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Unformatted text preview: 68. (a) The ﬁrst contribution to the overall deviation is at the ﬁrst refraction: δθ1 = θi − θr . The next
contribution(s) to the overall deviation is (are) the reﬂection(s). Noting that the angle between
the ray right before reﬂection and the axis normal to the back surface of the sphere is equal to θr ,
and recalling the law of reﬂection, we conclude that the angle by which the ray turns (comparing
the direction of propagation before and after [each] reﬂection) is δθr = 180◦ − 2θr . Thus, for k
reﬂections, we have δθ2 = kθr to account for these contributions. The ﬁnal contribution is the
refraction suﬀered by the ray upon leaving the sphere: δθ3 = θi − θr again. Therefore,
θ dev = δθ1 + δθ2 + δθ3 = 2 (θi − θr ) + k (180◦ − 2θr ) = k (180◦ ) + 2θi − 2(k + 1)θr .
(b) For k = 2 and n = 1.331 (given in problem 67), we search for the second-order rainbow angle
numerically. We ﬁnd that the θ dev minimum for red light is 230.37◦ , and this occurs at θi = 71.90◦ .
(c) Similarly, we ﬁnd that the second-order θ dev minimum for blue light (for which n = 1.343) is
233.48◦ , and this occurs at θi = 71.52◦ .
(d) The diﬀerence in θ dev in the previous two parts is 3.11◦ .
(e) Setting k = 3, we search for the third-order rainbow angle numerically. We ﬁnd that the θ dev
minimum for red light is 317.53◦ , and this occurs at θi = 76.88◦ .
(f) Similarly, we ﬁnd that the third-order θ dev minimum for blue light is 321.89◦ , and this occurs at
θi = 76.62◦ .
(g) The diﬀerence in θ dev in the previous two parts is 4.37◦ . ...
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- Fall '08