Question

3.) An optical fiber is covered by a layer of glass, to protect its surface. The index of refraction of the core is n1 = 1.8, and that of the protective layer is n2 = 1.5. The diameter of the core is d= 0.2mm. a. Calculate the critical incident angle at the interface between the core and the protective layer, at which total reflection occurs! (Give your result in degrees!) b. What is the minimal radius to which the fiber can be bent, without light leaving the core? (Assume that light enters the optical fiber parallel to its axis!) c. Calculate the minimal radius of curvature if the diameter of the core is 10

Answer 1

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ni=1.8 n2=1.5 d= 0.2mm Oc= critical = bin angle 1.8 O = 56.443° And sin 12 Sino-R (Rtd) b) Por light to not leave the core, oz, ac Sino > Sina 3 So R 21 Sin56.443 R> 0-8333(R+0-2) Rt0.2 R7, 0.9999 mm or (R7 Imm Ramin = 1mm) And minimum Radius of fiber C) if d = 10 mm. Rinin = 49.998 mm And then, >0.83333 Rtio