10.2 Single-Slit Diffraction

Transcription

1 10. Single-Slit Diffraction If you shine a beam of light through a ide-enough opening, you might expect the beam to pass through ith very little diffraction. Hoever, hen light passes through a progressively narroer opening, the ave properties of light often produce unusuallooking patterns, such as the diffraction pattern shon in Figure 1. In this section, you ill learn about the formation of such diffraction patterns. Single-Slit Diffraction Figure 1 The interference of light passing through a single slit from a circular opening produces a pattern of concentric rings. Fraunhofer diffraction an interference pattern that shos distinctive differences beteen the bright central fringe and darker flanking fringes central maximum the bright central region in the interference pattern of light and dark lines produced in diffraction secondary maxima the progressively less-intense bright areas, outside the central region, in an interference pattern To understand ho a narro opening (for example, 100 mm ide) can affect light passing through it, consider the behaviour of light passing through a single slit as shon in Figure (a). While investigating this configuration, assume that the folloing conditions exist: 1. The idth,, of the slit is narro enough that a substantial amount of diffraction can occur, but not so narro that it acts as a single point source of aves.. The light source is monochromatic ith avelength l. 3. The light source is far from the slit, so that light rays that reach the slit are all travelling approximately parallel to each other. Such configurations produce Fraunhofer diffraction, hich shos a bright central fringe called the central maximum. The central maximum is flanked by dark fringes, called minima, and less-intense bright fringes, called secondary maxima. Why do these fringes occur, and ho can you predict here bright fringes and dark fringes ill appear on the? The key to ansering these questions is Huygens principle. Recall from Chapter 9 that Huygens principle states that each point on a ave front acts as a ne source of aves. Figure (b) is a magnified vie of the slit looking don on it from above, shoing its idth. According to Huygens principle, all points across the slit idth in Figure (b) act as ave sources, and these different aves create an interference pattern at the. When the slit is divided in half (Figure (b)), coherent aves that interfere ith each other appear as the light passes through the slit. The points in one half are labelled A, and the points in the other half are labelled B. Each of these points is a source of Huygens aves (spherical aves), including point A 1 at one edge of the slit and the corresponding source point B 1 at the centre of the slit. incoming light idth, A 1 A A 3 B 1 B B 3 slit A 1 B 1 L L sin P bright fringes intensity (a) single slit (b) (c) Figure (a) Light passes through a single slit and hits a. (b) A vie from above the slit looking don at its idth. Each point in the slit acts as a source of spherical aves. (c) Looking don at the slit and, the single-slit pattern is apparent. Assume that point P is far aay. 51 Chapter 10 Applications of the Wave Nature of Light NEL

2 Figure (c) shos the distances from these points to a point P on the the path lengths and the path length difference L. Although Figure shos the close to the slit, assume that the is quite far aay, so all the angles denoted by θ are approximately equal. Suppose the aves from A 1 and B 1 interfere destructively hen they reach a particular point P on the. The path length difference is DL 5 l. When this condition for destructive interference is met for points A 1 and B 1, the aves from A and B ill also interfere destructively, as ill the aves from points A 3 and B 3, and so on, for all similar pairs of points ithin the slit. Destructive interference produces a dark fringe on the. From Figure (c), you can see that DL 5 sin u n here is the idth of the slit. Substituting l for DL leads to sin u n 5 l, hich gives the angle of the first dark fringe on one side of the centre of the. On the other side, the corresponding value gives the angle of the first dark fringe to that side of the centre. The first dark fringes are thus at the angles sin u n 5 l single-slit diffraction: first dark fringes 1destructive interference Figure 3 shos the method of determining the position of the second dark fringe. In Figure 3(a), the slit is divided into four regions. The points in these four regions are denoted by A, B, C, and D. Figure 3(b) shos the path lengths from the farthest point in each region (A 1, B 1, C 1, and D 1 ) to a point P on the. Suppose the path length difference for the Huygens aves from A 1 and B 1 is DL 5 l, so these aves interfere destructively. The path length difference for C 1 and D 1 has the same value, so these aves also interfere destructively. The path length difference is the same for aves from other pairs of points A and B, C and D, and so on, so these aves also interfere destructively, resulting in no light, or a total intensity of zero, on the. Thus, for the second dark fringe, DL 5 4 sin u. Setting this expression equal to l, the interference condition in this case is the folloing, here the angle again corresponds to fringes left and right of the centre of the : sin u 5 l single-slit diffraction: second dark fringes 1destructive interference P 4 A 1 A B 1 B C 1 C D 1 D slit 4 A 1 B 1 C 1 D 1 L L L sin 4 sin bright fringes intensity (a) (b) Figure 3 (a) To determine the locations of the second dark fringes, e divide the idth into four regions, A, B, C, and D. (b) The locations of the second dark fringes can then be determined. In a scale draing, the slit idth ould be much smaller than the idth of the central diffraction peak. NEL 10. Single-Slit Diffraction 513

