(Ill) Consider a ray of sunlight incident from air on a

The full step-by-step solution to problem: 32.88 from chapter: 32 was answered by , our top Physics solution expert on 11/10/17, 05:57PM. This textbook survival guide was created for the textbook: Physics for Scientists & Engineers with Modern Physics, edition: 4. Since the solution to 32.88 from 32 chapter was answered, more than 254 students have viewed the full step-by-step answer. Physics for Scientists & Engineers with Modern Physics was written by and is associated to the ISBN: 9780131495081. This full solution covers the following key subjects: Incident, Angles, angle, Rays, ray. This expansive textbook survival guide covers 44 chapters, and 3904 solutions. The answer to “(Ill) Consider a ray of sunlight incident from air on a sphericalraindrop of radius r and index of refraction n. Defining0 to be its incident angle, the ray then follows the pathshown in Fig. 32-67, exiting the drop at a scattering angle <f>compared with its original incoming direction, (a) Show that(f) = 180 + 20 - 4 siiT1 (sin 0/n). (b) The parallel rays ofsunlight illuminate a raindrop with rays of all possibleincident angles from 0 to 90. Plot 4> vs. 0 in the range0 < 0 < 90, in 0.5 steps, assuming n = 1.33 as is appropriatefor water at visible-light wavelengths. (c) From yourplot, you should find that a fairly large fraction of theincident angles have nearly the same scattering angle.Approximately what fraction of the possible incident anglesis within roughly 1 of (f> = 139? [This subset of incidentrays is what creates therainbow. Wavelengthdependentvariations inn cause the rainbow toform at slightly different<f> for the various visiblecolors.]FIGURE 32-67” is broken down into a number of easy to follow steps, and 159 words.