Physics · Ch 6 — Optics
Optical Fibre
Optical Fibre
An optical fibre consists of an inner part called the core, surrounded by an outer part called the cladding (or sleeving); the refractive index of the core material must be higher than that of the cladding for total internal reflection to work. A signal, carried as light, is launched into the core at an angle greater than the critical angle for the core-cladding boundary, so it undergoes repeated total internal reflection along the entire length of the fibre without ever refracting out through the cladding -- even while the fibre is gently bent, provided the bend is not so sharp that the TIR condition at the boundary is broken. This means the light signal travels the full length of the fibre with essentially no loss of intensity from leakage through the sides. An endoscope, used by doctors to see (and even perform minor procedures) inside a patient's body, is built from a …
What this figure shows. A zig-zagging light ray is shown travelling down the length of a thin optical fibre, bouncing repeatedly off the boundary between the fibre's central core (drawn with a higher refractive index) and its surrounding cladding layer (drawn with a lower refractive index). Because the light strikes this core-cladding boundary at an angle greater than the critical angle every single time, it undergoes total internal reflection at every bounce rather than escaping through the cladding, letting the signal travel the full length of the fibre -- even while the fib …
Worked out. A fibre has a core of refractive index 1.68 and a cladding of refractive index 1.44, sitting in air. Using the acceptance-angle formula i(a) = sin inverse[sqrt(n1^2 - n2^2)] with n1 = 1.68 and n2 = 1.44 gives sqrt(1.68^2 - 1.44^2) = sqrt(2.822 - 2.074) = sqrt(0.748) = 0.865, so i(a) = sin inverse(0.865), approximately 60 degrees -- any ray entering the flat end face within 60 degrees of the fibre's own axis will be guided by total internal reflection all the way along the core. If the cladding is removed entirely (bare core in air, n2 = 1), the same formula would require sin(i(a)) = sqrt(1.68^2 - 1) = 1.35, which is impossible since sine can never exceed 1 -- meaning every ray entering the flat end face over the full 0 to 90 degree range is guided, and more generally a bare core surrounded only by air is guaranteed to totally internally reflect any ray only if its own …