9.3. Diffraction and Interference of Water Waves


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1 Diffraction an Interference of Water Waves 9.3 Have you ever notice how people relaxing at the seashore spen so much of their time watching the ocean waves moving over the water, as they break repeately an roll onto the shore? Water waves behave similarly to other kins 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 sprea out on the other sie of the opening, an 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 create an the two waves combine in preictable 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 epth is constant an no obstacles are in the way. If, however, the waves pass by an ege of an obstacle or through a small opening, the waves sprea out. Diffraction is the bening of a wave as the wave passes through an opening or by an obstacle. The amount of iffraction epens on the wavelength of the waves an the size of the opening. In Figure, an obstacle iffracts shorter wavelengths slightly. In Figure, the same obstacle iffracts longer wavelengths more. iffraction the bening an spreaing of a wave when it passes through an opening; epenent on the size of the opening an the wavelength of the wave short wavelengths long wavelengths Figure When waves travel by an ege, shorter wavelengths iffract less than longer wavelengths. 9.3 Diffraction an Interference of Water Waves 459
2 Although iffraction also occurs for soun an light waves, we will stuy the phenomenon first for water waves because their long wavelength allows for easier observation of the effects of iffraction. In Figure 3, you can see a wave encountering a slit with a certain with, w. In Figure 3 an Figure 3(c), the with of the opening is the same but the wavelength of the incient waves has change. Notice how the amount of iffraction changes. Therefore, as the wavelength increases but the with of the slit oes not, the iffraction also increases. In Figure 3, the wavelength is approximately onethir 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 sie. In Figure 3, the wavelength is about half of w, which means that significantly more iffraction occurs, but areas to the ege exist where no waves are iffracte. In Figure 3(c), the wavelength is approximately twothirs of w, an the sections of the wave front that pass through the slit are almost all converte to circular wave fronts. (c) Figure 3 As the wavelength increases, the amount of iffraction increases. What happens if you change the with of the slit but keep the wavelength fixe? As Figure 4 an Figure 4 show, as the size of the slit ecreases, the amount of iffraction increases. If waves are to unergo more noticeable iffraction, the wavelength must be comparable to or greater than the slit with (λ $ w). For small wavelengths (such as those of visible light), you nee to have narrow slits to observe iffraction. Figure 4(c) shows what happens to the iffraction pattern when the wavelength ecreases but the size of the slit oes not change. incient waves nearly straightline propagation incient waves incient waves Investigation Properties of Water Waves (page 487) You have learne how observing water waves helps in unerstaning the properties of waves in general. This investigation will give you an opportunity to test conitions for iffraction to occur. large opening small opening shorter wavelength, no change in opening (c) Figure 4 an As the size of the slit (aperture) ecreases, iffraction increases. (c) With a shorter wavelength an no change in the size of the opening, there is less iffraction. The relationship between wavelength, with of the slit, an extent of iffraction is perhaps familiar for soun waves. You can hear soun through an open oor, even if you cannot see what is making the soun. The primary reason that soun waves iffract aroun the corner of the oor is that they have long wavelengths compare to the with of the oorway. Low frequencies (the longer wavelengths of the soun) iffract more than high frequencies (the shorter wavelengths of the soun). So if a soun system is in the room next oor, you are more likely to hear the lower frequencies (the bass). In the following Tutorial, you will learn how to solve problems relating to iffraction. 460 Chapter 9 Waves an Light
