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(C) For total internal reflection to occur,the angle of incidence $	heta$ at the core-cladding interface must be greater than the critical angle $	h

Vedclass · Accepted Answer

$0.78$

Vedclass · Answer

(C) For total internal reflection to occur,the angle of incidence $	heta$ at the core-cladding interface must be greater than the critical angle $	heta_{c}$.Thus,$\sin 	heta > \sin 	heta_{c}$.Given the refractive index of the core $\mu_{1} = 1.6$ and cladding $\mu_{2} = 1.5$,the critical angle is $\sin 	heta_{c} = \frac{\mu_{2}}{\mu_{1}} = \frac{1.5}{1.6} = 0.9375$.From the geometry of the bent fiber,if $r$ is the inner radius and $d = 0.05 \, mm$ is the core diameter,the angle of incidence $	heta$ at the outer interface is related by $\sin 	heta = \frac{r}{r+d}$.Therefore,$\frac{r}{r+d} > 0.9375$.$r > 0.9375(r + 0.05)$.$r > 0.9375r + 0.046875$.$0.0625r > 0.046875$.$r > \frac{0.046875}{0.0625} = 0.75 \, mm$.The closest value among the options is $0.78 \, mm$.

What is the least radius through which an optical fiber of core diameter $0.05 \, mm$ may be bent (as shown in figure) without serious loss of light (in $, mm$)? The refractive index of the core is $1.6$ and that of cladding is $1.5$.

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