The glass core of an optical fiber has an index of refraction1.65. The index of

Question

The glass core of an optical fiber has an index of refraction1.65. The index of refraction of the cladding is 1.50. 1. What is the maximum angle a light ray canmake with the wall of the core if it is to remain inside thefiber? The glass core of an optical fiber has an index of refraction1.65. The index of refraction of the cladding is 1.50. 1. What is the maximum angle a light ray canmake with the wall of the core if it is to remain inside thefiber? 1. What is the maximum angle a light ray canmake with the wall of the core if it is to remain inside thefiber?

Explanation / Answer

There is an angle referred to as the critical angle. Ifthe light ray inside the fiber contacts the fiber/cladding boundaryat any angle greater than this angle, total internal reflectionoccurs and no refracted ray appears. This equation is crit = sin-1(n2/n1) In this case, n2 equals the cladding, 1.5 and n1 is the core,1.65. Substituting into the equation we get crit = 65.38o So any angle larger than this will undergo total internalreflection, and any angle smaller than this will have somerefracted ray leave the core. So any angle larger than this will undergo total internalreflection, and any angle smaller than this will have somerefracted ray leave the core.

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