The index of refraction of the core of a piece of fiber optic cable Is 1.75. If
ID: 1562750 • Letter: T
Question
The index of refraction of the core of a piece of fiber optic cable Is 1.75. If the Index of the surrounding cladding Is 1.29, what is the critical angle for total internal reflection for a light ray in the core incident on the core-cladding interface? Express the answer (numerical value only) in degrees with one decimal place. A light ray is incident upon a plane interface between two materials. The ray makes an incident angle with the normal to the surface of 38 degrees in a material with index of refraction 1.2. If the second material has an index of refraction 1.7. what is the refracted angle in degrees in that material? Express the answer (numerical value only) in degrees with one decimal place. The speed of light in a material is 0.64 x c. What is the critical angle in degrees of a light ray at the interface between the material and a vacuum? Express the answer (numerical value only) with one decimal place.Explanation / Answer
a) critica angle = sin^ -1 (1.29/ 1.75) = 47.488 degree apprx
b)1.2 sin 38 = 1.7 sin r
r= sin^-1 (0.434) =25.7588 degree apprx
c)r
v = c/n
0.64 x 3x 10^8 = 3x 10^8/ n
n= 1/0.64= 1.5625
critical angle= sin^-1 ( 1/ 1.5625) = 39.79 degree apprx-
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