Physics 012 1st Edition Lecture 24 Outline of Last Lecture I. 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/nII. Snell’s Lawa. n1sinθ1 = n2sinθ2i. n1 = index of refraction for medium 1ii. θ1 = incident angle (angle between incoming light ray and normal line to plane of medium 1 surface)iii. n2 = index of refraction for medium 2iv. θ2 = refracted angle (angle between outgoing light ray and normal line to plane of medium 2 surface)b. If n1 > n2, then θ1 > θ2.c. If n1 < n2, then θ1 < θ2.III. 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.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.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 λn in this material.i. λn = λair/n = 550/1.47 = 374 nm (UV light)IV. 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 degreesOutline of Current Lecture V. Total Internal Reflectiona. As θ1 increases, θ2 also increases.b. At the critical angle, refracted light will follow the surface between the two media (θrefracted = 90 degrees).c. Any angle of incidence greater than the critical angle will result in total internal reflection (all light reflected, none entering second medium).i. Only happens when light beam goes from higher index of refraction to lower index of refraction (n1 > n2).ii. θc = sin-1(n2/n1)d. Problem: A beam of light is incident on the circular face of a glass cylinder at an angle of θ1. Find the maximum value of θ1 to get total internal reflection in the glass cylinder.i. θ1 will be at max value for T.I.R. when θ3 is at critical angle.ii. ngsinθ3 = nasin(90) = naiii. θc = θ3 = sin-1(na/ng) = 47.3 degreesiv. θ2 = 90 – θ3 = 42.6 degreesv. nasinθ1 = ngsinθ1vi. θ1 = sin-1([ngsinθ2]/na) = 67.1 degreesVI. Brewster’s Angle: angle at which light becomes completely polarized parallel to the reflecting surface (electric field will oscillate parallel to reflecting surface).a. At this angle, the reflected ray and refracted ray will be perpendicular.b. tanθB = n2/n1VII. Dispersiona. The index of refraction for a given material varies slightly with wave length of light passing through that material. The index of refraction increases with smallerwavelengths.b. Ex. Glass Prismi. Different colored light within white light will have different angles of refraction, causing colors to be separated when exiting prism.VIII. Thin
View Full Document