Answer to Question #260537 in Optics for ajay

Question

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.

Answer 1

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.

$\Delta={n_1-n_2 \over n_1}\\n_1^2-n_2^2=(n_1+n_2)\\ =({n_1+n_2 \over2})({n_!-n_2\over n_1})2n_1$

${n_1+n_2 \over 2} \approx n_1$

 and

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

Therefore

$N.A=n_1{\sqrt2\Delta}$

V-Number or Normalized Frequency

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

$V={\pi d\over \lambda}NA$

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

  

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

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

$V={2 \pi \over 1.55 \times10^{-6} }\times40 \times10^{-6}(2 \times0.013)^{1\over 2}$

V=26

$\lambda_c= {V.\lambda\over2.405}={26\times1.55\mu\over 2.405}=16.57\mu$

$F_c={C\over \lambda_c}={3 \times 10^{8}\over16.57 \times 10^{-6}}=18.10\times10^{12}H_z$