Diffraction and Interference of Water Waves PDF

Diffraction and Interference 9.3 of Water Waves Have you ever noticed how people relaxing at the seashore spend so much of their time watching the ocean waves moving over the w...

Diffraction and Interference 9.3 of Water Waves Have you ever noticed how people relaxing at the seashore spend so much of their time watching the ocean waves moving over the water, as they break repeatedly and roll onto the shore? Water waves behave similarly to other kinds of waves in many ways. Figure 1 illustrates one characteristic behaviour of waves when a part of the wave enters through a narrow opening. The section of the wave that gets through acts as a source of new waves that spread out on the other side of the opening, and the waves from the two sources combine in specific ways. You will learn about these characteristic behaviours of waves in this section. Figure 1 When part of a wave enters a narrow opening, such as this bay, new waves are created and the two waves combine in predictable ways. Diffraction If you observe straight wave fronts in a ripple tank, you can see that they travel in a straight line if the water depth is constant and no obstacles are in the way. If, however, the waves pass by an edge of an obstacle or through a small opening, the waves spread out. Diffraction is the bending of a wave as the wave passes through an diffraction the bending and spreading opening or by an obstacle. The amount of diffraction depends on the wavelength of the of a wave when it passes through an waves and the size of the opening. In Figure 2(a), an obstacle diffracts shorter wave- opening; dependent on the size of the lengths slightly. In Figure 2(b), the same obstacle diffracts longer wavelengths more. opening and the wavelength of the wave short wavelengths long wavelengths (a) (b) Figure 2 When waves travel by an edge, (a) shorter wavelengths diffract less than (b) longer wavelengths. NEL 9.3 Diffraction and Interference of Water Waves   459 8160_CH09_p434-469.indd 459 4/30/12 9:47 AM  Although diffraction also occurs for sound and light waves, we will study the phenom- enon first for water waves because their long wavelength allows for easier observation of the effects of diffraction. In Figure 3(a), you can see a wave encountering a slit with a certain width, w. In Figure 3(b) and Figure 3(c), the width of the opening is the same but the wavelength of the incident waves has changed. Notice how the amount of diffraction changes. Therefore, as the wavelength increases but the width of the slit does not, the dif- fraction also increases. In Figure 3(a), the wavelength is approximately one-third of w. Notice that only the part of the wave front that passes through the slit creates the series of circular wave fronts on the other side. In Figure 3(b), the wavelength is about half of w, which means that significantly more diffraction occurs, but areas to the edge exist where no waves are diffracted. In Figure 3(c), the wavelength is approximately two-thirds of w, and the sections of the wave front that pass through the slit are almost all converted to circular wave fronts. (a) (b) (c) Figure 3 As the wavelength increases, the amount of diffraction increases. What happens if you change the width of the slit but keep the wavelength fixed? As Figure 4(a) and Figure 4(b) show, as the size of the slit decreases, the amount of dif- fraction increases. If waves are to undergo more noticeable diffraction, the wavelength must be comparable to or greater than the slit width (λ $ w). For small wavelengths (such as those of visible light), you need to have narrow slits to observe diffraction. Figure 4(c) shows what happens to the diffraction pattern when the wavelength decreases but the size of the slit does not change. incident waves incident waves incident waves nearly straight-line propagation shorter wavelength, large opening small opening no change in opening (a) (b) (c) Investigation 9.3.1 Figure 4 (a) and (b) As the size of the slit (aperture) decreases, diffraction increases. (c) With a shorter wavelength and no change in the size of the opening, there is less diffraction. Properties of Water Waves (page 487) The relationship between wavelength, width of the slit, and extent of diffraction is You have learned how observing perhaps familiar for sound waves. You can hear sound through an open door, even if water waves helps in understanding you cannot see what is making the sound. The primary reason that sound waves dif- the properties of waves in general. fract around the corner of the door is that they have long wavelengths compared to the This investigation will give you an width of the doorway. Low frequencies (the longer wavelengths of the sound) diffract opportunity to test conditions for more than high frequencies (the shorter wavelengths of the sound). So if a sound system diffraction to occur. is in the room next door, you are more likely to hear the lower frequencies (the bass). In the following Tutorial, you will learn how to solve problems relating to diffraction. 