3 Tutorial 1 Diffraction The following Sample Problem provies an example an sample calculations for the concepts of iffraction. Sample Problem 1: Diffraction through a Slit Determine an explain the ifference between the iffractions observe in Figure 5 an Figure cm 0.5 cm w a cm w b 0.5 cm Figure 5 Given: l cm; w a 5 cm; w b cm Require: iffraction analysis Analysis: l w $ 1 Solution: For Figure 5, l 0.5 cm 5 w a c m l, 1 w a For Figure 5, l 0.5 cm 5 w b 0.5 cm l 5 1 w b Statement: Since the ratio in Figure 5 is less than 1, little iffraction occurs. Since the ratio in Figure 5 is 1, more noticeable iffraction occurs. Practice 1. Determine whether iffraction 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]. A laser shines re light with a wavelength of 630 nm onto an ajustable slit. Determine the maximum slit with that will cause significant iffraction of the light. T/I [ans: 6.3 * 107 m] 9.3 Diffraction an Interference of Water Waves 461
4 Interference interference the phenomenon that occurs when two waves in the same meium interact constructive interference the phenomenon that occurs when two interfering waves have isplacement in the same irection where they superimpose When two waves cross paths an become superimpose, they interact in ifferent ways. This interaction between waves in the same meium is calle interference. You experience interference when you listen to music an other types of soun. For example, soun waves from two speakers may reach the listener s ears at the same time, as illustrate in Figure 6. If the crest of one wave coincies with the crest of the other, then the waves are in phase an combine to create a resultant wave with an amplitue that is greater than the amplitue of either iniviual wave resulting in a louer soun. This phenomenon is calle constructive interference. Waves are in phase. Waves are 180 out of phase. speaker 1 L 1 L speaker estructive interference the phenomenon that occurs when two interfering waves have isplacement in opposite irections where they superimpose coherent compose of waves having the same frequency an fixe phases listener constructive interference estructive interference (c) Figure 6 If L 1 an L are the same, the waves arrive at the listener in phase. Waves interfere constructively if they arrive in phase. (c) If L 1 an L iffer by, say, l, the waves arrive 1808 out of phase, an they interfere estructively. For two waves that iffer in phase by l, shown in Figure 6(c), the crest of one wave coincies with the trough of the other wave. This correspons to a phase ifference of The two waves combine an prouce a resulting wave with an amplitue that is smaller than the amplitue of either of the two iniviual waves. This phenomenon is calle estructive interference. The following conitions must be met for interference to occur: 1. Two or more waves are moving through ifferent regions of space over at least some of their way from the source to the point of interest.. The waves come together at a common point. 3. The waves must have the same frequency an must have a fixe relationship between their phases such that over a given istance or time the phase ifference between the waves is constant. Waves that meet this conition are calle coherent. Figure 7 shows water waves interfering. Figure 7 Interference of water waves in a ripple tank 46 Chapter 9 Waves an Light
5 In Figure 7, two point sources have ientical frequencies. The sources prouce waves that are in phase an have the same amplitues. Successive wave fronts travel out from the two sources an 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 amplitue.) Symmetrical patterns sprea out from the sources, proucing line locations where constructive an estructive interference occur. A noe is a place where estructive interference occurs, resulting in zero amplitue (a net isplacement of zero). As shown in Figure 8, the interference pattern inclues lines of maximum isplacement, cause by constructive interference, separate by lines of zero isplacement, cause by estructive interference. The lines of zero isplacement are calle noal lines. noe a point along a staning wave where the wave prouces zero isplacement noal line a line or curve along which estructive interference results in zero isplacement areas of constructive interference lines of estructive interference (noal lines) amplitue here is A crests troughs amplitue here is A noe S 1 S Figure 8 Interference pattern between two ientical sources