Science, asked by madhu4147, 11 months ago

optical fiber can be satisfactorily operated if, refractive index of​

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Answered by cosmiccreed
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Answer:

Applying Snell's law, the complement of the critical angle is 0.085 radian (NA 0.12), so predicted minimum bend radius with an 8.2 micron core diameter would be around 0.1 mm. In practice, this fibre shows significant attenuation for bend radii of a few millimetres.

Geometric optics predictions are independent of wavelength, and suggest that smaller core radii will tolerate tighter bends - counter to practical experience, and the predictions of a more complete analysis.

A wave analysis shows that the bend loss has an exponential dependence on the ratio of bend radius to core radius and on the core-cladding index difference. There is a strong wavelength dependence, with bend loss increasing with wavelength.

Small core-clad index differences are usable if the fibre is operated at high normalised frequency (V-number). For single wavelength operation, the fibre should be designed with a single mode cut-off wavelength close to (or possibly longer than) the operating wavelength. If the core-clad refractive index is reduced, the core diameter should be increased to maintain or increase the V-number and cut-off wavelength.

If microbending loss is a concern, good cabling practice will help, as will use of a large silica cladding diameter such as 250 microns, rather than the 125 micron commonly used for standard single mode fibres.

If the requirement is a low value for the absolute refractive index, then a low cladding index will allow the core index to be reduced while still maintaining an index difference which gives acceptable bend performance. Doping a silica cladding with fluorine and/or boron oxide will help, though boron does degrade the attenuation at 1300 nm and longer wavelengths.

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