460 Chapter 9 Waves and Light NEL 8160_CH09_p434-469.indd 460 4/30/12 9:47 AM  Tutorial 1 Diffraction The following Sample Problem provides an example and sample calculations for the concepts of diffraction. Sample Problem 1: Diffraction through a Slit Determine and explain the difference between the diffractions observed in Figure 5(a) and Figure 5(b).   0.5 cm   0.5 cm w a  2 cm w b  0.5 cm (a) (b) Figure 5 Given: l 5 0.5 cm; wa 5 2 cm; wb 5 0.5 cm Required: diffraction analysis l Analysis: $ 1 w Solution: For Figure 5(a), l 0.5 cm 5 wa 2 cm 5 0.25 l ,1 wa For Figure 5(b), l 0.5 cm 5 wb 0.5 cm l 51 wb Statement: Since the ratio in Figure 5(a) is less than 1, little diffraction occurs. Since the ratio in Figure 5(b) is 1, more noticeable diffraction occurs. Practice 1. Determine whether diffraction will be noticeable when water waves of wavelength 1.0 m pass through a 0.5 m opening between two rocks. T/I [ans: yes] 2. A laser shines red light with a wavelength of 630 nm onto an adjustable slit. Determine the maximum slit width that will cause significant diffraction of the light. T/I [ans: 6.3 * 10-7 m] NEL 9.3 Diffraction and Interference of Water Waves   461 8160_CH09_p434-469.indd 461 4/30/12 9:47 AM  Interference When two waves cross paths and become superimposed, they interact in different interference the phenomenon that occurs ways. This interaction between waves in the same medium is called interference. when two waves in the same medium You experience interference when you listen to music and other types of sound. For interact example, sound waves from two speakers may reach the listener’s ears at the same constructive interference the time, as illustrated in Figure 6. If the crest of one wave coincides with the crest of the phenomenon that occurs when two other, then the waves are in phase and combine to create a resultant wave with an interfering waves have displacement in the amplitude that is greater than the amplitude of either individual wave—resulting in a same direction where they superimpose louder sound. This phenomenon is called constructive interference. Waves are in phase. Waves are 180° out of phase.   2 speaker 1 L1      L2 speaker 2 listener constructive interference destructive interference (a) (b) (c) Figure 6 (a) If L1 and L2 are the same, the waves arrive at the listener in phase. (b) Waves interfere l constructively if they arrive in phase. (c) If L1 and L2 differ by, say, , the waves arrive 1808 out of 2 phase, and they interfere destructively. l For two waves that differ in phase by , shown in Figure 6(c), the crest of one wave 2 coincides with the trough of the other wave. This corresponds to a phase difference of 1808. The two waves combine and produce a resulting wave with an amplitude that is destructive interference the smaller than the amplitude of either of the two individual waves. This phenomenon is phenomenon that occurs when two called destructive interference. interfering waves have displacement The following conditions must be met for interference to occur: in opposite directions where they 1. Two or more waves are moving through different regions of space over at least superimpose some of their way from the source to the point of interest. 2. The waves come together at a common point. 3. The waves must have the same frequency and must have a fixed relationship coherent composed of waves having the between their phases such that over a given distance or time the phase difference same frequency and fixed phases between the waves is constant. Waves that meet this condition are called coherent. Figure 7 shows water waves interfering. Figure 7 Interference of water waves in a ripple tank 462   Chapter 9 Waves and Light NEL 8160_CH09_p434-469.indd 462 4/30/12 9:47 AM  In Figure 7, two point sources have identical frequencies. The sources produce waves that are in phase and have the same amplitudes. Successive wave fronts travel out from the two sources and interfere with each other. Constructive interference occurs when a crest meets a crest or when a trough meets a trough. Destructive interference occurs when a crest meets a trough (resulting in zero amplitude.) Symmetrical patterns spread out from the sources, producing line locations node a point along a standing wave where constructive and destructive interference occur. A node is a place where where the wave produces zero destructive interference occurs, resulting in zero amplitude (a net displacement of displacement zero). As shown in Figure 8, the interference pattern includes lines of maximum nodal line a line or curve along which displacement, caused by constructive interference, separated