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Optical fiber has a numerical aperture of 0.2 and a cladding refractive index of 1.59. Determine the acceptance angle for the fiber in water which has a refractive index of 1.33.
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Solution:

We know that,

NA =n21n22n20

Given,

NA=0.2 in air (n0=1) and n2=1.59

0.2=n21(1.59)2n1=1.602.

Now we have to find the acceptance angle in water i.e.

no=1.33

NA=sinα=n21n22n2o or sinα=0.21.33=0.15

α=8.64

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