Physics 012 1st Edition Lecture 21 Outline of Last Lecture I. Problem 28 from ch. 22a. coil: N = 105, r = 4x10-2, R = 0.48 Ωb. ΔB/Δt = 0.783 T/sc. Bcenter = (Nμ0I)/2rd. ΦB = ΣBΔAcosθ = Bπr2e. εind = -Nπr2 (ΔB/Δt)f. Iind = |εind|/R = [Nπr2 (ΔB/Δt)]/Rg. B = (Nμ0I)/2r = [(N2πrμ0)/2R][ΔB/Δt] = II. Problem 53 from ch. 24a. A = 135m2, Pavg = 1.2 x 104 Wb. Savg = Pavg/A = E2max/2μ0cc. Erms = Emax/√(2)d. Pavg/A = 2E2rms/2μ0c = E2rms/μ0ce. Erms = [(Pavgμ0c)/A]1/2 f. Brms = Erms/cIII. Problem 40 from ch.21a. ΣFx = FB – 2Tsinφ = 0b. ΣFy = 2Tcosφ –mg = 0c. φ = tan-1[(ILB)/(mg)]Outline of Current Lecture IV. Lighta. particle-wave dualityb. from a point source:i. far from point, waves become essentially planar:ii. If λ << d, waves pass straight through a small opening.iii. If λ ≈ d, waves spread out past opening.iv. If λ >> d, waves spread out a lot past opening (diffraction).V. Reflectiona. specular reflectioni. reflection off of a smooth surfaceThese notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.ii. parallel incident light rays will be parallel after reflectionb. diffuse reflectioni. reflection off of a rough surfaceii. parallel incident light rays will not be parallel after reflectionc. Law of Reflection: When light is reflected off a surface, the incident angle (θi) is equal to the reflected angle (θr). i. Incident ray, reflected ray, and normal line will all be in the same plane.ii. The path of the light is reversible.VI. Image Formationa. virtual image: light rays only appear to emanate from the image location, the raysdo not actually diverge from this locationb. real image: light rays diverge from the image location; this kind of image can be displayed on a screenc. hi = image heightd. ho = object heighte. magnification = M = hi/hof. do = object distance (from mirror)g. di = image distance (from mirror)h. F = focal pointi. R = radius of curvaturej. C = center of curvaturek. for spherical mirrors: f =
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