by lines of zero dis- destructive interference results in zero placement, caused by destructive interference. The lines of zero displacement are displacement called nodal lines. areas of constructive interference lines of destructive interference amplitude (nodal lines) here is 2A crests amplitude here is 2A troughs node S1 S2 Figure 8 Interference pattern between two identical sources each with positive amplitude A When the frequency of the two sources increases, the wavelength decreases, Investigation 9.3.2 which means that the nodal lines come closer together and the number of nodal lines increases. When the distance between the two sources increases, the number of Interference of Waves in Two nodal lines also increases. The symmetry of the pattern does not change when these Dimensions (page 488) two factors are changed. However, if the relative phase of the two sources changes, So far, you have read about two-point- then the pattern shifts, as shown in Figure 9, but the number of nodal lines stays source interference in theory, with the same. In Figure 9(a), the sources are in phase, but in Figure 9(b), there is a phase diagrams and photographs. This investigation gives you an oppor­tunity difference of 1808. to see and measure interference for yourself. S1 S2 S1 S2 (a) (b) Figure 9 Effect of phase change on interference pattern for (a) zero phase difference and (b) 1808 phase difference NEL 9.3 Diffraction and Interference of Water Waves   463 8160_CH09_p434-469.indd 463 4/30/12 9:47 AM  Mini Investigation Interference Mini from Two Speakers Investigation SKILLS Skills: Performing, Observing, Analyzing, Communicating HANDBOOK A2.1 At the start of the discussion on interference, you considered 4. Walk around the different areas of the classroom and sound waves coming from two speakers. In this investigation, mark on your plan where constructive and destructive your teacher will set up two speakers in the classroom, which interference occur. will emit a single-frequency sound. You will examine how A. How should your distances to the two speakers differ to get the sound intensity varies from place to place because of constructive interference? K/u interference. B. How should your distances to the two speakers differ to get Equipment and Materials: speakers; plan of classroom destructive interference? K/u 1. Before you begin, decide how to identify areas of constructive C. Use your results marked on your class plan to identify two and destructive interference. How will these manifest locations for either constructive or destructive interference, themselves with sound? estimate their distances from the speakers, and use these 2. Mark the location of the speakers on the classroom plan. estimates to estimate the wavelength of the sound. A 3. Your teacher will turn on the speakers. WEB LINK Mathematics of Two-Point-Source Interference You can measure wavelength using the interference pattern produced by two point sources and develop some mathematical relationships for studying the interference of other waves. Figure 10 shows an interference pattern produced by two point sources in a ripple tank. n2 n1 n1 n2 P1 P2 S1 S2 Figure 10 Ripple tank interference patterns can be used to develop relationships to study interference. The two sources, S1 and S2, separated by a distance of three wavelengths, are vibrating in phase. The bisector of the pattern is shown as a white line perpendicular to the line that joins the two sources. You can see that each side of the bisector has an equal number of nodal lines, which are labelled n1 and n2 on each side. A point on the first nodal line (n1) on the right side is labelled P1, and a point on the second path length the distance from point to nodal line (n2) on the right side is labelled P2. The distances P1S1, P1S2, P2S1, and P2S2 point along a nodal line are called path lengths. 464 Chapter 9 Waves and Light NEL 8160_CH09_p434-469.indd 464 4/30/12 9:47 AM  7 If you measure wavelengths, you will find that P1S1 5 4λ and P1S2 5 λ. The path length difference, Ds1, on the first nodal line is equal to 2 path length difference the difference between path lengths, or distances Ds1 5 0 P1S1 2 P1S2 0 7 5 `4l 2 l ` 2 1 Ds1 5 l 2 This equation applies for any point on the first nodal line. We use the absolute value because only the size of the difference in the two path lengths matters, not which one is greater than the other. A node can be found symmetrically on either side of the perpen- dicular bisector. The path length difference, Ds2, on the second nodal line is equal to Ds2 5 0 P2S1 2 P2S2 0 3 Ds2 5 l 2 Extending this to the nth nodal line, the equation becomes Dsn 5 0 Pn S1 2 Pn S2 0 1 Dsn 5 an 2 bl 2 where Pn is any point on the nth nodal line. Thus, by identifying a specific point on a nodal line and measuring the path lengths, you can use this relationship