Optical Fibers and Refraction

Q.
What is the Index of Refraction of the optical fibers, including the core and cladding?

A.
Since the numerical aperture (NA) is given in the catalog, as 0.22±0.02, then applying the following equation at a certain wavelength.
NA=(nf2-nc2)1/2
The index of refraction of the cladding (nc) can be found, if the index of refraction of the core (nf) is given at the same wavelength that is being worked with.

Example:

NA=(nf2-nc2)1/2

Given:
NA = 0.22
nf = 1.46 @ 630nm

0.22 = ((1.46)2-nc2)1/2

Solving for nc:
nc = 1.44 @ 630nm

Likewise:
NA = 0.22
nf = 1.50 @ 300nm

0.22 = ((1.50)2-nc2)1/2

Solving for nc:
nc = 1.48 @ 300nm