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An Optical fiber has a numerical aperture of 0.15 in air and a cladding refractive index of 1.50. Find the numerical aperture of the fiber in liquid with a refractive index of 1.30.
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Solution:

We know that the numerical aperture NA is given by,

NA=n21n22n20=(NA)ait n0

For air n0=1. Given n2=1.50 and NA=0.15

0.15=n21(1.50)2

n=1.5075, the refractive index of the core. 

Also, Given,

n0=1.30 for water and (NA)sit =0.15

We know, NA=(NA)air n0

(NA)water =0.151.30=0.1154

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