S 1 S 2 A B C. 7/25/2006 Superposition ( F.Robilliard) 1

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1 P S S S 0 x S A B C 7/5/006 Superposition ( F.Robilliard)

2 Superposition of Waes: As we hae seen preiously, the defining property of a wae is that it can be described by a wae function of the form - y F(x - t) y a sin k(x - t) for any wae for an harmonic wae where, y is a wae displacement. For string waes, y is the mechanical displacement of a string particle from its centre position; for sound waes in a gas, y could be either the mechanical displacement of a molecule from its centre position, or an associated gas pressure difference from ambient pressure; for radio waes, the displacement would be an electric, or magnetic field ector. When two different waes, of the same type, pass, simultaneously, through the same point in space, the indiidual wae displacements add - where and y y + y () y the wae displacement at the point due to the first wae only. y the wae displacement at the point due to the second wae only y the total resultant wae displacement at the point due to both waes Equation () is called the principle of superposition. 7/5/006 Superposition ( F.Robilliard)

3 Linear Superposition: y y + y () Waes propagate through media, by deforming the medium. If the amplitude of a wae is small, this deformation will be within the elastic limits of the medium, and proportional to the wae forces causing it. When more than one such wae passes a point in such a medium, their indiidual amplitudes add linearly, as in equation (). Howeer, in extreme cases, where wae amplitudes are large, the elastic limit of the medium can be exceeded, and amplitudes will add non-linearly (equation () does not hold). We will consider only the linear superposition case (equation ()). We will assume the superposing waes to be harmonic (sinusoidal). These waes can differ in amplitude, frequency, wae elocity, and direction of trael. Whilst any two waes can superpose, the most important cases are where the two waes are identical, except in one wae parameter. We will deal with the following cases, which hae the following single difference -. Standing Waes Beats 3. Interference waes trael in opposite directions waes hae slightly different frequencies waes differ in phase 7/5/006 Superposition ( F.Robilliard) 3

4 .Standing Waes - Introduction: This is our first case of superposition. Standing waes occur, when two identical waes traerse a medium in opposite directions. This happens, when a wae is continuously reflected back along its path, so that the original and reflected waes pass through each other. Examples: +y original wae 7/5/006 Superposition ( F.Robilliard) 4 +x reflected wae In stringed musical instruments, such a piano, guitar, or iolin, a string is plucked, or otherwise disturbed, so that a wae propagates along the string. On reaching the end of the string, this wae is reflected back, so that a standing wae is established on the string. The standing wae is the source of the musical note. Wind instruments are similar, except that the wae is a sound wae in air. In lasers a light wae is sent through a caity bounded by two mirrors. The light wae reflects back and forth between the two mirrors, so an optical standing wae is set up in the caity. The properties of the laser light depend largely on this standing wae.

5 Standing Waes - Graphical : To understand standing waes, we will look at the details of their deelopment, graphically. We start our clock (t 0), when the two oppositely-traeling waes are in phase with each other. Their sum (the standing wae) is shown in red ( ). y t0 +x direction -x direction sum wae x Let T period, and λ waelength of each of the traeling waes. Time (T/8) later, the blue ( ) wae has moed distance (λ/8) in the (+x)- direction, and the purple ( ) wae has traeled (λ/8) in the ( x)-direction. The red sum wae ( ) has not moed, but its magnitude has been reduced. y t(t/8) The full deelopment of the standing wae, oer a time T, is shown in the next slide. x 7/5/006 Superposition ( F.Robilliard) 5

6 y Graph Sequence t0 to T: t0 T4T/8 tt/8 T5T/8 TT/8 T6T/8 T7T/8 T3T/8 T8T/8 7/5/006 Superposition ( F.Robilliard) 6 x

7 Standing Wae Only: y Let us extract the standing wae (the sum wae), from the graphs of the preious slides, oer a time interal t 0 to T/. (Note: this animation will not render in pdf) x y Drawing all these sum waes onto the same graph, using different colours: x 7/5/006 Superposition ( F.Robilliard) 7

8 y Nodes & Antinodes: antinodes P Q R S x nodes Along the x-axis, there are points, such as P, Q, R, & S, where the amplitude of ibration is zero. These points are called nodes. Points midway between nodes hae a maximum amplitude. These points are called antinodes. Between adjacent nodes (eg.between P & Q), all points moe together are in phase. Points between P & Q, are 80 deg out of phase with points between Q & R. Points between P & Q, are in phase with points between R & S. These waes are called standing, or stationary because their node, and antinode positions remain fixed. 7/5/006 Superposition ( F.Robilliard) 8