each with positive amplitue A When the frequency of the two sources increases, the wavelength ecreases, which means that the noal lines come closer together an the number of noal lines increases. When the istance between the two sources increases, the number of noal lines also increases. The symmetry of the pattern oes not change when these two factors are change. However, if the relative phase of the two sources changes, then the pattern shifts, as shown in Figure 9, but the number of noal lines stays the same. In Figure 9, the sources are in phase, but in Figure 9, there is a phase ifference of Investigation 9.3. Interference of Waves in Two Dimensions (page 488) So far, you have rea about twopointsource interference in theory, with iagrams an photographs. This investigation gives you an oppor tunity to see an measure interference for yourself. S 1 S S 1 S Figure 9 Effect of phase change on interference pattern for zero phase ifference an 1808 phase ifference 9.3 Diffraction an Interference of Water Waves 463
6 Mini Investigation Interference from Two Speakers Mini Investigation Skills: Performing, Observing, Analyzing, Communicating At the start of the iscussion on interference, you consiere soun waves coming from two speakers. In this investigation, your teacher will set up two speakers in the classroom, which will emit a singlefrequency soun. You will examine how the soun intensity varies from place to place because of interference. Equipment an Materials: speakers; plan of classroom 1. Before you begin, ecie how to ientify areas of constructive an estructive interference. How will these manifest themselves with soun?. Mark the location of the speakers on the classroom plan. 3. Your teacher will turn on the speakers. SKILLS HANDBOOK A.1 4. Walk aroun the ifferent areas of the classroom an mark on your plan where constructive an estructive interference occur. A. How shoul your istances to the two speakers iffer to get constructive interference? K/u B. How shoul your istances to the two speakers iffer to get estructive interference? K/u C. Use your results marke on your class plan to ientify two locations for either constructive or estructive interference, estimate their istances from the speakers, an use these estimates to estimate the wavelength of the soun. A WEB LINK Mathematics of TwoPointSource Interference You can measure wavelength using the interference pattern prouce by two point sources an evelop some mathematical relationships for stuying the interference of other waves. Figure 10 shows an interference pattern prouce by two point sources in a ripple tank. n n 1 n 1 n P 1 P S 1 S Figure 10 Ripple tank interference patterns can be use to evelop relationships to stuy interference. path length the istance from point to point along a noal line The two sources, S 1 an S, separate by a istance of three wavelengths, are vibrating in phase. The bisector of the pattern is shown as a white line perpenicular to the line that joins the two sources. You can see that each sie of the bisector has an equal number of noal lines, which are labelle n 1 an n on each sie. A point on the first noal line (n 1 ) on the right sie is labelle P 1, an a point on the secon noal line (n ) on the right sie is labelle P. The istances P 1 S 1, P 1 S, P S 1, an P S are calle path lengths. 464 Chapter 9 Waves an Light
7 If you measure wavelengths, you will fin that P 1 S 1 5 4λ an P 1 S 5 7 λ. The path length ifference, Ds 1, on the first noal line is equal to Ds 1 5 0P 1 S 1 P 1 S 0 path length ifference the ifference between path lengths, or istances 5 `4l 7 l ` Ds l This equation applies for any point on the first noal line. We use the absolute value because only the size of the ifference in the two path lengths matters, not which one is greater than the other. A noe can be foun symmetrically on either sie of the perpenicular bisector. The path length ifference, Ds, on the secon noal line is equal to Ds 5 0P S 1 P S 0 Ds 5 3 l Extening this to the nth noal line, the equation becomes Ds n 5 0P n S 1 P n S 0 Ds n 5 an 1 bl where P n is any point on the nth noal line. Thus, by ientifying a specific point on a noal line an measuring the path lengths, you can use this relationship to etermine 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 ifference is too small to measure accurately. If either or