Question

If Numerical Aperture (NA) = 0.22 and Δ = 0.012 where Δ is the fractional refractive index change. The refractive indices of the core and cladding of a fibre is ----- respectively. *​

Answer 1

• the refractive indices of core and cladding materials of an optical fiber if its numerical aperture is 0.22

and

• relative refractive index difference is 0.012. ...

= 1.558 x (1-0.015)

= 1.534

• The refractive index of the fiber core is 1.558 and cladding is 1,534.

Answer 2

Given,

Numerical aperture = 0.22

Fractional refractive index change = 0.012

To find,

The refractive indices of the core and the cladding of a fiber

Solution,

The refractive indices of the core and the cladding of fiber are 1.424 and 1.407.

We can simply solve the numerical question by the following process.

We know that,

Numerical aperture is mathematically defined as √μ₁²-μ₂², where

μ₁ = Refractive index of the core

μ₂ = Refractive index of the cladding

Thus,

⇒ 0.22 = √μ₁²-μ₂²

⇒ 0.22*0.22 = μ₁²-μ₂²  

⇒ 0.0484 = (μ₁+μ₂)(μ₁-μ₂)    (Equation 1)

Now,

Fractional refractive index change = $\frac{u1-u2}{u1}$

Thus,

⇒ 0.012 = $\frac{u1 - u2}{u1}$

⇒ 0.012μ₁ = μ₁-μ₂

⇒ μ₂ = 0.988μ₁   (Equation 2)

On substituting the value of μ₂ from equation 2 to equation 1.

⇒ 0.0484 = (μ₁+ 0.988μ₁)(μ₁- 0.988μ₁)

⇒ 0.0484 = 1.988 μ₁ x 0.012 μ₁

⇒ 2.029 = μ₁²

⇒ μ₁ = 1.424

Now,

On putting the value of μ₁ in equation 2.

⇒ μ₂ = 0.988μ₁

⇒ μ₂ = 0.988 x 1.424

        = 1.407

Thus,

The core's refractive indices and fiber cladding are 1.424 and 1.407.