9 y Separation Between Nodes : P Q R x λ/ t0 By comparing the standing wae, with the two component traeling waes, it is clear, that the distance between adjacent nodes is half a waelength of the traeling waes λ/. 7/5/006 Superposition ( F.Robilliard) 9

10 Standing Waes Wae Function: Next we deelop the wae function for a standing wae. Standing waes occur, when two identical waes traerse a medium in opposite directions. We will assume harmonic waes. For both waes: a amplitude k wae number ω angular frequency We firstly write the wae function for each traeling wae If the initial wae is: y a sin (kx - ωt) () the reflected wae will be: y a sin (kx + ωt) () where y and y are the indiidual displacements for each wae. At any point, x, in the medium, at any time, t, the resultant displacement, y, will be: +y +x y y + y a sin (kx - ωt) + a sin (kx + ωt) a sin (kx). cos (ωt) since: + sin + sin sin cos Standing-Wae Wae Function: y [a sin (kx)]. cos (ωt) ---(3) 7/5/006 Superposition ( F.Robilliard) 0

11 Interpretation of Wae Function: y P Q R x Standing-Wae Wae Function: y [a sin (kx)]. cos (ωt) ---(3) If we fix t, (3) becomes y [a cos (ωt)] sin (kx) A sin (kx) where A [a cos (ωt)] acts as the amplitude of the standing wae pattern, at particular moments, as represented graphically, in the indiidual sine cures plotted aboe. If we fix x, (3) becomes y [a sin (kx)] cos (ωt)] A cos (ωt) where A [a sin (kx)] acts as an amplitude term, giing the amplitude of ibration of the particular particle located at the gien x alue. 7/5/006 Superposition ( F.Robilliard)

12 Node Positions from Wae Function: We can find the positions of nodes, from the wae function. Standing-Wae Wae Function: y [a sin (kx)]. cos (ωt) A cos (ωt) -----(3) At nodes, the amplitude of ibration of particles is zero, at all times, t: A ( kx ) x [ a sin ( kx )] 0,, 0 0, π, π, 3π,..., 3,... since Nodes are equally spaced, and separated by a distance of (λ/), where λ is the waelength of either of the component traeling waes. 0 (λ/) (λ/ 3(λ/)... k y P Q R x x Which corresponds to our preious graphical result. 7/5/006 Superposition ( F.Robilliard)

13 Antinode Positions from Wae Function: We can also find the positions of antinodes, from the wae function. Standing-Wae Wae Function: y [a sin (kx)]. cos (ωt) A cos (ωt) -----(3) At antinodes, the amplitude of ibration of particles is max., at all times, t: A ( kx ) x [ a sin ( kx )] a sin ( kx) π π π π, 3, 5, 7,... 4, 3 4, 5 4, 7,... since 4 Antinodes are equally spaced, separated by a distance of (λ/), and located midway between the nodes. 0 (λ/4) 3(λ/4) 5(λ/4) 7(λ/4)... k y P Q R x x Which also corresponds to our preious graphical result. 7/5/006 Superposition ( F.Robilliard) 3

14 Example: A standing wae is represented by the following wae function: y sin (00x). cos (300t) (SI units) Find the amplitude, frequency, waelength, and elocity of the component traeling waes. Find also, the separation of nodes. Compare this wae-function with the general case. General case: y a sin (kx) cos (ωt) (3) This wae: y sin (00x). cos (600t) (4) Amplitude: a a m.0 mm Waelength: k 00 (π/λ) 00 λ (π)/ m 6.3 cm Frequency: ω 600 πf 600 f 600/(π) 96 Hz Velocity of traeling waes f λ (96) (0.063) 6.05 m/s Separation of nodes λ/ (6.3)/ 3. cm 7/5/006 Superposition ( F.Robilliard) 4

