Number of reflections in an Optical fibre

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The discussion centers on calculating the number of reflections in a 2m 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 calculated number of reflections is approximately 36,368.48, based on the formula L/(d tan θ). There is a debate about the relevance of the refractive index, with some arguing it is only necessary for ensuring total internal reflection. The angle of incidence is confirmed to be in degrees. The calculation and assumptions regarding the refractive index and angle are critical for accurate results.
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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
 
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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?
 
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yes...3.43 degrees
 
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