4. Optical Fibers and Evanescent Waves . Shown below is a step-index optical fib

Question

4. Optical Fibers and Evanescent Waves. Shown below is a step-index optical fiber, consisting of a core with refractive index n1 and a cladding with a smaller refractive index n2. The numerical aperture (NA) of an optical fiber is the sine of the maximum angle of an incident ray (from the surrounding air) with respect to the fiber axis such that the transmitted beam is confined within the core.

(a) Derive an expression for the NA=sin of a step index fiber, in terms of n1 and n2.

(b) For a typical n1=1.47 and n2=1.45, what are and NA?

(c) Suppose that infrared light of wavelength 1100 nm is incident on the core at an angle that is 5 degrees smaller than the maximum angle for confinement, sin-1 (NA). How thick must the cladding be to ensure that the evanescent electric field within it has decayed by a factor of 1/e4 or to roughly 1% of its in-core value at its outer surface? (This is important to minimize losses in the medium surrounding the cladding.)

Explanation / Answer

let incident angle = phi
then form snell's law
refracted angle r = ?
sin(phi)/sin(r) = n1 ( refractive index of core)

now, in the core cladding interface
sin((90 - r))/sin(90) = n2/n1

so as phi increases, r increases, 90 - r decreases
hencefor maximum angle phi, the angle of refraction for cladding = 90 deg
cos(r) = n2/n1
a. sin(phi) = n1*sqroot(1 - (n2/n1)^2) = sqroot(n1^2 - n2^2)
b. n1 = 1.47
n2 = 1.45
sin(phi) = sqroot(1.47^2 - 1.45^2)
NA = sin(phi) = 0.24166
phi = 13.98 deg

c. evanascent wave electric field is given by
E = Eo*e^(-lambda*t)
E/Eo = e^-4 = e^(-lambda*t)
thickness = t = 4/lambda ( where lambda is attenruation constant for the material at the given wavelength)

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