The figure below shows an optical fiber in which a central plastic core of index
ID: 2156544 • Letter: T
Question
The figure below shows an optical fiber in which a central plastic core of index of refraction n1 = 1.58 is surrounded by a plastic sheath of index of refraction n2 = 1.50. Light can travel along different paths within the central core, leading to different travel times through the fiber. This causes an initially short pulse of light to spread as it travels along the fiber, resulting in information loss. Consider light that travels directly along the central axis of the fiber and light that is repeatedly reflected at the critical angle along the core-sheath interface, reflecting from side to side as it travels down the central core. If the fiber length is 350 m, what is the difference in the travel times along these two routes?
Explanation / Answer
Consider that the reflected path length is equal to the length of the hypotenuse of a right triangle formed by the centerline ray, the extended reflecting ray, and a line perpendicular to the fiber. the critical angle: ?=sin-1(n2/n1)=sin-1(1.5/1.58)=71.690 reflected path length = fiber length / sin(critical angle) =350/sin(71.69)=368.66m Path length difference=18.66m n of the core = 1.58, then the speed of light in the fiber is c/1.58 c = 299792458 m/s v = c / 1.58 v = 189742062 m/s v = distance / time time = distance / v time = (18.66 s) / (189742062 m/s) time = 9.834x10-8 seconds
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.