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In 1996, the Fiberoptic Link Around the Globe (FLAG) was started. It initially involves placing a 27,000 km fiber optic cable at the bottom of the Mediterranean Sea and the Indian Ocean. Suppose the index of refraction of this fiber is 1.56. If all reflections occur at the critical angle, what is the total distance a ray of light will travel along the fiber?

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  1. The critical angle of incidence is
    arcsin(1/N) = 39.9 degrees, which means the ray is inclined 50.1 degrees to the side of the fiber. For each meter of fiber length, the ray travels sec50.1 = 1.56 meters.

    So, multiply the cable length by 1.56.

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