Derive the expression of numerical aperture for a step index fiber.

Derive the expression of numerical aperture for a step index fiber. A light ray enters an optical fiber from air. The fiber has core refractive index 1.52 and cladding refractive index 1.41. Find the Critical angle and Numerical aperture.

\(\text{NA} = \sqrt{n_1^2 - n_2^2}\). For \(n_1 = 1.52,\ n_2 = 1.41\): critical angle ≈ 68.07° and NA ≈ 0.568.

Derivation

A ray enters the fibre end face from air (\(n_0 = 1\)) at angle \(\theta_0\) to the axis and refracts into the core at angle \(\theta_r\). It then meets the core–cladding boundary at angle \(\phi = 90^\circ - \theta_r\).

1. Snell's law at the end face:

\[n_0 \sin\theta_0 = n_1 \sin\theta_r\]

2. For total internal reflection the ray must strike the core–cladding surface at least at the critical angle \(\phi_c\), where

\[\sin\phi_c = \frac{n_2}{n_1}\]

3. The largest acceptance angle \(\theta_{max}\) corresponds to \(\phi = \phi_c\), i.e. \(\theta_r = 90^\circ - \phi_c\):

\[n_0\sin\theta_{max} = n_1\cos\phi_c\]

\[= n_1\sqrt{1 - \frac{n_2^2}{n_1^2}}\]

\[= \sqrt{n_1^2 - n_2^2}\]

Numerical aperture is defined as \(\text{NA} = n_0\sin\theta_{max}\), so with \(n_0 = 1\):

\[\text{NA} = \sqrt{n_1^2 - n_2^2}\]

Numerical

Critical angle:

\[\sin\phi_c = \frac{1.41}{1.52} = 0.9276\]

\[\phi_c = 68.07^\circ\]

Numerical aperture:

\[\text{NA} = \sqrt{1.52^2 - 1.41^2}\]

\[= \sqrt{2.3104 - 1.9881} = \sqrt{0.3223}\]

\[\text{NA} \approx 0.568\]

(The acceptance angle is \(\sin^{-1}(0.568) \approx 34.6^\circ\).)