15 Modes of Vibration: Here, we will consider finite media. These are media, that hae boundaries. Examples are the strings of musical instruments, which are bounded by their ends, the bounded air columns within wind instruments, and the membrane of a drum, which is bounded by its rim. When a bounded medium is ibrated by some external force, waes are set up in the medium. These waes are reflected back from the boundaries. The interaction of the waes and their reflections produce stationary waes in the medium. Howeer, conditions at the boundary will determine the ibration at the boundary. For example, the boundary may be rigid, so that the ibration there must hae zero amplitude, and must consequently be a displacement node. Such boundary conditions limit the ways in which standing waes can be set up in the medium, and therefore limit the ways in which the medium can ibrate. The discrete ways in which a bounded medium can ibrate are called its modes of ibration. Each of these modes will be associated with certain standing wae frequencies. Thus, there will be only certain, characteristic, discrete frequencies of ibration possible other frequencies will be suppressed. 7/5/006 Superposition ( F.Robilliard) 5

16 Stretched Strings: This is our first example of a bounded ibrating medium. Assume that the string is stretched between two rigid support points. An oscillating force acts on the string so as to cause standing waes to be set up on the string. Because the string is fixed at both ends, both ends must be nodes. Only standing waes that comply with this requirement can be established on the string. Each of these standing waes corresponds to a ibratory mode of the string. Let the arious modes be characterised by a mode number, i, where i,, 3, 4, 5,... Let L length of the string speed of the traeling waes along the string λ i the waelength of the i th mode of ibration of the string the waelength of the component traeling waes for the i th mode f i the frequency of the i th mode of ibration the frequency of the component traeling waes for the i th mode Remember: for traeling waes f λ thus: for the traeling waes forming the i th mode standing wae therefore: f i λ i f i f/λ 7/5/006 Superposition ( F.Robilliard) 6

17 7/5/006 Superposition ( F.Robilliard) 7 String Modes: Fundamental (st) mode nd ibratory mode 3 rd ibratory mode 4 th ibratory mode ( )... L f L L L x y ( ) f L 4 f L 4 L ( )... f L f L L ( ) f L 3 f L 3 3 L 3 3 3

18 f Modal frequencies: The string can ibrate in many modes, each one of which has a characteristic frequency. The i th mode has frequency f i The simplest mode (fewest nodes) has the lowest frequency, f, which is called the fundamental frequency of the string. Higher modes hae frequencies that are multiples of this fundamental. The fundamental frequency, f, of a stretched string is gien by: L L T f 4 since :f : f3 : f... : : 3: /5/006 Superposition ( F.Robilliard) 8 ( ) from From equations (), (), (3), (4), the modal frequencies for the string are in the ratios: fn n n n f (n,,3...) L When excited, the string will ibrate simultaneously in all these possible modes. Howeer, amplitude of the mode decreases, as their frequency increases. T T tension in string m string s mass per unit length For a gien string, under fixed tension, the frequency of the fundamental is inersely proportional to the length of the string.

19 µ 0.40 gram/m L0 cm W50 N Example: A mass of weight W 50 N is hung from the end of a string of mass per unit length m 0.40 gram/metre, as shown. Find the fundamental frequency with which the string will ibrate, when plucked. Find also, the frequencies of the second, and third ibrational modes. Fundamental: y L x Second & third modes: L f f 884 Hz L & L f f 3 L T 0.0 x khz 3x khz Hz Fundamental, second and third modes hae frequencies 0.884,.77, &.65 khz 7/5/006 Superposition ( F.Robilliard) 9

20 Acoustic caities: This is our second example a bounded medium. The medium is a gas which is bounded by the ends of the caity which contains it. When the caity is excited by some ibrating external force, the resultant traeling sound waes are reflected from the ends of the caity, producing standing waes. The end of an acoustic caity can be closed (a solid boundary), or open (no boundary) If the end is closed, the displacement of air particles, there, must be zero, constituting a displacement node. If the end is open, the ibrating air particles there, hae maximum freedom to ibrate, and the end is a displacement antinode. The particular boundary conditions will determine the standing wae patterns, that can be established in the caity. Each of these patterns is a ibrational mode of the caity, and each will hae a characteristic frequency. There are three cases, depending on whether the two ends are open or closed:. Closed-Open Open-Open Closed-Closed Let: L the caity length. Other terms are as for stretched strings. 7/5/006 Superposition ( F.Robilliard) 0

