Total Internal Reflection — Question 2

Question 2

A fiber entrance angle is limited by reflection at its side wall

A straight step-index fiber has core index n1=1.50 and cladding index n2=1.47. Light enters a flat end from index n0=1.00 at angle alpha from the fiber axis. Consider meridional rays; their core angle to the axis is beta.

1. Derive the maximum external acceptance angle using end-face refraction and side-wall total internal reflection.

2. Calculate the numerical aperture and acceptance half-angle.

Question 2 - Solution

A fiber entrance angle is limited by reflection at its side wall

(a) Method and calculation

$$n_0\sin\alpha=n_1\sin\beta,\quad\theta_{\mathrm{wall}}=90^\circ-\beta,\quad\sin\theta_c=n_2/n_1.$$$$\cos\beta>n_2/n_1\Rightarrow n_0\sin\alpha<\sqrt{n_1^2-n_2^2}.$$

(b) Interpretation and further calculation

$$\boxed{\mathrm{NA}=\sqrt{n_1^2-n_2^2}=0.298496}.$$$$\boxed{\alpha_{\max}=\sin^{-1}(\mathrm{NA}/n_0)=17.3673^\circ}.$$

Physical check

The maximum is the limiting ray; strict total internal reflection occurs inside the acceptance cone. The formula assumes the square-root ratio is at most one.