3 The same method can predict the third and other dark fringes by dividing the slit into 6, 8, 10, or any even number of regions. The result for the nth dark fringe is sin u n 5 nl single-slit diffraction: nth dark fringes 1destructive interference here n 5 1,, 3,.... The values of θ that satisfy this equation give the angles of all dark fringes. To determine the angles for the maxima, e use a similar method of dividing up the slit. For maxima, hoever, e divide the slit into an odd number of regions. Light from all but one region ill interfere destructively, as in the discussion for minima, and the light from the remaining region provides the maxima. To calculate the first maximum after the central peak, divide the slit into three regions, for example. In the equation for destructive interference of to of the regions, the effective slit idth for each region is, and the path length difference is 3 a half-avelength as alays. The location of the fringe is then 3 sin u 5 l, or sin u 5 3 l single-slit diffraction: first bright fringes 1constructive interference Investigation Diffraction of Light Using a Single Slit (page 545) No that you have learned about single-slit diffraction, perform Investigation to study ho changes in slit idth, avelength, and source distance affect the interference patterns. You can calculate additional bright fringes by breaking the slit into 5, 7, 9, or any odd number of regions. The result is that maxima occur in general hen sin u m 5 am 1 1 b l single-slit diffraction: mth bright fringes 1constructive interference here m 5 1,, 3,, and the values of θ that satisfy this equation give the angles of all bright fringes. Figure 4 shos the intensity curve for single-slit diffraction. For each successive bright fringe, more of the slit acts as point sources of light that interferes destructively. The intensity of bright fringes therefore decreases aay from the central peak. intensity y sin Figure 4 A full calculation of the intensity curve shos that the central bright fringe is about 0 times as intense as the bright fringes on either side. 3 y y 0 y y y 1 y Chapter 10 Applications of the Wave Nature of Light NEL

4 You can approximate the idth of the central bright fringe by using the angular separation of the first dark fringes on either side. You can calculate the angles for these first dark fringes by setting n 5 1 to obtain sin u 1 5 l Figure 5 shos the projection of the fringes on the. The distance y 1 of the first minimum from the central maximum (using sin θ tan θ for small angles) is then y 1 5 Ll The idth of the central maximum is the distance beteen first minima, y 1, or Dy 5 Ll The separation of the central maximum from the first minimum, Dy, is half of this, or just y 1, hich gives l 5 Dy. From the equation for the location of the L minima, the distance from each minimum to the next one is also y 1. This is useful for predicting the positions of the fringes and the avelength of light from the diffraction pattern. In particular, a longer avelength or a smaller idth produces a larger value of Dy and a ider diffraction pattern. Figure 5 also shos the angles for the single-slit diffraction pattern. By determining the angle and distance from the, you can calculate the distances beteen fringes. The folloing Tutorial shos ho to calculate the idth of the central maximum of a single-slit diffraction pattern. L 1 y 1 y 1st minimum nd minimum Figure 5 You can determine the value of sin u 1 using the approximation sin u 1 < tan u 1 5 y 1 L. Tutorial 1 Analyzing Single-Slit Interference In the folloing Sample Problems, you ill analyze interference patterns produced during single-slit diffraction. Sample Problem 1: Determining the Distance beteen Dark Fringes and the Central Maximum Light ith a avelength of nm is incident on a slit of idth 11 mm and produces a diffraction pattern on a located 80.0 cm behind the slit. Calculate the distance of the first dark fringe from the central maximum on the. Given: l nm m; 5 11 mm m; L cm m Required: y 1 Analysis: The distance from the central maximum to the first minimum is Dy 5 y 1. l 5 Dy L Dy 5 Ll Solution: Dy 5 Ll m m m Dy m Statement: The first dark fringe is 3.9 cm from the central maximum. NEL 10. Single-Slit Diffraction 515