21 y L Closed-Open: x Fundamental mode f 4L 4L... ( 4) nd ibratory mode 4 L 3 f 3 4L 3 f... ( 5) 3 rd ibratory mode 4 3 L 5 f L 5 f... ( 6) n n n f (n,3,5...) n 4L f Mode frequencies are in ratio f :f :f 3... :3:5... 7/5/006 Superposition ( F.Robilliard)

22 y L Open-Open: x Fundamental ( st ) mode f L L... ( 7) nd ibratory mode f L L f...( 8) 3 rd ibratory mode f L 3 L 3 f...( 9) n n n f (n,,3...) n L f Mode frequencies are in ratio f :f :f 3... ::3... 7/5/006 Superposition ( F.Robilliard)

23 Closed-Closed: When both ends of the acoustic caity are closed, the modal patterns set up correspond to those of the stretched string. The modal frequencies are in the ratio ::3... Fundamental: Nomenclature: The fundamental mode is the one with the smallest frequency. Harmonic: A harmonic is a frequency that is an integer multiple of the fundamental frequency. The i-th harmonic has a frequency i times the fundamental. Oertone: An oertone is any standing-wae frequency aboe the harmonic. The j-th oertone is the j-th mode aboe the fundamental. 7/5/006 Superposition ( F.Robilliard) 3

24 Nomenclature: To clarify notation, the numbers of the mode, harmonic, and oertone are tabled below, firstly for a ::3 system, and secondly for a :3:5 system. f fundamental frequency ::3 system: frequency mode f f 3f 3 4f 4 Example: 3 rd mode 3 rd harmonic nd oertone harmonic 3 4 All harmonics are present oertone 3 :3:5 system: frequency mode f 3f 5f 3 7f 4 Example: 3 rd mode 5 th harmonic nd oertone harmonic oertone Only the odd harmonics are present een harmonics are missing. 7/5/006 Superposition ( F.Robilliard) 4

25 Physics of Music: All acoustic musical instruments consist of a bounded medium, in which standing waes are set up. For example, in a iolin, guitar, and piano, standing waes are set up on a string. In a flute, clarinet, and organ, standing waes are set up in an air column. In a drum, a -dimensional standing wae is set up on a membrane. As we hae seen, many standing wae modes, with increasing frequencies, occur simultaneously, but with decreasing amplitude, as the frequency of the mode increases. The total wae, in the medium, is the instantaneous sum of the many modes present. A discrete sound played on a musical instrument is called a note. The sound of a musical note, played on an instrument, is determined by its particular pattern of standing waes the standing waes generate the sound. There are three important properties of a musical note: loudness, pitch, and tone colour Loudness: Loudness (or olume ) of a note is determined by the amplitude of the standing waes (in particular, the fundamental) producing it. This can be controlled by the igor with which the player of the instrument generates the standing waes. 7/5/006 Superposition ( F.Robilliard) 5

26 Pitch and Musical Scales: The pitch of a musical note is determined by the frequency of the fundamental. Other harmonics are not recognized in the ear s determination of pitch. For example: middle C on the musical scale has a fundamental frequency of 6.63 Hz. In music, notes are generally organized into pleasing sequences according to pitch. These sequences are called musical scales. The fundamental frequencies of the notes in a musical scale hae particular ratios to the fundamental frequency of the key note of that scale. There are seeral types of scales, depending on the particular set of ratios. For example, the 8 notes, in the major key, of the Just (or Helmholtz) scale are in the following frequency ratios: note nr. (n) f n / f.00 9/8. 5/4.0 4/3.33 3/.50 5/3.60 5/ The n note is called the key note of the scale. 7/5/006 Superposition ( F.Robilliard) 6

27 The C Major Scale : note nr. (n) f n / f.00 9/8 5/4 4/3 3/ 5/3 5/ If the key note is middle C (6.63 Hz 6 Hz), the scale is C major, and the names gien to the notes, and their frequencies are: note name C D E F G A B C f n (Hz) If the frequency of the note is a simple ratio to the frequency of the key note (n ), then that note tends to be consonant (or pleasant when sounded) with the key note. n 8 is the most consonant note, called the octae, and is twice the frequency of the key note. The second most consonant note is n 5, called the dominant. Tone Colour: The characteristic sound ( tone colour ) of a particular instrument is determined by the characteristic mixture of harmonics it produces, and their relatie amplitudes. For example: middle C played on a iolin sounds different to middle C played on a flute. 7/5/006 Superposition ( F.Robilliard) 7

