Question

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 1.3um.​

Answer 1

Answer:

The refractive index is inversely proportional to the wavelength.  

Let the refractive index of cladding be n' .  

Hence, we know

Δn=   n-n'/n  

0.025 = 1.36-n'/1.36

n' = 1.326  = Refractive index of cladding.  

We know,  

Numerical aperture, Nₐ =√(n₁²-n'²)  

n₁= 1.36

n' = 1.326

Nₐ=√1.36²-1.326²

Nₐ=0.302

Answer 2

Given:

Refractive index = 1.36

Diameter = 49pm

Fractional index change = 0.025

Wavelength = 1.3μm

To find:

Numerical aperture

Solution:

Let refractive index of cladding be n'

As we know wavelength is inversely proportional to refractive index.

hence,

$\Delta n = \frac{n-n'}{n}$

$0.025 = \frac{1.36-n'}{1.36}$

n' = 1.326

We know that, we have to calculate numerical aperture.

formula used-

Numerical aperture, $N_a = \sqrt{n_1^{2}- n_2^2 }$

n₁ =1.36

n₂ = 1.326

$N_a = \sqrt{1.36^{2}- 1.326^2 }$

Nₐ = 0.302

Hence, numerical aperture is 0.302.