Question

40 : The refractive index of core and cladding are 1.563
and 1.498 respectively, and then NA is​

Answer 1

Answer:  0.446055

Explanation:

The refractive index of core and cladding are 1.563 and 1.498 respectively, and then we have to find the NA.

we know that, numerical aperture (NA) of optical fibre is defined as,

$NA = \sqrt{n_{core}^2- n_{clad}^2}$

where $n_{core} , n_{clad}$  is the refractive index of core and cladding respectively.

Given:

$n_{core} = 1.563 , n_{clad} = 1.498$

So, we have

$NA = \sqrt{(1.563)^2 - (1.498)^2} = 0.446055$

​

Answer 2

Answer:

The NA is 0.446

Explanation:

The ratio of the velocity of light within a vacuum to that in a second object with a higher density is used to compute the refractive index (also known as the index of refraction). In mathematical algorithms and descriptive text, the letter n or n' is most frequently used to represent the refractive index parameter.

The twisting of a light ray as it moves from one medium into the other is measured by the refractive index. The Refractive Index, to put it simply, gauges the difference in speed of light as it enters a medium from the air. The Index of Refraction is another name for the index of refraction.

The refractive index of the core is given to be 1.563, and the cladding's refractive index is stated to be 1.498.

Numerical Aperture (N.A) $=\sqrt{n_{1} ^{2} -n_{2}^{2} }$

Core's refractive index $= n1 = 1.563$

Index of Refraction of the Cladding $= n2 = 1.498$

                                  $N.A=\sqrt{(1.563)^2-(1.498)^2 }$

                                          $=\sqrt{2.443-2.244}$

                                          $=\sqrt{0.199}$

                                          $= 0.446$

The numerical aperture is therefore discovered to be $0.446.$

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