5 Sample Problem : Determining the Width of the Central Maximum Light ith a avelength of 670 nm is incident on a slit of idth 1 mm and produces a diffraction pattern on a that is 30.0 cm behind the slit. (a) Calculate the angular idth and the absolute idth, in centimetres, of the central maximum. (b) Calculate the distance beteen the first and second minima. (c) Determine the distance beteen the second and third maxima. Compare this anser to your anser in (b). Does this anser make sense? Explain hy or hy not. Solution (a) Given: n 5 1; l nm m; 5 1 mm m; L cm m Required: u 1 ; Dy, the idth of the central maximum Analysis: The distance from the central maximum to the first minimum on each side is Dy 5 y 1. The angle for the first minimum is sin u 1 5 l, here n 5 1 and the angular idth is u 1. To calculate y 1, use l 5 Dy L. Solution: sin u 1 5 l m m u to extra digits carried l 5 Dy L Dy 5 Ll m m m Dy m 1one extra digit carried Statement: The angular idth of the central maximum is 6.48, corresponding to 3.4 cm on the. (b) Given: l nm m; 5 1 mm m; L cm m Required: distance beteen the first and second minima Analysis: This is half the idth of the central maximum in (a), since Dy 5 y 1 Dy 5 y 1 y 1 5 Dy Solution: y 1 5 Dy m y cm Statement: The distance beteen the first and second minima is 1.7 cm. (c) Given: l nm m; 5 1 mm m; L cm m Required: distance beteen the second and third maxima Analysis: This is half the idth of the central maximum in (a); y 1 5 Dy Solution: y 1 5 Dy m y cm Statement: The distance beteen the second and third maxima is 1.7 cm. Ansers (b) and (c) are the same. This makes sense because the distance beteen adjacent maxima is identical to the distance beteen adjacent minima because the pattern is symmetrical. Practice 1. Would the central maximum in Sample Problem 1 be ider or narroer if the slit and the ere submerged in ater? K/U. Helium neon laser light, ith a avelength of m, passes through a single slit ith a idth of 43 mm onto a 3.0 m from the slit. Calculate the separation of adjacent minima, other than those on either side of the central maximum T/I [ans: 5.1 cm] 3. Monochromatic light falls onto a slit m ide. The angle beteen the first dark fringes on either side of the central maximum is Calculate the avelength of the light. T/I [ans: m] 4. The first dark fringe in a single-slit diffraction pattern is located at an angle of u a With the same light, the first dark fringe formed ith another single slit is located at u b Calculate the ratio a b. T/I [ans: 0.67] 516 Chapter 10 Applications of the Wave Nature of Light NEL

6 Resolution The resolution of an optical device is the device s ability to separate closely spaced objects into distinctly different images. Light passing through a small opening is diffracted, and smaller openings produce greater diffraction. When light from to objects passes simultaneously through the same opening, the light from both sources is diffracted. It produces overlapping diffraction patterns and the images appear fuzzy. If the opening is quite small and the objects are close to each other, distinguishing beteen the to images can be quite challenging (Figure 6). resolution the ability of an optical device to separate close objects into distinct and sharp images slight diffraction of light S 1 S large opening I Images are fuzzy but resolved. I 1 region here diffracted light from each source overlaps S 1 S small opening significant diffraction I I 1 Images overlap because of diffraction and are unresolved. Figure 6 With a large opening, the images from to light sources are fuzzy but distinguishable. With a smaller opening, the images overlap and are difficult to distinguish. In many optical devices, a small lens replaces the opening. The resolution of the device is then the ability of the lens to form to sharp images. When light passes through the lens, the diffraction leads to a single bright spot surrounded by concentric circles in a diffraction pattern called an Airy disc (Figure 7). To small objects ill each produce an image surrounded by diffraction fringes. If the bright parts of the diffraction pattern for the first image fall on the dark areas of the other image, the to images may be inseparable. Lenses and openings that are small relative to the avelength of light usually produce loer-resolution images by increasing the effects of diffraction, as seen in Figure 8 on the next page. But factors other than diffraction affect the resolution of lenses such as those on cameras. intensity y sin (a) (b) 3 y y y 0 y y 1 y 3 Figure 7 (a) The classical Airy disc is a pattern of concentric circles of light around a bright central spot. (b) The light intensity pattern of an Airy disc is similar to a single-slit diffraction pattern. NEL 10. Single-Slit Diffraction 517

