Page 396 - Schaum's Outline of Theory and Problems of Applied Physics
P. 396

CHAP. 30]                                 LIGHT                                       381



        TOTALINTERNALREFLECTION
        The phenomenon of total internal reflection can occur when light goes from a medium of high index of refraction
        to one of low index of refraction, for example, from glass or water to air. The angle of refraction in this situation is
        greater than the angle of incidence, and the refracted light ray is bent away from the normal, as in Fig. 30-10(a),
        at the interface between the two media. At the critical angle of incidence, the angle of refraction is 90 ◦
        [Fig. 30-10(b)], and at angles of incidence greater than this the refracted rays are reflected back into the original
        medium in accordance with the law of reflection [Fig. 30-10(c)]. If the critical angle is i c ,
                                                       n 2
                                                sin i c =
                                                       n 1

                n  > n 2
                 1

                         r
                                                r =
                 n 2                            90°
                 n 1
                                                                             i  r′   r′ = i
                      i                       i c





                                                 Fig. 30-10


        SOLVED PROBLEM 30.17
              Derive the formula sin i c = n 2 /n 1 .
                  At the critical angle i c the angle of refraction r is 90 , so sin r = sin 90 = 1 when i = i c . Substituting sin r = 1
                                                      ◦
                                                                    ◦
              into Snell’s law
                                                 n 1 sin i = n 2 sinr

                                                                 n 2
              yields                        n 1 sin i c = n 2  sin i c =
                                                                 n 1
        SOLVED PROBLEM 30.18
              Find the critical angle for light going from crown glass (n = 1.52) to air (n = 1.00) and for light going
              from crown glass to water (n = 1.33).
                  For light going from glass into air,
                                                1.00
                                          sin i c =  = 0.658   i c = 41 ◦
                                                1.52
              For light going from glass into water,
                                                1.33
                                          sin i c =  = 0.875   i c = 61 ◦
                                                1.52

        SOLVED PROBLEM 30.19
              Prisms are used in optical instruments instead of mirrors to change the direction of light beams by 90 ◦
              because total internal reflection better preserves the sharpness and brightness of light beams. What is the
              minimum index of refraction of the glass used in such prisms?
   391   392   393   394   395   396   397   398   399   400   401