Lect02 - Lecture 2: Interference P y S3 d S2 S1 2 Incident...

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Lecture 2: Interference S 3 S 2 P Incident wave (wavelength λ ) y L d δ S 1 2 Overview: z Interference of Sound waves z Two-Slit Interference of Light z Phasors z Multiple-Slit Interference
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Interference of Waves (from last lecture): z When two waves are present at the same point in space and time they lead to interference. For the single ω case z Add amplitudes (e.g., pressures or electric fields). z What we observe however is Intensity (absorbed power). I = A 2 Stereo speakers: Listener: y 2 = A 1 cos(kx - ω t + φ ) y 1 = A 1 cos(kx - ω t) 12 1 y +y = 2A cos( / 2) cos(kx t / 2) φ −ω +φ A In this lecture we confine ourselves to waves with the same wavelengths. φ= 0:waves add “in phase” (“constructive”) Æ I = |2 A 1 | 2 = 4|A 1 | 2 = 4I 1 φ = π :waves add “out of phase” (“destructive”) Æ I = |2 A 1 *0|2 = 0
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Spatial Interference The relative phase of (two or more) waves also depends on the relative distances to the sources: The two waves at this point are “out of phase”. Their phase difference φ depends on the path difference δ≡ r 2 -r 1 relative to wavelength λ . r 2 r 1 φ = 2π(δ/λ) = 360 ˚ (δ/λ) λ δ = π φ 2 = “ # wavelengths out of phase” (for in-phase sources) δφ Ι 0 λ/4 λ/2 λ Path difference Phase difference A = 2A 1 cos( φ /2)
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Spatial Interference The relative phase of (two or more) waves also depends on the relative distances to the sources: δφ Ι 04 Ι 1 λ/4 1 λ/2 0 λ4 Ι 1 002 Α 1 λ/4 π/2 1 π 0 λ2 π2 Α 1 Path difference Phase difference The two waves at this point are “out of phase”. Their phase difference φ depends on the path difference δ≡ r 2 -r 1 relative to wavelength λ . r 2 r 1 A = 2A 1 cos( φ /2) φ = 2π(δ/λ) = 360 ˚ (δ/λ) λ δ = π φ 2 = “ # wavelengths out of phase” (for in-phase sources) Here we use equal intensities.
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Amplitude vs. Intensity (for 2 interfering waves) Amplitude vs. Intensity (for 2 interfering waves) cos( cos( φ /2) /2) cos cos 2 ( φ /2) Plot here as a function of φ. For equal intensities. 0 λ φ A = 2A 1 cos( φ /2) I = 4A 1 2 cos 2 ( φ /2) φ 0 10π δ Constructive Interference Destructive Interference 2A 1 4A 1 2 What is the spatial average intensity? I ave = 4I 1 *0.5 = 2I 1
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Sound wave example: Sound velocity: v = 330 m/s Each speaker alone produces intensity I 1 = 1 W/m 2 at the listener, and f = 900 Hz. Drive the speakers in phase. Compute the intensity I at the listener: r 2 r 1 3 m 4 m φ = 2π(δ/λ) with δ = r 2 –r 1 λ δ = π φ 2 www.falstad.com/interference Procedure: 1) Compute path-length difference: δ = r 2 -r 1 = 2) Compute wavelength: λ = 3) Compute phase difference (in degrees): φ = 4) Write a formula for the resultant amplitude: 5) Compute the resultant intensity, I =
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Sound wave example: Procedure: 1) Compute path-length difference: δ = r 2 -r 1 = 1 m 2) Compute wavelength: λ = v/f = (330 m/s)/(900 Hz) = 0.367 m 3) Compute phase difference (in degrees): φ = 360 °(δ / λ) = 360 ° (1/0.367) = 981 ° 4) Write a formula for the resultant amplitude: A = 2A 1 cos( φ /2) , A 1 = I 1 5) Compute the resultant intensity, I = 4 I 1 cos 2 ( φ /2) = 4 (1 W/m 2 ) (0.655) 2 = 1.72W/m 2 Sound velocity: v = 330 m/s r 2 r 1 3 m 4 m φ = 2π(δ/λ) with δ = r 2 –r
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Lect02 - Lecture 2: Interference P y S3 d S2 S1 2 Incident...

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