28 Resonance: If a system, in which standing waes can be set up, is disturbed, and then left to oscillate by itself, it will oscillate only at the frequencies of its natural standing wae modes. These are the natural frequencies of the system. Howeer, if such a system is drien continuously, by an external periodic force, it can be forced to oscillate at other frequencies. Let s say, that the frequency of the driing force ( f ) is aried, and the resulting amplitude of ibration of the system is measured. The amplitude ( a ) with which the system ibrates, will be greatest when the driing frequency equals, or is near, a natural frequency ( f 0 ) of the system a f 0 f This phenomenon is called resonance, and the natural frequencies of ibration are the resonant frequencies of the system. 7/5/006 Superposition ( F.Robilliard) 8

29 Damping: When first drien at the resonant frequency, the system will absorb energy and, as a consequence, the amplitude of the system will increase. As the amplitude increases, losses due to friction increase, until all the input energy is dissipated in friction. At this point, the amplitude has reached its maximum. We say that the system is damped by its internal friction. The greater the damping, the smaller will be the final resonant amplitude of the system. a f 0 weakly damped strongly damped f Damping is often measured in terms of a quantity called the Q-factor of the system. Weakly damped systems hae a high Q; strongly damped, a low Q. If a system is weakly damped, the large amplitude at resonance can lead to damage to the system for example: shattering a champagne glass with a sustained note. Examples of Resonance: a child s swing, champagne glass, the tuned circuit of a radio receier. 7/5/006 Superposition ( F.Robilliard) 9

30 .Beats: This is our second case of superposition. Beats occur when two waes, which are identical, except for a (slight) difference in frequency, pass through the same point. f f P Say the two waes start in step. Because their frequencies are slightly different, the waes will progressiely get further out of step, until they hae a phase difference of 80 deg, and consequently annul each other, at the point. They will then progressiely get back into step with each other, until they are back in phase, and consequently reinforce each other. When the waes are in step, their amplitudes add, and a maximum occurs; when out of step, their amplitudes cancel, and a minimum results. If the waes are acoustic, this produces a waxing and waning in the olume of the resultant sound, which is called beats. This can be seen graphically as follows. Note, that we are representing the displacements of the waes oer time, at a fixed point in space. 7/5/006 Superposition ( F.Robilliard) 30

31 y Beats Graphically: t Displacement, y, at a point, P, due to single wae. Indiidual displacements due to two waes of different frequency The sum wae total.displacement Enelope of sum wae showing the beats. A beat occurs at times of maximum sum wae amplitude. 7/5/006 Superposition ( F.Robilliard) 3

32 The Enelope Wae: The enelope of the sum wae represents its amplitude. y t Amplitude t Beats occur where the amplitude is a maximum. 7/5/006 Superposition ( F.Robilliard) 3

33 Beats Analytically : Here we deelop the wae function for beats. Beats occur, when two waes of different frequencies traerse a medium in the same direction. We will assume harmonic waes along the x-axis. We firstly write the wae function for each traeling wae For both waes: a amplitude k wae number ω and ω angular frequencies +y ω ω +x ω wae: y a sin (kx ω t) () ω wae: y a sin (kx ω t) () where y and y are the displacements of the two waes For conenience, let s choose the point P to be at the origin, where x 0. Thus: ω wae: y a sin (ω t) (3) ω wae: y a sin (ω t) (4) 7/5/006 Superposition ( F.Robilliard) 33

34 y a y + sin( y t) + a sin( + + a sin t) Beats Wae Function : At Point P (x 0), in the medium, at any time, t, the resultant displacement, y, will be: Beat wae Function: { from ( 3) & ( 4) } t cos t a cos + since: sin + sin sin + sin t... 7/5/006 Superposition ( F.Robilliard) 34 t y The resultant wae at P consists of an harmonic wae ( 5) cos Time-dependent amplitude term + sin whose angular frequency is the aerage of the two component waes, with a time-dependent amplitude a cos of lower frequency, equal to half the difference between the components t t +