both of those conitions apply, you nee to use another metho to calculate the path length ifference. For any point P n, the path length ifference is AS 1 (Figure 11): 0P n S 1 P n S 0 5 AS 1 When P n is very far away compare to the separation of the two sources,, then the lines P n S 1 an P n S are nearly parallel (Figure 11). Unit TASK BOOKMARK You can apply what you have learne about interference to the Unit Task on page 556. S 1 A S 1 S S n A Figure 11 When eveloping the equations to calculate path length ifference, it is important to consier the istance to P n to be far enough away that P n S 1 an P n S can be consiere parallel. P n 9.3 Diffraction an Interference of Water Waves 465
8 In Figure 11, the lines AS, S 1 S, an AS 1 form a rightangle triangle, which means that the ifference in path length can be written in terms of the angle u n, since sin u n 5 AS 1 Rearranging, sin u 1 5 AS 1 However, on the previous page we establishe that AS 1 5 0P n S 1 P n S 0 B x n P n AS 1 5 an 1 bl Combining this with the expression for AS 1 in terms of sinu 1 leas to sin u n 5 an 1 b l The angle for the nth noal line is u n, the wavelength is l, an the istance between the sources is. Using this equation, you can approximate the wavelength for an interference right bisector pattern. Since sin u n cannot be greater than 1, it follows that an 1 b l cannot be greater than 1. The number of noal lines on the right sie of the pattern is the maximum number of lines that satisfies this conition. By counting this number an measuring, you can etermine an approximate value for the wavelength. For example, if is.0 cm an n is 4 for a particular pattern, then A S 1 C S n n Figure 1 Use the triangle P n BC to measure sin u9 n. L an 1 b l L 1 a4 1 b l.0 cm L cm L 1 l l L 0.57 cm Although it is relatively easy to measure u n for water waves in a ripple tank, for light waves, the measurement is not as straightforwar. The wavelength of light is very small an so is the istance between the sources. Therefore, the noal lines are close together. You nee to be able to measure sin u n without having to measure the angle irectly. Assuming that a pair of point sources is vibrating in phase, you can use the following erivation. For a point P n that is on a noal line an is istant from the two sources, the line from P n to the mipoint of the two sources P n C can be consiere to be parallel to P n S 1. This line is also at right angles to AS. The triangle P n BC in Figure 1 is use to etermine sin u9 n. In this erivation, is the istance between the sources, x n is the perpenicular istance from the right bisector to the point P n, L is the istance from P n to the mipoint between the two sources, an n is the number of the noal line. sin ur n 5 x n L 466 Chapter 9 Waves an Light
9 Figure 13 is an enlarge version of the triangle with base CS in Figure 1. Since an 1 bl sin u n 5 an, since the right bisector, CB, is perpenicular to S 1 S, you can see in Figure 13 that u9 n 5 u n. So we can say that an 1 x n L 5 bl right bisector n n 90 n 90 n n or, n n n C S Figure 13 Enlargement of the base triangle from Figure 1 The following Tutorial will emonstrate how to solve problems involving twoimensional interference. Tutorial Interference in Two Dimensions This Tutorial provies examples an sample calculations for the concepts of twopointsource interference. Sample Problem 1: Interference in Two Dimensions The istance from the right bisector to a point P on the secon noal line in a twopoint interference pattern is 4.0 cm. The istance from the mipoint between the two sources, which are 0.5 cm apart, to point P is 14 cm. Calculate the angle u for the secon noal line. Calculate the wavelength. Solution Given: n 5 ; x cm; L 5 14 cm Solution: u 5 sin x 1 L 4.0 cm 5 sin 1 a 14 cm b u Statement: The angle of the secon noal line is 178. Require: sin u Analysis: sin u n 5 x n L u 5 sin 1 x L 9.3 Diffraction an Interference of Water Waves 467