to determine the wavelength of the interfering waves. This technique will not work if the wavelengths are too small or the point P is too far away from the sources, because the path length difference is too small to measure accurately. If either or both of those conditions apply, you need to use another method to calculate the path length difference. For any point Pn, the path length difference is AS1 (Figure 11(a)): Unit TASK BOOKMARK 0 Pn S1 2 Pn S2 0 5 AS1 You can apply what you have learned When Pn is very far away compared to the separation of the two sources, d, then the about interference to the Unit Task on lines PnS1 and PnS2 are nearly parallel (Figure 11(b)). page 556. S1 A d S1 S2 d S2 n A Pn (a) (b) Figure 11 When developing the equations to calculate path length difference, it is important to consider the distance to Pn to be far enough away that Pn S1 and Pn S2 can be considered parallel. NEL 9.3 Diffraction and Interference of Water Waves   465 8160_CH09_p434-469.indd 465 4/30/12 9:47 AM  In Figure 11(b), the lines AS2, S1S2, and AS1 form a right-angled triangle, which means that the difference in path length can be written in terms of the angle un, since AS1 sin u n 5 d Rearranging, d sin u 1 5 AS1 However, on the previous page we established that AS1 5 0 Pn S1 2 Pn S2 0 1 AS1 5 an 2 bl xn Pn 2 B Combining this with the expression for AS1 in terms of sinu1 leads to 1 l sin u n 5 an 2 b 2 d The angle for the nth nodal line is un, the wavelength is l, and the distance between the sources is d. Using this equation, you can approximate the wavelength for an interference 1 l pattern. Since sin un cannot be greater than 1, it follows that an 2 b cannot be right 2 d bisector greater than 1. The number of nodal lines on the right side of the pattern is the maximum number of lines that satisfies this condition. By counting this number and measuring d, you can determine an approximate value for the wavelength. For example, if d is 2.0 cm and n is 4 for a particular pattern, then 1 l an 2 b L 1 L 2 d 1 l a4 2 b L1 2 2.0 cm 1 1.75 cm L l l L 0.57 cm Although it is relatively easy to measure un for water waves in a ripple tank, for light waves, the measurement is not as straightforward. The wavelength of light is n very small and so is the distance between the sources. Therefore, the nodal lines are close together. You need to be able to measure sin un without having to measure the angle directly. Assuming that a pair of point sources is vibrating in phase, you can use the following derivation. For a point Pn that is on a nodal line and is distant from the two sources, the line from Pn to the midpoint of the two sources PnC can be considered to be parallel n to PnS1. This line is also at right angles to AS2. The triangle PnBC in Figure 12 is A used to determine sin u9n. In this derivation, d is the distance between the sources, S1 C S2 xn is the perpendicular distance from the right bisector to the point Pn, L is the d distance from Pn to the midpoint between the two sources, and n is the number Figure 12 Use the triangle Pn BC to of the nodal line. measure sin u9n. xn sin urn 5 L 466   Chapter 9 Waves and Light NEL 8160_CH09_p434-469.indd 466 4/30/12 9:47 AM  Figure 13 is an enlarged version of the triangle with base CS2 in Figure 12. Since 1 an 2 bl 2 sin un 5 d and, since the right bisector, CB, is perpendicular to S1S2, you can see in Figure 13 that u9n 5 un. So we can say that 1 an 2 bl xn 2 5 L d n    90°  n    90° n    n   right bisector or, n  n n n C S2 d 2 Figure 13 Enlargement of the base triangle from Figure 12 The following Tutorial will demonstrate how to solve problems involving two- dimensional interference. Tutorial 2 Interference in Two Dimensions This Tutorial provides examples and sample calculations for the concepts of two-point-source interference. Sample Problem 1: Interference in Two Dimensions The distance from the right bisector to a point P on the second Solution: nodal line in a two-point interference pattern is 4.0 cm. x2 The distance from the midpoint between the two sources, u 2 5 sin21 L which are 0.5 cm apart, to point P is 14 cm. 4.0 cm (a) Calculate the angle u2 for the second nodal line. 5 sin21 a b 14 cm (b) Calculate the wavelength. u 2 5 178 Solution Statement: The angle of the second nodal line is 178. (a) Given: n 5 2; x2 5 4.0 cm; L 5 14 cm Required: sin u2 xn Analysis: sin un 5 L x2 u 2 5 sin21 L NEL 9.3 Diffraction and Interference of Water Waves   467 8160_CH09_p434-469.indd 467 4/30/12 9:47 AM  (b) Given: n 5 2; x2 5 4.0 cm; L 5 14; d 5 0.5 cm x2d Solution: l 5 Required: l 1 an 2 bL 1 2 an 2 bl 14.0 cm2 10.5 cm2 xn 2 Analysis: 5 l5 L d 3 a b 114 cm2 xn d 2 l5 1 l 5 0.095 cm an 2 bL 2 Statement: The wavelength is 9.5 3 10-2 cm. Sample Problem 2: Determining Wave Pattern Properties Using Differences in Path Length Two identical point sources are 5.0 cm apart, in phase, and (b) Given: f 5 12 Hz; l 5 1.0 cm vibrating at a frequency of 12 Hz. They produce an interference Required: v pattern. A point on the first nodal line is 5 cm from one source Analysis: v 5 fl and 5.5 cm from the other. Solution: v 5 fl (a) Determine the wavelength. 