Week 7 Phys 5b Notes
Week 7 Phys 5b Notes PHYS 5B
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This 3 page Class Notes was uploaded by Shanee Dinay on Monday February 22, 2016. The Class Notes belongs to PHYS 5B at University of California - Santa Cruz taught by A.Steinacker in Fall 2015. Since its upload, it has received 10 views. For similar materials see Intro to Physics II in Physics 2 at University of California - Santa Cruz.
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Date Created: 02/22/16
Day 18 2/17/2016 Phys 5b Midterms Back anyone with a score above 45 is in the running for an A that does not mean that if you have a 30 in the exam, then you have a chance provided that you show improvement she is interested in detailed steps she will drop the 1c) problem points Phase, Waves ϕ ϕ 1 2 y(x, t) = 2acos( Δϕ )sin(kxagvt ϕ )avg Δϕ 2 A = | 2acos( 2 ) | Δϕ = 2nπ → you get a perfectly constructive interference Δϕ = 2π Δx + Δϕ λ o A = |2a| Δϕ = 2(n+1)π → perfect destructive A = 0 if Δϕ = 0 identical speakers sources o if Δϕ o Δx P.C.I 2πΔx = 2nπ Δx = 2n λ λ λ 2 P.D.I Δx = (2n + 1) 2 Example. Speakers Two speakers, one meter away. Δϕ o 2 f = 500 Hz a = 0.1mm Δϕ A = |2a cos( 2 )| Δx = x2− x 1= 1m λ = cs f λ = 343 m/ = .686 m 500 Hz Δϕ = λx + Δϕ =o10.73 rad 10.73 A = | 2 • 0.1mm • cos 2 | = 0.121mm Amplitude is larger than original amplitude (but not twice the original amplitude) so we have partial constructive interference. Destructive Interference when amplitude is between 0 and 1 Sound Waves, propagate away from source Two Sources Δϕ = 2λΔr + Δϕ o Δϕ o 0 2π 2π Δϕ A λ (r2A 1A λ (2λ 3λ ) = 2π perfect constructive interference Δϕ = 2π (r r) = 2π(2.5λ 3λ) = π A λ 2B 1B λ perfect destructive interference Δϕ A λπ(r2C 1C 2λ (3λ .5λ ) = 5π perfect destructive Where we have symmetry antinode such as where Δr = 0, Δr = λ, Δr = − λ, … we can draw nodal lines between the antinodal lines Light light as a particle Day 19 2/19/2016 Reading 34.3, 34.4, 35.7 Last Class Review: Narrow Beam of Light Wave Front: ← sources of spherical waves ← planar front Double Split Δϕ = Δr λ Δϕ Δϕ A = | 2a cos 2 | 2 = nπ PCI → bright fringes Δϕ = nπ Δϕ = Δr = 2nπ Δr = 2n n = 0, 1, 2, … 2 λ 2 PDI Dark Fringes Δϕ 2 =(2n+1) 2 Δϕ = (2n + 1)π = λπΔr λ Δr = (2n + 1)2 n = 1, 3, 5, … Angles of Bright Fringes λ θn= 2n 2d Location of Bright Fringes tanθn= yn yn Ltanθ n yn~ Lθ n= 2n λL L 2d Location of Dark λL ydark = n(n + 1) 2d Two consecutive fringes Δy = y y = 2(n+1) λL 2n λL n+1 n 2d 2d Δy = λL 2d Electromagnetic Waves 2 I = CA 2 2Δϕ 2 2Δϕ = C4a cos 2 4Ca cos 2 Δϕ A = | 2acos 2 | a → amplitude of individual wave a Δϕ = 2πΔr Δr = dsinθ λ =λπdsinθ 2 π dy I(y) = 2Loos ( λ L Imin0 Imax4I o
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