Physics 012 1st Edition Lecture 22 Outline of Last Lecture I. 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).II. Reflectiona. specular reflectioni. reflection off of a smooth surfaceii. 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.III. Image FormationThese 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.a. 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 = R/2Outline of Current Lecture IV. Curved Mirrorsa. When light passes through the center of curvature C, it will reflect back along thesame path it took to the mirror, but in the opposite direction.b. When light passes through the focal point F, it will reflect back parallel to the principal axis.c. When light reflects off of the point where the mirror and the principal axis meet, it reflects back at the same angle θ relative to the principal axis as when it was it was first incident upon the mirror.d. Mirror equation: 1/f = 1/ do + 1/dii. f = distance along principal axis from mirror to focal point F1. f = ½ radius (distance along principal axis to center of curvature)2. object at focal point will appear infinitely far awayii. do = distance along principal axis from mirror to objectiii. di = distance along principal axis from mirror to imagee. ho/do = - hi/di i. ho = height of objectii. hi = height of imagef. magnification = m = hi/ho = -di/doV. Problem: Concave Mirrorsa. If f = 10 cm and do = 25 cm, find di and m.i. 1/di = 1/f – 1/do = 1/10 – 1/25 = 3/50 cm-1ii. di = 1/(3/50) = 50/3 cmiii. m = - di/do = -(50/3)/25 = -2/3 xiv. The image will be real.b. If f = 10 cm and do = 5 cm, find di and m.i. 1/di = 1/f – 1/do = 1/10 – 1/5 = -1/10 cm-1ii. di = 1/(1/10) = -10 cmiii. m = - di/do = -(-10)/5 = 2 xiv. The image will be virtual.VI. Problem: Convex Mirrorsa. ho = 3 cm, do = 20 cm, and f = - 8 cm, find di and hi.i. 1/di = 1/f – 1/do = - 1/8 – 1/20 = - 7/40 cm-1ii. di = 1/(-7/40) = -40/7 cmiii. hi = - (hodi)/do = - (3 * -40/7) / 20 = 6/7 cmiv. The image will be
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