35 Beat Frequency: + y a cos t sin t... ( 5) We say that the resultant wae consists of a high-frequency carrier wae, with an amplitude that is modulated by a lower frequency wae. Definition: The beat frequency, f B, is the number of beats that occur per second. From (5), the angular frequency of the amplitude term is (ω ω )/.. The corresponding frequency will be (f f )/ But a beat occurs when the amplitude is either +a or -a, that is, twice per cycle of the amplitude. Thus the beat frequency, f B, will be (f f ) The beat frequency is equal to the frequency difference between the two component waes But we hae assumed that f > f. It could be that f < f. Therefore: f B f f 7/5/006 Superposition ( F.Robilliard) 35

36 Example: When a tuning fork of frequency 440 Hz is sounded with a second tuning fork, beats per second are heard. What is the frequency of the second fork? Let and f 440 frequency of the first fork f frequency of the second fork Beat frequency f B f f Therefore f f 440 +/- f B 440 +/- 44 Hz or 438 Hz (There are two possible answers) Beats In Music: When two notes are sounded together, the sound can seem pleasant (consonant), or harsh (dissonant). This is determined by the beat pattern that is produced. To hear some examples check the following url. 7/5/006 Superposition ( F.Robilliard) 36

37 3. Interference : This is our third, and final, example of superposition. Interference occurs when two waes that differ only in phase, pass through the same point. This difference in phase is typically due to a difference in path length traeled, from a common source, S, to some point P, where the interference happens. S y P y reflector If the two waes are in phase, at P, they will add together, producing a resultant wae of maximum amplitude; if they are 80 deg out of phase, they will cancel each other out, producing a null, or minimum. Thus, as we shift the point P, in space, we get a spatial pattern of maxima, and minima, called an interference pattern. 7/5/006 Superposition ( F.Robilliard) 37

38 Interference Wae Function: Say that wae traels to point P by path SP, while wae traels by longer path SRP. Let the path difference (SRP - SP). If SP x, the wae function for wae, at P, is: y a sin (kx - ωt)...() and for wae : y a sin (k [x + ] ωt)...() S y y R reflector P Thus the resultant wae at P will be: y y + y y a sin ( kx -t ) + a sin( k[ x + ] -t ) kx - t + a sin a cos k k cos sin kx t + k k...( 3) since: + sin sin + sin cos amplitude term 7/5/006 Superposition ( F.Robilliard) 38

39 y a cos k sin kx t + k...( 3) Intensity: Interference can occur for waes of any type for example, sound waes can interfere, as can radio waes, and light waes. In fact, interference is a fundamental, and characteristic property of waes in general. In the case of light waes, the frequency is too high for current technology to directly obsere wae displacements or amplitudes. Only wae intensities can be measured. As we hae seen earlier, intensity is proportional to the square of the amplitude of a wae. Thus the intensity of an interference pattern, I, can be got from the amplitude- From ( 3) amplitude a cos I I a, using k ( amplitude ) cos π 4a : k + a cos cos... ( 5)... using cos cos hence cos 7/5/006 Superposition ( F.Robilliard) 39 ( 4) - ( cos θ + ) Intensity of the interference pattern

40 a a cos I a cos π... + ( 4)...( 5) Graphically: We can plot the amplitude and intensity functions together, as the path difference increases. maxima For comparison, we only plot the cosine parts. Intensity: cos(π /λ) + /λ Amplitude: cos(π /λ) minima Note that the minimum intensity is zero can t be negatie. Note that the intensity is a maximum, where amplitude is (+a) or (-a). Minima occur midway between maxima. 7/5/006 Superposition ( F.Robilliard) 40

41 Maxima & Minima: Maxima: A maximum intensity occurs where ( /λ) 0,,, 3,... 0, λ, λ, 3λ... /λ A maximum occurs where there is a path difference equal to an een number of waelengths Minima: A minimum amplitude occurs where ( /λ) /, 3/, 5/, 7/,... λ/, 3λ/, 5λ/, 7λ/,... A minimum occurs where there is a path difference equal to an odd number of half waelengths Increasing the path difference,, between the two waes by λ/, causes the sum wae at P, to go from a maximum to the next minimum, Interference produces a pattern of maxima, and minima in space the interference pattern. 7/5/006 Superposition ( F.Robilliard) 4

42 λ/ λ In Terms of the Component Waes: x 0 Constructie λ/ Destructie λ/ Constructie 3λ/ Destructie 4λ/ Constructie Consider the two interfering waes, at a gien instant, at different points near P. y y As the path difference,, between the waes, increases in steps of λ/, one wae shifts in phase, in steps of π, relatie to the other, and the pattern at P goes from a maximum (constructie interference), to the next minimum (destructie interference). Constructie interference is where the waes add to a maximum; destructie interference is where they add to zero. 7/5/006 Superposition ( F.Robilliard) 4