5 112 Hz2 11.0 cm2 (b) Determine the speed of the waves. v 5 12 cm/s Solution Statement: The speed of the waves is 12 cm/s. (a) Given: n 5 1; P1S1 5 5.5 cm; P1S2 5 5.0 cm; d 5 5.0 cm Required: l 1 Analysis: 0 Pn S1 2 Pn S2 0 5 an 2 bl 2 1 Solution:   0 Pn S1 2 Pn S2 0 5 an 2 bl 2 1 0 P1S1 2 P1S2 0 5 a1 2 bl 2 1 0 5.5 cm 2 5.0 cm 0 5 a1 2 bl 2 1 0.5 cm 5 a bl 2 l 5 1.0 cm Statement: The wavelength is 1.0 cm. Practice 1. Two point sources, S1 and S2, are vibrating in phase and produce waves with a wavelength of 2.5 m. The two waves overlap at a nodal point. Calculate the smallest corresponding difference in path length for this point. T/I A [ans: 1.2 m] 2. A point on the third nodal line from the centre of an interference pattern is 35 cm from one source and 42 cm from the other. The sources are 11.2 cm apart and vibrate in phase at 10.5 Hz. T/I (a) Calculate the wavelength of the waves [ans: 2.8 cm] (b) Calculate the speed of the waves [ans: 29 cm/s] 3. Two point sources vibrate in phase at the same frequency. They set up an interference pattern in which a point on the second nodal line is 29.5 cm from one source and 25.0 cm from the other. The speed of the waves is 7.5 cm/s. T/I (a) Calculate the wavelength of the waves. [ans: 3.0 cm] (b) Calculate the frequency at which the sources are vibrating. [ans: 2.5 Hz] 468   Chapter 9 Waves and Light NEL 8160_CH09_p434-469.indd 468 4/30/12 9:48 AM  9.3 Review Summary Diffraction is the bending and spreading of a wave whose wavelength is comparable to or greater than the slit width, or where l $ w. Interference occurs when two waves in the same medium meet. Constructive interference occurs when the crest of one wave meets the crest of another wave. The resulting wave has an amplitude greater than that of each individual wave. Destructive interference occurs when the crest of one wave meets the trough of another wave. The resulting wave has an amplitude less than that of each individual wave. Waves with shorter wavelengths diffract less than waves with longer wavelengths. A pair of identical point sources that are in phase produce a symmetrical pattern of constructive interference areas and nodal lines. The number of nodal lines in a given region will increase when the frequency of vibration of the sources increases or when the wavelength decreases. When the separation of the sources increases, the number of nodal lines also increases. The relationship that can be used to solve for an unknown variable in a 1 two-point-source interference pattern is 0 PnS1 2 PnS2 0 5 an 2 bl. 2 Questions 1. Under what condition is the diffraction of waves 6. Two identical point sources are 5.0 cm apart. through a slit maximized? K/U T/I A metre stick is parallel to the line joining the two 2. Two loudspeakers are 1.5 m apart, and they vibrate in sources. The first nodal line intersects the metre phase at the same frequency to produce sound with stick at the 35 cm and 55 cm marks. Each crossing a wavelength of 1.3 m. Both sound waves reach your point is 50 cm away from the middle of the line friend in phase where he is standing off to one side. joining the two sources. K/U T/I C A You realize that one of the speakers was connected (a) Draw a diagram illustrating this. incorrectly, so you switch the wires and change its (b) The sources vibrate at a frequency of 6.0 Hz. phase by 1808. How does this affect the sound volume Calculate the wavelength of the waves. that your friend hears? K/U (c) Calculate the speed of the waves if the frequency 3. (a) Determine the maximum slit width that will of the sources is the same as in part (a). produce noticeable diffraction for waves of 7. A student takes the following data from a ripple wavelength 6.3 3 1024 m. tank experiment where two point sources are in (b) If the slit is wider than the width you calculated phase: n 5 3, x3 5 35 cm, L 5 77 cm, d 5 6.0 cm, in (a), will the waves diffract? Explain your θ3 5 258, and the distance between identical points answer. K/U T/I on 5 crests is 4.2 cm. From these data, you can work 4. Two speakers are 1.0 m apart and vibrate in phase to out the wavelength in three different ways. T/I produce waves of wavelength 0.25 m. Determine (a) Carry out the relevant calculations to determine the angle of the first nodal line. T/I the wavelength. 5. What conditions are necessary for the interference (b) Which piece of data do you think has been pattern from a two-point source to be stable? K/U T/I incorrectly recorded? NEL 9.3 Diffraction and Interference of Water Waves   469 8160_CH09_p434-469.indd 469 4/30/12 3:21 PM