Question

An optical fiber has core refractive
index 1.5 and refractive index of
cladding is 3% less than the core index
Calculate the numerical aperture and
acceptance angle. *​

Answer 1

Answer:

Numerical aperture of the optical fiber is

NA = 0.36

Acceptance angle of the fiber is 21.4 degree

Explanation:

As we know that the refractive index of core is

$n_1 = 1.5$

now refractive index of cladding is 3% less than core

so it is given as

$n_2 = 0.97(1.5) = 1.455$

now we know that numerical aperture of the optical fiber is given as

$NA = \sqrt{n_1^2 - n_2^2}$

$NA = \frac{1.5^2 - 1.455^2}$

$NA = 0.36$

Now for acceptance angle we know that

$sin\theta = NA$

$sin\theta = 0.36$

$\theta = sin^{-1}(0.36)$

$\theta = 21.4 ^o$

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Topic : Optical Fiber

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Answer 2

Explanation:

calculate the refractive index core material of a fiber from the following data Na =0.22. ∆ =0.02