Number of reflections in an Optical fibre

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SUMMARY

The discussion centers on calculating the number of reflections in a 2-meter long optical fiber cable with a core diameter of 60μm and a refractive index of 1.48, given an angle of incidence of 3.43°. The calculation provided, which results in approximately 36,368 reflections, is derived using the formula No. of reflections = L/(d tan θ), where θ is determined from the refractive indices. The relevance of the refractive index is confirmed as necessary for achieving total internal reflection.

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  • Basic trigonometry, including sine and tangent functions
  • Knowledge of refractive index and its implications in optics
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Rahulr2k
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How many reflections will be made by a light entering a 2m long optical fibre cable. If the angle of incidence with the axis of the core is 3.43° . Given that the core diameter and refractive index are 60μm and 1.48 respectively

I got the answer as 36368.48 is this correct?

L=2m

d=60μm

sinθ= n(air)/n(fibre) = 0.67567

∴ θ= 42.50664°

tanθ= 0.9165
∴No. of reflections = L/(d tan θ )
= 36368.48
 
Last edited:
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Hi Rahulr2k. http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif

I would have thought that the refractive index is irrelevant, other than needing to be sufficient to ensure that there is total internal reflection?

Does that 3.43 figure have units of degrees?
 
Last edited by a moderator:
yes...3.43 degrees
 

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