7 (a) (b) Figure 8 (a) When the resolution of an optical device is not sharp enough, stars that are in close proximity appear blurry. (b) When the resolution is sufficient, more individual stars can be clearly seen. In an ideal set of optics, 84 % of the incident light is concentrated in the central maximum. If there are obstructions in the optical path, such as a mirror, more light ill end up being spread into the rings, making the image less clear. This principle makes a refracting telescope the sharpest theoretical optical instrument for a given diameter. A refracting telescope uses a lens ith no obstructions in the optical path. Reflecting telescopes use mirrors instead of lenses, but they often contain more than one mirror in the optical path. Optical microscopes have small-diameter objective lenses. Due to diffraction, the best optical microscopes can typically achieve a magnification no greater than 1000 times. One ay to improve resolution is to use shorter-avelength light. Therefore, an ultraviolet microscope can achieve a much greater resolution than one that uses visible light. Most observatory telescopes use curved mirrors (Figure 9). Astronomers use the telescopes ith the largest mirrors to vie the most distant objects, not only because they collect more light, but also because the large apertures reduce diffraction effects and offer greater resolution. Figure 9 The telescope inside the Canada France Haaii Observatory (left) detects visible light and infrared radiation. 518 Chapter 10 Applications of the Wave Nature of Light NEL

8 10. Revie Summary Monochromatic light passing through a single slit produces a diffraction pattern that consists of a bright central region surrounded by alternating light and dark bands (maxima and minima) resulting from constructive and destructive interference. Fraunhofer diffraction is a special case of diffraction that shos distinctive differences beteen the central fringe and darker flanking fringes. In single-slit diffraction, the dark bands (minima) occur at angles u n that satisfy sin u n 5 nl 1n 5 1,,... In single-slit diffraction, the bright bands (maxima) occur at angles u m that satisfy sin u m 5 1m 1 1 l 1m 5 1,,... The distance beteen successive minima, and successive maxima, is given by Dy 5 ll, and the idth of the central maximum is Dy. The resolution of an optical instrument is its ability to separate closely spaced objects into distinctly different images limited by diffraction. Questions 1. Light ith a avelength of 794 nm produces a single-slit diffraction pattern in hich the ninth dark fringe lies at 6.48 cm from the direction of the central maximum. The distance to the is 1.0 m. Determine the idth of the slit. T/I. Red light (l nm) is diffracted by a single slit, and the first dark fringe occurs at u Determine the idth of the slit. T/I 3. A slit of idth 0.15 mm is located 10.0 m from a. Light ith a avelength of 450 nm passes through this slit. Determine the distance beteen the first and the third dark fringes on the. T/I 4. Light ith a avelength of 550 nm passes through a single slit and illuminates a.0 m aay. The distance from the first dark fringe to the centre of the interference pattern is 5.5 mm. Determine the idth of the slit. T/I 5. Light from a helium neon laser (l nm) passes through a single slit to a 3.0 m from the slit. The slit is 0.5 mm ide. What is the idth of the central maximum on the? T/I 6. Yello light from a vapour lamp passes through a single slit cm ide to a 60.0 cm aay. A first-order dark fringe is 0.10 cm from the centre of the central maximum. Determine the avelength of the yello light. K/U T/I 7. Suppose monochromatic light incident on a single slit produces a diffraction pattern. K/U (a) Ho ould the pattern differ if you doubled the avelength of the light? (b) Ho ould the pattern differ if you doubled both the avelength and slit idth at the same time? 8. Blue light passing through a single narro slit forms an interference pattern. Determine ho the spacing of the maxima ould be different if you replaced blue light ith green light. Explain your reasoning. K/U T/I C 9. A dooray acts as a single slit for light that passes through it, but a dooray is much larger than the avelength of visible light, so the diffraction angles are very small. Calculate the angle of the tenth minimum for a typical dooray. K/U T/I 10. Digital images consist of pixels, or small coloured dots. If an image is 84 pixels ide by 6 pixels high, ho could you achieve better resolution? (Hint: Consider aperture and distance.) K/U T/I A 11. The star Mizar is actually a double star that only appears as to distinct stars hen resolved by a telescope. Explain hy. K/U T/I C 1. Explain ho a single-slit interference pattern differs from a double-slit interference pattern. K/U T/I C NEL 10. Single-Slit Diffraction 519