Phase difference plays an important role in interference. Recalling the phases in (3.32) and (3.33), the phase difference, φ, is

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1 Phase Difference Phase difference plays an important role in interference. Recalling the phases in (3.3) and (3.33), the phase difference, φ, is φ = (kx ωt + φ 0 ) (kx 1 ωt + φ 10 ) = k (x x 1 ) + (φ 0 φ 10 ) = π x λ + φ 0. (3.34) We see that there are two contributions to the phase difference: The path-length difference, x = x x 1, is the distance between the sources. It is the extra distance traveled by wave on the way to the point where the two waves are combined. The inherent phase difference, φ 0 = φ 0 φ 10, is the actual difference between the phase constants of the sources. Condition for being in phase, i.e. where the crest crest and trough trough align: is that φ = 0, π, 4π,. Therefore, we have the following condition for maximum constructive interference: φ = π x λ + φ 0 = nπ rad, where n = 0, 1,, 3,. (3.35) So for identical sources, in which φ 0 = 0, maximum constructive interference occurs when x = nλ, i.e. 117

2 Two identical sources produce maximum constructive interference when the path-length difference is an integer number of wavelengths. So in figure 68 55, we have x = λ i.e. traveled by wave. this is the extra distance Speaker moves forward exactly one wavelength during a time interval of one period. Speaker 1 emits a crest exactly as a crest of wave passes by. The two waves are thus in phase, and interfere constructively to give an amplitude of a. Condition for perfect constructive interference, i.e. where there is crest trough alignment: is when two waves are out-of-phase, i.e. φ = π, 3π, 5π or any odd multiple of π. Therefore, we have the following condition for perfect destructive interference: φ = π x λ + φ 0 ( = n + 1 ) π rad, where n = 0, 1,, 3,. (3.36) So for identical sources, in which φ 0 = 0 rad, perfect destructive interference occurs when x = (n + 1/)λ, i.e. Two identical sources produce perfect destructive interference when the path-length difference is a half-integer number of wavelengths. 55 Knight, Figure 1.19, page

3 Figure 68: Two identical sources one wvelength apart. So two waves can be out-of-phase because: The sources are located at different positions (figure ). The sources themselves are out-of-phase (figure ). A combination of the sources being located at different positions and out-of-phase (figure ) Mathematics of Interference 56 Knight, Figure 1.0(a), page Knight, Figure 1.0(b), page Knight, Figure 1.0(c), page

4 Figure 69: Destructive interference: waves out-of-phase. Figure 70: Destructive interference: sources out-of-phase. Looking more closely at the superposition of two traveling waves, we can use (3.30) and (3.31) to calculate the net displacement of the medium via the principle of superposition: 10

5 Figure 71: Destructive interference: waves and sources out-of-phase. u = u 1 + u = a sin (kx 1 ωt + φ 10 ) + a sin (kx ωt + φ 0 ). (3.37) A useful trigonometric identity is and so in (3.37), [ ] [ ] α β α + β sin α + sin β = cos sin, (3.38) sin (kx ωt + φ 0 ) + sin (kx 1 ωt + φ 10 ) { } (kx ωt + φ 0 ) (kx 1 ωt + φ 10 ) = cos { } (kx ωt + φ 0 ) + (kx 1 ωt + φ 10 ) sin. (3.39) However, from (3.34), we have 11

6 φ = φ 0 φ 10 = (kx ωt + φ 0 ) (kx ωt + φ 10 ), (3.40) and also in (3.39), (kx ωt + φ 0 ) + (kx 1 ωt + φ 10 ) = k (x + x 1 ) ωt + (φ 10 + φ 0 ). (3.41) Consequently in the right-hand side of (3.39) may be rewritten as ( ) [ φ k (x1 + x ) cos sin ωt + (φ ] 10 + φ 0 ). (3.4) Defining the average distance x avg to the two sources, and the average phase constant (φ 0 ) avg to be x avg = x 1 + x, (φ 0 ) avg = φ 10 + φ 0 then the net displacement in (3.37) may be expressed as, (3.43) ( ) φ ] u (x, t) = a cos sin [kx avg ωt + (φ 0 ) avg. (3.44) From (3.44), The sine term indicates the superposed wave is still a traveling wave, moving along the +x-axis with the same wavelength λ = π/k and frequency ν = ω/π as the component waves. Whereas the component waves both had an amplitude a, the superposed wave has an amplitude A given by A = a cos So A depends upon the phase difference φ: ( ) φ. (3.45) 1

7 Maximum amplitude: A = a when cos( φ/) = ±1, i.e. when φ = nπ, where n = 0, 1,, 3,. (3.46) Minimum amplitude: A = 0 when cos( φ/) = 0, i.e. when ( φ = n + 1 ) π, where n = 0, 1,, 3,. (3.47) We notice that conditions (3.46) and (3.47) are identical to (3.35) and (3.36), respectively, for constructive and destructive interference. So initially we found (3.35) and (3.36) by considering crest/trough alignments. The expressions were confirmed in (3.46) and (3.47) mathematically. Of course, two traveling waves need not be in-phase are out-of-phase. What happens for an arbitrary phase difference? Equation (3.45) allows us to calculate the superposed amplitude for any phase difference. For example, figure 7 59 shows the superposition for different values of φ. It can be confirmed that φ = 40 A = a cos (0 ) = 1.88a φ = 90 A = a cos (45 ) = 1.41a φ = 160 A = a cos (80 ) = 0.35a Example: two loudspeakers in a 0 C room emit 650 Hz sound waves along the x-axis. Supposing that the loudspeakers are in phase, calculate the smallest 59 Knight, Figure 1., page

8 Figure 7: The interference of two waves for three different phase differences. 14

9 distance between the speakers for which the interference between the waves is destructive. Supposing that the loudspeakers are out-of-phase, what is the smallest distance for which the interference is constructive? Solution: o interference between the two waves depends upon their phase difference φ. In the first part, if speakers are in phase, φ = 0 rad. Letting the path-length difference be d, and using n = 0 for the smallest d and the condition for destructive interference, if then φ = π x ( λ + φ 0 = n + 1 ) rad, where n = 0, 1,,, and so π x λ + φ 0 = π rad π d λ + 0 = π, d = λ = 1 v ν = 1 ( ) 343 = 0.64 m. 650 Similarly for the second part, when the speakers are out-of-phase, φ 0 = π rad. Using n = 1 for the smallest d and the condition for constructive interference, if then φ = π x λ + φ 0 = nπ, where n = 0, 1,, 3,, (3.48) π d λ + π = π d = λ = 1 v ν = 1 ( ) 343 = 0.64 m

10 4 Ray Optics 4.1 Introductory Concepts We are aware that light waves appear to travel in straight lines from observer things like laser beams and shadows. This is the basis of the so-called ray model of light: This is an oversimplification of reality, but it is very useful in many situations. In particular, the ray model is valid as long as the apertures through which light passes (i.e. lenses, mirrors, holes, etc.) are very large compared to the wavelength of light (e.g. for apertures > 1 mm in diameter). We define a light ray as a line in the direction along which light energy is flowing. Any narrow beam of light is, in reality, a collection of light rays. We can think of a single light ray as the limiting case of a laser beam whose diameter approaches zero. Properties of light rays: Light rays travel in straight lines. Light rays do not interact with each other two rays can cross without either being affected in any way. 16