43 Young s Double Slit Experiment: Eer since the time of Newton, a controersy existed as to whether light was a stream of particles, or a wae motion. In 80 a piotal experiment was carried out by Thomas Young, which demonstrated conclusiely that light behaed like a wae. He showed that light from two sources could produce interference effects. The setup he used was as follows: P S S S 0 x S A B C S monochromatic (ie. single frequency) physical light source A, B, & C are screens S 0 slit in screen A S & S pair of slits in screen B 7/5/006 Superposition ( F.Robilliard) 43

44 The Experiment: Waefronts are emitted from P S physical light source S. These waefronts, which may be irregular in shape, arrie at S S x 0 screen A. Slit S S 0 selects a tiny sample of each of these total waefronts, and transmits it on to A B C screen B. Because only a small section of the original irregular waefront from S, was transmitted by S 0, the waefronts arriing at S & S will be regular (smooth). At screen B, two sections of this regular waefront are selected, simultaneously, by slits S & S and transmitted on to screen C. Thus S & S, become two irtual sources of waes, that are in phase at S & S, and are transmitted to screen C. Consider a typical point, P, on screen C. Because the path lengths S P, and S P are generally different, the waes arriing at a particular point, P, will hae a fixed phase difference. The particular phase difference will depend on where point P is located on the screen. Thus we hae the conditions for interference to occur at P two waes that differ only in phase, passing through a gien point in space. 7/5/006 Superposition ( F.Robilliard) 44

45 At Screen C: P S S S 0 S x A B C Optical interference will take place on screen C. Where there is constructie interference between the two waes, we get a maximum light intensity (a bright fringe); where there is destructie interference we get a minimum light intensity (a dark fringe). We will firstly, find an expression for the path difference between the two beams, from the geometry of the experiment. We will then be able to find the positions for maxima, and minima. Finally, we will find an expression for the intensity of the interference pattern. 7/5/006 Superposition ( F.Robilliard) 45

46 Geometry: P S y S S 0 d S θ L x A B C θ Let x-axis be the symmetry axis passing d mid-way between slits S & S. Let: sin θ /d slit separation S S d d sin θ...() distance from B to C L angle of P to x-axis θ θ height of P aboe x-axis y L path difference S P S P tan θ y/l...() 7/5/006 Superposition ( F.Robilliard) 46 y

47 Approximation: P S y S S 0 d S θ T d sin θ...() L x A B C In deriing equation (), we hae assumed that if S P TP, then the angle at T is 90 deg. This will only be precisely correct if S P is parallel to S P, which is clearly not so. Therefore we must settle for the assumption that S P is approximately parallel to S P. This approximation will be alid if L >>d. (or, that screen C is far from the slits, or that angle θ is small.) (Conditions that satisfy this assumption are called Fraunhoffer conditions.) 7/5/006 Superposition ( F.Robilliard) 47

48 tan θ y/l...() sin θ /d d sin θ...() Constructie Interference: We saw earlier, for an intensity maximum (constructie interference), that the path difference,, needs to be a whole number of waelengths. 0, λ, λ, 3λ,... or d sin θ mλ (m 0,,, 3,...) from () Max. Using (), we can express this in terms of y: d sin θ d tan θ (d/l) y mλ (m 0,,, 3,...)...(3) Max. sin θ ~ tan θ, since θ is small Thus, from (3), y positions of intensity maxima are: y m (L/d) λ (m 0,,, 3,...)...(4) Max. Thus the intensity maxima occur as bright parallel bands of light ( bright fringes ), equally-spaced up the y-axis (parallel to the z-axis). From (4), the distance up the y-axis from one fringe to the next (the bright fringe spacing) (L/d)λ ) 7/5/006 Superposition ( F.Robilliard) 48

