel each other out. Let us consider many point sources, say 2k, equally spaced within the slit opening so that there are 1 to k above the midpoint and k+1 to 2k below. Then when a sin T = l, waves 1 and k+1 are out of phase and so cancel each other out, as do waves 2 and k+2, and 3 and k+3 ... through to waves k and 2k. And so all waves cancel, and thus the resultant wave has an amplitude of zero. An amplitude of zero means zero intesity, and so the first diffraction minimum occurs at a sin T = l, or sin T = l / a, which was our (*) above. This argument can be extended to explain the second and third and other diffraction minima. At the angle when a sin T = 2l, we can divide the slit into four regions of point sources, two above the midpoint and two below. Then, using the argument above, the total intensity of the top two regions is zero due to cancellation of pairs of sources, and the same goes for the bottom two regions. And so, the general equation for the points of zero intensity in the diffraction pattern of a single slit is: a sin T = ml m = 1,2,3, ... But usually we are just interested in the location of the first minimum, when m = 1, because most of the light enery is located in the central diffraction maximum. last Let y be the distance from the center of the central diffraction maximum to the first diffraction minimum. The angle T is related to this distance y and the distance to the screen, L, by the equation: tan T = y / L Since the angle T is very small, cos T » 1. Thus, tan T » sin T. Then, combining the above equation with equation (*), we have sin T = l / a »y / L, or y = L l / a And so, given the distance to the screen, the width of the slit, and the wavelength of the light, we can use the equation y = L l / a to calculate where the first diffraction minimum will occur in the single slit diffraction pattern. And we have learned that this is the point where the waves from point sources in the slit all cancel in pairs that are out of phase.

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