10 Given: n 5 ; x cm; L 5 14; cm Require: l Solution: l 5 x an 1 bl Analysis: x an 1 n L 5 bl x n l 5 an 1 bl 14.0 cm 10.5 cm l 5 a 3 b 114 cm l cm Statement: The wavelength is cm. Sample Problem : Determining Wave Pattern Properties Using Differences in Path Length Two ientical point sources are 5.0 cm apart, in phase, an vibrating at a frequency of 1 Hz. They prouce an interference pattern. A point on the first noal line is 5 cm from one source an 5.5 cm from the other. Determine the wavelength. Determine the spee of the waves. Solution Given: n 5 1; P 1 S cm; P 1 S cm; cm Require: l Given: f 5 1 Hz; l cm Require: v Analysis: v 5 fl Solution: v 5 fl 5 11 Hz 11.0 cm v 5 1 cm/s Statement: The spee of the waves is 1 cm/s. Analysis: 0P n S 1 P n S 0 5 an 1 bl Solution: 0P n S 1 P n S 0 5 an 1 bl 0P 1 S 1 P 1 S 0 5 a1 1 bl 05.5 cm 5.0 cm0 5 a1 1 bl 0.5 cm 5 a 1 bl l cm Statement: The wavelength is 1.0 cm. Practice 1. Two point sources, S 1 an S, are vibrating in phase an prouce waves with a wavelength of.5 m. The two waves overlap at a noal point. Calculate the smallest corresponing ifference in path length for this point. T/I A [ans: 1. m]. A point on the thir noal line from the centre of an interference pattern is 35 cm from one source an 4 cm from the other. The sources are 11. cm apart an vibrate in phase at 10.5 Hz. T/I Calculate the wavelength of the waves [ans:.8 cm] Calculate the spee of the waves [ans: 9 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 secon noal line is 9.5 cm from one source an 5.0 cm from the other. The spee of the waves is 7.5 cm/s. T/I Calculate the wavelength of the waves. [ans: 3.0 cm] Calculate the frequency at which the sources are vibrating. [ans:.5 Hz] 468 Chapter 9 Waves an Light
11 9.3 Review Summary Diffraction is the bening an spreaing of a wave whose wavelength is comparable to or greater than the slit with, or where l $ w. Interference occurs when two waves in the same meium meet. Constructive interference occurs when the crest of one wave meets the crest of another wave. The resulting wave has an amplitue greater than that of each iniviual wave. Destructive interference occurs when the crest of one wave meets the trough of another wave. The resulting wave has an amplitue less than that of each iniviual wave. Waves with shorter wavelengths iffract less than waves with longer wavelengths. A pair of ientical point sources that are in phase prouce a symmetrical pattern of constructive interference areas an noal lines. The number of noal lines in a given region will increase when the frequency of vibration of the sources increases or when the wavelength ecreases. When the separation of the sources increases, the number of noal lines also increases. The relationship that can be use to solve for an unknown variable in a twopointsource interference pattern is 0P n S 1 P n S 0 5 an 1 bl. Questions 1. Uner what conition is the iffraction of waves through a slit maximize? K/U T/I. Two louspeakers are 1.5 m apart, an they vibrate in phase at the same frequency to prouce soun with a wavelength of 1.3 m. Both soun waves reach your frien in phase where he is staning off to one sie. You realize that one of the speakers was connecte incorrectly, so you switch the wires an change its phase by How oes this affect the soun volume that your frien hears? K/U 3. Determine the maximum slit with that will prouce noticeable iffraction for waves of wavelength m. If the slit is wier than the with you calculate in, will the waves iffract? Explain your answer. K/U T/I 4. Two speakers are 1.0 m apart an vibrate in phase to prouce waves of wavelength 0.5 m. Determine the angle of the first noal line. T/I 5. What conitions are necessary for the interference pattern from a twopoint source to be stable? K/U T/I 6. Two ientical point sources are 5.0 cm apart. A metre stick is parallel to the line joining the two sources. The first noal line intersects the metre stick at the 35 cm an 55 cm marks. Each crossing point is 50 cm away from the mile of the line joining the two sources. K/U T/I C A Draw a iagram illustrating this. The sources vibrate at a frequency of 6.0 Hz. Calculate the wavelength of the waves. (c) Calculate the spee of the waves if the frequency of the sources is the same as in part. 7. A stuent takes the following ata from a ripple tank experiment where two point sources are in phase: n 5 3, x cm, L 5 77 cm, cm, θ , an the istance between ientical points on 5 crests is 4. cm. From these ata, you can work out the wavelength in three ifferent ways. T/I Carry out the relevant calculations to etermine the wavelength. Which piece of ata o you think has been incorrectly recore? 9.3 Diffraction an Interference of Water Waves 469
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