49 tan θ y/l...() sin θ /d d sin θ...() Destructie Interference: We also saw earlier, for an intensity minimum (destructie interference), that the path difference,, needs to be an odd number of half waelengths. λ/, 3λ/, 5λ/, 7l/,... (0+/)λ, (+/)λ, (+/)λ, (3+/)λ,... or d sin θ (m+/) λ (m 0,,, 3,...) from () Min. Using (), we can express this in terms of y: d sin θ d tan θ (d/l) y (m+/) λ (m 0,,, 3,...)...(3) Min. Using (3), y positions of intensity maxima: sin θ ~ tan θ, since θ is small y (m+/) (L/d) λ (m 0,,, 3,...)...(5) Min. Thus the intensity minima occur as dark parallel bands of light ( dark fringes ), equally-spaced up the y-axis, and midway between the bright fringes ( From (4): dark fringe spacing (L/d)λ ) 7/5/006 Superposition ( F.Robilliard) 49

50 d S Consider screens B & C. Interference Pattern: y x P (Bright) P (Dark) P (Bright) For m 0,,, 3,... y m (L/d) λ...(4) Max. y (m+/) (L/d) λ...(5) Min. y S L B C From (4): m gies a bright fringe at y (L/d) λ From (5): m 0 gies a dark fringe at y (/)(L/d) λ From (4): m 0 gies a bright fringe at y 0 z The screen C will be crossed by alternating, equally spaced, bright and dark fringes aligned parallel to the z-axis. 7/5/006 Superposition ( F.Robilliard) 50

51 Intensity Pattern: From our earlier general discussion of interference, we found that the intensity of an interference pattern is gien by: I ( amplitude) 4a cos I I0 cos...( 6) where I is the maximum intensity 0 ( at the centre of a bright fringe) I And, here: d sin θ (d/l)y (L/d)λ (L/d)λ y Hence, for the intensity pattern (we rotated it for conenience) we can plot the intensity, I. 7/5/006 Superposition ( F.Robilliard) 5

52 Example: Two slits separated by.00 mm, are illuminated by a laser, so that an interference pattern is produced, on a screen, distant.00 m from the slits. The separation between adjacent bright fringes on the screen is measured to be.6 mm. What was the waelength of the laser? Fringe separation is gien by: y (L/d) λ therefore λ ( y)(d/l) (.6x0-3 ) (.00x0-3 ) / (.00) 6.30x0-7 m 630 nm 7/5/006 Superposition ( F.Robilliard) 5

53 The Nature of Light: Young s experiment showed that light behaes like a wae. But what is it that ibrates, to produce the wae? what sort of wae is light? Light consists of a ibrating electric field. That is, at a point in space through which light passes, an electric field ector exists, which ibrates in length (that is, in strength), as the light wae propagates past the point. Associated with this electric field, there is a simultaneous magnetic field ector, at the same point, which is perpendicular to, and ibrates with, the electric ector. Both electric and magnetic ectors (E and B) are perpendicular to the direction of the wae elocity () of the light wae. (The direction of the cross product of the field ectors, (ExB), at any moment, is in the direction of the wae elocity,.) Because it is composed of ibrating electric, and magnetic, fields, we call light an electromagnetic wae. The energy of the wae is the total of the energies contained within the electric and magnetic fields. 7/5/006 Superposition ( F.Robilliard) 53

54 Electromagnetic Waes: For isible light (waelength range about nm), the frequency of the light wae is perceied, by the human eye, as colour. Light of lower frequency appears red; light of higher frequency appears blue. If we reduce the light frequency below red frequencies, or aboe blue frequencies, we get other types of electromagnetic wae, which are inisible to the human eye. Lower frequencies: infra-red, microwae, radio waes. Higher frequencies: ultra-iolet, X-rays, γ-rays. The frequencies of the Electromagnetic spectrum are illustrated in the following diagram: 7/5/006 Superposition ( F.Robilliard) 54

55 EM Spectrum: 7/5/006 Superposition ( F.Robilliard) 55

56 Waes or Particles?: Young s experiment demonstrated, for the first time, that light is a wae. Howeer, there are other experiments, such as the photoelectric effect, in which light behaes like a stream of particles, called photons. Photons are tiny discrete particles of electromagnetic energy. The question therefore arises: is light a wae, or a stream of particles which is the correct model? It turns out that light, together with other forms of electromagnetic radiation, has a strange, dual nature it shows both wae and particle characteristics. Quantum Physics had to be deeloped to account properly for this dual character of light. 7/5/006 Superposition ( F.Robilliard) 56

57 P S S S 0 x S A B C 7/5/006 Superposition ( F.Robilliard) 57