An optical fiber consists of a cylindrical core of refractive index 1.6 surrounded by a cladding of refractive index 1.28 . A light ray travelling in air enters the flat end of the fiber at an angle of incidence . If is the maximum angle of incidence for which the refracted ray undergoes total internal reflection at the core-cladding boundary, the value of 100 is ________.

Step-by-step solution

Let the critical angle at the core-cladding interface be _c . Applying Snell's law at the core-cladding boundary: _ core _c = _ cladding 90^ 1.6 _c = 1.28 _c = 1.28 1.6 = 0.8 From geometry, the angle of refraction r at the flat end of the fiber is related to _c by r = 90^ - _c . r = (90^ - _c) = _c = 1 - ^2 _c = 1 - 0.8^2 = 0.6 Now, applying Snell's law at the flat entrance (air to core): 1 = _ core r = 1.6 0.6 = 0.96 Therefore, the value of 100 is 96 . Answer: 96