For total internal reflection to occur inside anoptic fiber as shown in Fig. 22.

Question

For total internal reflection to occur inside anoptic fiber as shown in Fig.

22.31, the angle must begreater than the critical angle for the fiber–air interface.At the end

of the fiber, the incident light undergoes refraction toenter the fiber. If total internal reflection is

to occur for any angle of incidence,i outside the end of the fiber, whatis the minimum index of

refraction of the fiber?

Hint: The worst case scenario is the that the angle ofincidence at the end of the fiber is 90o.

Use geometry to relate the first refracted angle at theend to .

Sin(90o - ) =Cos()

Explanation / Answer

According to Snell's law n * sin = n1 * sin1 --------------(1) n refractive index of air = 1.0 angle of incidence at the end of the fiber =90o n1 refractive index of fiber 1 angle of refraction through the fiber 1 = arcsin(n1/n1) = arcsin(1) = 90o from (1) n * sin = n1 * sin1 or sin = (n1 * sin1/n) = (n1 *sin(90o)/n) = (n1/n) ------------(2) When the light goes out of the fiber n1 * sin = n * sin2 2 = arcsin(n/n1) or n1 * sin = n * sinarcsin(n/n1) = (n^2/n1) sin = (n^2/n1^2) from (2) (n1/n) = (n^2/n1^2) or n1^3 = n^3 or n1 = n n = 1.0008 or n1 = 1.0008 from (1) n * sin = n1 * sin1 or sin = (n1 * sin1/n) = (n1 *sin(90o)/n) = (n1/n) ------------(2) When the light goes out of the fiber n1 * sin = n * sin2 2 = arcsin(n/n1) or n1 * sin = n * sinarcsin(n/n1) = (n^2/n1) sin = (n^2/n1^2) from (2) (n1/n) = (n^2/n1^2) or n1^3 = n^3 or n1 = n n = 1.0008 or n1 = 1.0008

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