dispersion

Name: Grade and Section: Score: School: _ Teacher: Subject: General Physics 2 LAS Writer: JEROL JAY C. ARANO Content Evaluator: EMMA T. SURITA MICHAEL DAVE M. NALAGON Lesson Topic: Dlsrslon Quarter 4 Wk. 2 LAS 3 Learning Targetls: 1. Explain dispersion by relating to Snell's Law. STEM-GP120PTlVb-1 6 References: Semay and Faugn, J.. 2006. Hot: Physics. 10801 N. MoPac Expressway, Building Austin, Texas 78759: Holt, Rinehart and Winston, pp.509-510. Dispersion An important property of the index of refraction is that its value in anything but a vacuum depends on the wavelength of light because the index of refraction is a function of wavelength, Snell's law indicates that incoming light of different wavelengths is bent at different angles as it moves into a refracting material. This phenomenon is called dispersion. The index of refraction decreases with increasing wavelength. For instance, blue light [A a 4? nm bends m re than re li ht es nm when assin into a refractin material. White light passed through a prism produces a visible spectrum. Dlspersinn the process of To understand how dispersion can affect light, consider what happens when light separating polychromatic strikes a prism, as in Figure 1. Because of dispersion, the blue component of the \"SM into its component incoming ray is bent more than the red component, and the rays that emerge from wave'engths' the second face of the prism fan out in a series of colors known as a visible spectrum. These colors, in order of decreasing wavelength, are red, orange, yellow, green, blue, and violet. Refraction is responsible for dispersion in rainbows and many other situations. The angle of refraction depends on the index of refraction, as we know from Snell's law. We know that the index Fl 1 wm h_ r m _ . . . . : its me , of refraction (n) depends on the medium. But for a given medium, (n) also ,,,':,'.,, \"gm 51:,\" ME, mg\": fear?\" depends on wavelength. the prism dlspenses the while light into its You can use Snell's law to find out the angle of refraction. \"3'5\"\" mecm'wmmnems- n1 sin 91: n2 sin 92 Plugging in the values, you get n1 sin Elc = n2 sin 90 Because sin 90" = 1, you have the following for the critical angle for total internal reflection: Fltnt Glass I'lz . - .- 11: Sln S = 9 = 5m 1 c n, c :1, Note that this equation requires that n2