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UVM PHYS 012 - Refraction of Light
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Physics 012 1st Edition Lecture 23Outline of Last Lecture I. 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/dii. ho = height of objectii. hi = height of imagef. magnification = m = hi/ho = -di/doII. 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.These 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.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.III. 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 virtual.Outline of Current Lecture IV. Light traveling through various media will have different speeds, all less than c, the speed of light through a vacuum.a. The frequency of light will remain the same as it travels through various media.b. c = λvacuumf and v = λnf (same frequencies)c. f = c/λvac = v/λnd. λn = (v/c) λvac ;λn will be less than λvace. index of refraction = n = c/v ; λn = λvac/nV. Snell’s Lawa. n1sinθ1 = n2sinθ2i. n1 = index of refraction for medium 1ii. θ1 = angle between light ray and plane of medium 1 surfaceiii. n2 = index of refraction for medium 2iv. θ2 = angle between light ray and plane of medium 2 surfaceb. If n1> n2, then θ1> θ2.c. If n1< n2, then θ1< θ2.VI. Problem: A beam of light with a wavelength of 550 nm (green) traveling in air is incident upon a transparent material. The incident angle is 40 degrees. The refracted beam is 26 degrees to the normal. Assume n = 1 for air.a. Find n for transparent material.i. n2 = sinθ1/sinθ2 = sin(40)/sin(60) = 1.47b. Find speed of light through this material.i. n = c/v ; v = c/n = 2.04 x 108 m/sc. Find λnin this material.i. λn= λair/n = 550/1.47 = 374 nm (UV light)VII. Problem: Liquid carbon disulfide (nc = 1.63) surrounds glass (ng = 1.52). a. Find θ4.i. ncsinθ1 = ngsinθ2ii. θ2 = sin-1([nc/ng]sinθ1) = 32.4 degreesiii. θ3 = 90 – θ2 = 57.6 degreesiv. ngsinθ3 = ncsinθ4v. θ4 = sin-1([ng/nc]sinθ3) = 51.9


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UVM PHYS 012 - Refraction of Light

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