Answer to Question #260537 in Optics for ajay

Question #260537

A graded index fiber with a core axis refractive index of 1.5 has a characteristic index profile (α) of

1.90, a relative refractive index difference of 1.3% and a core diameter of 40 μm. Estimate the number

of guided modes propagating in the fiber when the transmitted light has a wavelength of 1.55 μm,

and determine the cutoff value of the normalized frequency for single-mode transmission in the fiber.


1
Expert's answer
2021-11-04T18:00:52-0400

V-Number and No. of Modes in Fibre


Fractional Index Difference: It is the ratio of the difference in the refractive index of core and cladding with respect to refractive index of core.

Δ=n1n2n1n12n22=(n1+n2)=(n1+n22)(n!n2n1)2n1\Delta={n_1-n_2 \over n_1}\\n_1^2-n_2^2=(n_1+n_2)\\ =({n_1+n_2 \over2})({n_!-n_2\over n_1})2n_1


n1+n22n1{n_1+n_2 \over 2} \approx n_1


 and


N.A.=n12.2ΔN.A.={\sqrt{ n_1^2}.2\Delta}


Therefore

N.A=n12ΔN.A=n_1{\sqrt2\Delta}

V-Number or Normalized Frequency

V – number determines how many modes a fiber can support,  It is given by

V=πdλNAV={\pi d\over \lambda}NA


,where d is the diameter of the core, l is the wavelength of light used and NA is the numerical aperture of the fibre.

  


V=πdλ(n12+n22)V={\pi d \over \lambda}{\sqrt (n_1^2+n_2^2)}


No. of guided modes =V=2πλa(2Δ)12V={2 \pi \over\lambda}a(2\Delta)^{1\over2}


V=2π1.55×106×40×106(2×0.013)12V={2 \pi \over 1.55 \times10^{-6} }\times40 \times10^{-6}(2 \times0.013)^{1\over 2}


V=26


λc=V.λ2.405=26×1.55μ2.405=16.57μ\lambda_c= {V.\lambda\over2.405}={26\times1.55\mu\over 2.405}=16.57\mu



Fc=Cλc=3×10816.57×106=18.10×1012HzF_c={C\over \lambda_c}={3 \times 10^{8}\over16.57 \times 10^{-6}}=18.10\times10^{12}H_z


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