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A silicon optical fibre with a core diameter large enough has a core refractive index of 1.50 and acladding refrcative index 1.47.Determine(i)the critical angle at the core cladding interface.(ii)the numerical aperture for the fibre.(iii)the acceptanceangle in air for the fibre. |
| Answer» Solution :(i) Here `mu_(1)=1.50,mu_(2)=1.47`, `mu_(0)=1`, CRITICAL angle `theta_(c)` at the core - cladding interface is GIVEN by, `theta_(c)= sin ^(-1)((1.47)/(1.50))=78.5^(0)`(ii) Here `mu_(1)=1.50,mu_(2)=1.47`, `mu_(0)=1`,Numerical Aperture, NA =`(mu_(1)^(2)-mu_(2)^(2))^((1)/(2))` =`[(1.50)^(2)-(1.47)^(2)]^((1)/(2))=(2.25-2.16)^((1)/(2))=0.30`(iii) Here `mu_(1)=1.50,mu_(2)=1.47`, `mu_(0)=1`, Acceptance angle `theta_(a) =sin^(-1)(NA)`, =`sin^(-1) (0.30) =17.4^(0)`(iii) Here `mu_(1)=1.50,mu_(2)=1.47`, `mu_(0)=1`, Acceptance angle `theta_(a) =sin^(-1)(NA)`, =`sin^(-1) (0.30) =17.4^(0)` | |