if the acceptance cone for the given fiber is 68.16 calculate the maximum entrance angle and numerical aperture if the cladding glass has a refractive index of a 1.52 calculate the refractive index of cone
NA=nsinα=1.41,NA=n\sin \alpha=1.41,NA=nsinα=1.41,
αmax=αNA=80.94°,\alpha_{max}=\alpha \sqrt{ NA}=80.94°,αmax=αNA=80.94°,
n′=NA=1.41.n'=NA=1.41.n′=NA=1.41.
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