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A step index fibre has a relative refractive index difference of 0.9%.Estimate the critical angle at the core -cladding interface,when the core index is 1.46.find the numerical aperture,if the source to fiber medium is air. |
| Answer» Solution :Here,`mu_(1)=1.46`,`(mu_(1)-mu_(2))/(mu_(1))=(0.9)/(100)=0.009`….or `1-(mu_(2))/(mu_(1))=0.009` or `(mu_(2))/(mu_(1))=1-0.009=0.991` CRITICAL angle, `theta _(C)=sin^(-1)((mu_(2))/(mu_(1)))=sin^(-1)0.991=82.3^(0)`,`mu_(2)=0.991xx mu_(1)=0.991xx1.46=1.45`NA=`sqrt(mu_(1)^(2) -mu_(2)^(2))=sqrt ((1.46)^(2)-(1.45)^(2))=0.17` | |