Question #182245

A step index fiber is made with a core of refractive index 1.36, a diameter of 49pm and a fractional refractive index change of 0.025. it is operated at a wavelength of 1.3. Then the numerical aperture of the cable is


1
Expert's answer
2021-04-19T17:26:56-0400

The refractive index is inversely proportional to the wavelength.

Let the refractive index of cladding be n' .

Hence, we know

Δn=nnn\Delta n = \dfrac{n-n'}{n}


0.025=1.36n1.360.025 = \dfrac{1.36-n'}{1.36}


n=1.326n' = 1.326 = Refractive index of cladding.

We know,

Numerical aperture, Na=(n12n2)N_a = \sqrt{(n_1^2-n'^2)}

n1=1.36n_1 = 1.36

n=1.326n' = 1.326

Na=1.3621.3262N_a = \sqrt{{1.36^2-1.326^2}}

Na=0.302N_a = 0.302


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