Study Guide for Final Exam - Foundations of Physics I | PHYS 2306, Study notes of Physics

Download Study Guide for Final Exam - Foundations of Physics I | PHYS 2306 and more Study notes Physics in PDF only on Docsity! Phys 2306 rev—F-YY-f07 1 Review for Final Exam Ch 29.3 Lenz’s Law Ch 30 Inductance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1 Ch 31 Alternating Currents (AC), RLC Circuits and Forced Oscillations . . 4 Can skip: Sect. 31.6 (transformers) Ch 32 Electromagnetic waves; Energy, Momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . .7 Ch 33 The Nature and Propagation of Light . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 Can skip: circular and elliptical polarization; scattering of light Ch 35 Interference, Double-Slit, Young’s Experiment . . . . . . . . . . . . . . . . . . . . . . . 10 Can skip: Sect. 35.5 (Michelson ...) Ch 36 Diffraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .11 Can skip: Sections 36.5, 6, 8 Ch 30. Inductance ¥ Inductors and inductance, self-induction N Φ ∝ i ⇒ N Φ = L i L def = N Φ i , 1henry = 1H = 1T ·m2/A Equivalent definition: self-induced emf EL = −L di dt the sign gives direction (from EL = −N dΦ/dt) (for solenoid L = μ0n2Al). Direction of induced emf can also be obtained from Lenz’s Law. ¥ Mutual induction E2 = −N2dΦB2 dt , N2ΦB2 =M21i1 E2 = −M21 di1 dt , E1 = −M12 di2 dt , M21 =M12 =M Phys 2306 rev—F-YY-f07 2 ¥ RL circuits. Ri = E − EL ⇒ L di dt +R i = E i(0) = 0 ⇒ i = E R ³ 1− e−t/τ ´ , τ = τL = L R ; current decay in an RL circuit i = I0e −t/τL from L di dt +R i = 0 • L stores kinetic energy while C stores potential energy. ¥ Energy stored in the electric and magnetic fields UE = q2 2C , UB = 1 2 L i2 • Energy densities uE = 1 2 ε0E 2, uB = UB Al = B2 2μ0 uE obtained by considering parallel plate condenser; uB - infinitely long solenoid. ¥ RL circuits. Results can be taken from our analysis of RC circuits, through the following transcription: RC ←→ RL Rdqdt + 1 C q = E L didt +R i = E R L 1/C R q = CE ¡1− e−t/τ¢ , τC = RC i = ER ¡1− e−t/τ¢ , τL = L/R ¥ LC circuits. Analogy with simple harmonic motion: electrical ←→ mechanical LC − oscillator ←→ block-spring oscillator L d2q dt2 + 1 C q = 0 ←→ md 2x dt2 + k x = 0 Phys 2306 rev—F-YY-f07 5 Circuit equation L d2q dt2 +R dq dt + 1 C q = E where E = v is sinusoidal: v = V cos (ωt+ ϕ) vR + vL + vC = v (1) ¥ In our case, if the phase constant ϕ is chosen in an appropriate way then (underdamped case) i = transient partz }| { I 0e− R 2L t cos ¡ ω0t+ ϕ0 ¢ + steady state partz }| { I cosωt after a while' I cosωt (steady state solution). I 0 and ϕ0 are of no concern here; they can be computed from initial conditions. We are interested here the case of i = I cosωt (steady state solution), • Main problem: given that E = V cos (ωt+ ϕ), • Find amplitude I of the current i • Find the phase difference ϕ between applied emf E and the current i • Analyze how physical quantities depend on V and ϕ ¥ Results: the amplitude I of the steady state oscillations is I = Vq R2 + (XL −XC)2 = V Z (2) Phys 2306 rev—F-YY-f07 6 where XL = ωL is the inductive reactance, XC = 1/(ωC) is the capacitive reactance, Z = q R2 + (XL −XC)2 is the impedance, and the phase constant ϕ is given by tanϕ = XL −XC R or, cosϕ = R Z . (3) • XL > XC ⇒ ϕ > 0, L dominates C and i lags behind E • XL < XC ⇒ ϕ < 0, L is dominated by C and i leads E Average (over time) energy dissipated in the resistor is Pav = 1 2 V I cosϕ . (4) Given V and I, maximum is obtained for cosϕ = 1, i.e., for XL = XC (resonance). • (2) and (3) are obtained as follows: vR, vL and vC are computed in terms of i and substituted into the circuit equation (1) yielding VR cosωt+ VL cos (ωt+ π/2) + VC cos (ωt− π/2) = V cos(ωt+ ϕ) , (5) where VR = I R, VL = I XL and VC = I XC . Then equation (5) is solved using phasor diagram. ¥ Power and average power. Instantaneous rate of energy dissipation in the resistor P = i2R = (I cosωt)2R = I2R cos2 ωt . Phys 2306 rev—F-YY-f07 7 This yields for average power Pav = 1 2 I2R = I2rmsR , and then (4) and Pav = VrmsIrms cosϕ . The subscript rms stands for root mean square; here Vrms = V√ 2 , Irms = I√ 2 . Given Vrms and Irms, maximum is obtained for cosϕ = 1, i.e., for XL = XC ¥ Resonance I = Vq R2 + (XL −XC)2 , maximal when XL −XC ⇒ ω = 1√ LC (6) t 543210 6 5 4 3 2 1 0 I as a function of ω;R is doubled from one graph to the next Ch 32. Electromagnetic waves; Energy, Momentum ¥Maxwell’s Equations and EM Waves • Sect. 29.7 Displacement current (iD) and Maxwell-Ampère lawI B·dl = μ0iC + μ0iD where iD = ²0 dΦE dt Phys 2306 rev—F-YY-f07 10 • Ordinary beam of light is not polarized Malus’ law : I = I0 cos2 θ • Polarization by reflection; Brewster’s Law • The refracted beam is only partially polarized, unless angle of incidence (and re- flection) is equal to polarizing (θp) (sometimes called Brewster’s angle and denoted θB). Ray reflected at angle θp is completely polarized parallel to the surface and perpendicular to the refracted ray. θp = tan −1 nb na ¥ Light as a wave: Huygens’ Principle; Huygens’ Wavelets. • Huygens principle: every point of a wave-front serve as point sources of spherical secondary wavelets. At time t the new position of the wave front will be that of a surface tangent to these secondary wavelets. (Simplification (approximation) of Maxwell’s theory: scalars instead of vectors) • Wave fronts, and rays • wavelength and index of refraction λn = λ n Ch 35. Interference, Double-Slit, Young’s Experiment ¥ Superposition and Interference: plane waves: two coherent sources Recall. Two waves of equal amplitude, the result of their interference depends on their phase difference (ϕ) or, equivalently, path difference (PD, ∆L) ∆ϕ = 2π ∆L λ , PD = ∆L = ∆ϕ 2π λ Interference Phase Difference ∆ϕ P.D. ∆L Constructive integer × 2π m · 2π m · λ Destructive odd multiple of π (2m+ 1) · π (2m+ 1) · λ2 Phys 2306 rev—F-YY-f07 11 ¥ Young’s double-slit experiment. ((1801) - decisive experiment in favor of wave theory of Huygens against corpuscular theory of Newton). Here ∆L = r2 − r1 = d sin θ small θ approx.∼= d · tan θ = d · y L ⇒ m · λ = d · y L ⇒ ybright = λL d m ¥ Intensity distribution in the interference pattern. Average (over the period) intensity Iav = I0 cos 2 ∆ϕ 2 = I0 cos 2 π d sin θ λ small θ∼= I0 cos2 π d λL y where I0 is the intensity at the center and ∆ϕ = 2π d sin θ λ ¥ Thin films interference λn = λ n Phase difference on passing path L in two media: ∆ϕ = 2π L λn − 2πL λ = 2π L λ (n− 1) • Phase Change Due to Reflection. (as in the case of mechanical waves): phase changes by 180◦ on reflection of light (EM wave) from material with higher index of refraction (optically more dense material); no phase change in reflection from material with lower index of refraction. • “Nonreflective” coating Ch 36 Diffraction Please go to my class notes–posted on Bb, for more details. ¥ Diffraction and coherence. • Diffraction: result of interference of light from many coherent sources • Fresnel and Fraunhofer: near and far field diffraction. Phys 2306 rev—F-YY-f07 12 ¥ Fraunhofer’s Diffraction: slit case. Single slit of width a: minima at angle θ such that sin θ = m · λ a , m = ±1,±2, . . . • Intensity distribution I(θ) = Im ∙ sinβ/2 β/2 ¸2 = I0 ∙ sin (π a sin θ/λ) π a sin θ/λ ¸2 , β = 2π λ a sin θ where β is the phase difference between the top and bottom rays from the slit x 1.510.50-0.5-1-1.5 1 0.8 0.6 0.4 0.2 0 Relative intensity I/Im as a function of θ for π a/λ = 10 ¥ Fraunhofer’s Diffraction: circular apertures • Rayleigh’s criterion for resolution (barely): The angular separation θ of two light sources should be such that the central maximum of the diffraction pattern of one source is centered on the first minimum of the other: sin θ = 1.22 λ d ; d is the diameter of the aperture For a slit, the first dark fringe is at sin θ = λ a ; a is the width of the slit