Question

Derive the equation for acceptance angle and numerical aperture, of optical fiber. Acceptance angle in optical fibre:

Answer 1

To ensure the critical angle incidence in the core-cladding boundary inside the optical fibre, the light should be incident at a certain angle at the end of the optical fiber while entering in to it. This angle is called acceptance angle. It depends on the refractive indices of the core n₁, cladding n₂ and the outer medium n₃. Assume the light is incident at an angle called acceptance angle i at the outer medium and core boundary at A.

The Snell’s law in the product form, equation for this refraction at the point A.

n₃sin i_a = n₁sin r_a   …(1)

To have the total internal reflection inside optical fibre, the angle of incidence at the corecladding interface at B should be atleast critical angle i_c. Snell’s law in the product form, equation for the refraction at point B is,

n₁ sin i_c = n₂ sin 90° …(2) 

n₁ sin i_c = n ∵ sin 90° =1 

∴sin i_c = \(\frac{n_2}{n_1}\) …(3)

From the right angle triangle ∆ABC,

i_c = 90°- r_a

Now, equation (3) becomes, sin (90° – r_a) = \(\frac{n_2}{n_1}\)

Using trigonometry’ , cos ra =r_a = \(\frac{n_2}{n_1}\) ....... (4)

sin r_a = \(\sqrt{1-cos^2r_a}\)

Substituting for cos r_a

If outer medium is air, then n₃ = 1. The acceptance angle i_a becomes,

Light can have any angle of incidence from 0 to ia with the normal at the end of the optical fibre forming a conical shape called acceptance cone. In the equation (6), the term (n₃ sin i_a) is called numerical aperture NA of the optical fibre.

If outer medium is air, then n = 1. The numerical aperture NA becomes,

NA = sin i_a = \((\sqrt{n^2_1-n^2_2})\) ............. (11)