Question

JUST E THROUGH H

Problem 88. (20 pts) Consider a fiber like the one sketched below. Take the index of refraction of the core to be ni = 1.5 and the index of refraction of the cladding to be n2 = 1.46. A) What is the critical angle for the core/cladding interface? B) If the input angle from air is 6= 8° Calculate 01. C) If the input angle from air is fo=8° Calculate y. D) If the input angle from air is 0o = 8° will the ray be guided by the fiber with low loss? E) If the input angle from air is 0o = 45° Calculate 01. F) If the input angle from air is 0o = 45° Calculate p. G) If the input angle from air is 0o = 45° will the ray be guided by the fiber with low loss? H) What is the maximum input angle from air for a ray that will be guided with low loss by this fiber? 112

Answer 1

Mp=1.5 M₂ = 1.46 Mar1 Uling snelle law Main Singom, sino - 1 xsin 45º = 1.5 SIN G4 3 91= 2810] Ø =(90-01) - (20-28.1) = 61.80 . =61.80) (6) For low loves TIR(Total Internal Relation) must take place Using snell's law mi sin Ø - Masin 02 3 10bx sin 61.8" =1046 song, = 02-64.90 – No TIR since tinez </ 1. So, may will be not be guided will low loss by this fiber. (14) For TIR to occur, Oz, min=90° i musings Macin go° = $3514:46 = 76.740 ? 01 = 90-=13626° uling enelle low , IXLIN 80 = 105 Sin 04 = $1n00 = (05.549 13:26 3 00 = Bint(0.344) 1 80=20.10 Maximum input angle from air - for way to be guided with low - lose by this fiber