Acceptance angle derivation

Acceptance angle ($\theta_a$) is the maximum angle, measured from the fibre axis, at which a light ray can enter the fibre end and still be guided along the core by total internal reflection. Rays inside the cone of half-angle $\theta_a$ (the acceptance cone) are guided; rays outside it are refracted into the cladding and lost.

Derivation

Let the core index be $n_1$, cladding index $n_2$ ($n_1 > n_2$), and the outside medium index $n_0$ (air, $n_0 = 1$). A meridional ray hits the core end at angle $\theta_0$.

         n0 |      cladding n2
    ray \   |____________________
         \  |    /\ phi
  axis ---*-+---/--\------------ core n1
       th0  | th1   \
            |____________________
                   cladding n2

Step 1: refraction at the air–core face (Snell's law):

$$n_0\sin\theta_0 = n_1\sin\theta_1 \qquad (1)$$

Step 2: geometry. The refracted ray strikes the core–cladding boundary at angle $\phi$ to the normal. Since the normal to the boundary is perpendicular to the axis:

$$\theta_1 = 90^\circ - \phi \quad\Rightarrow\quad \sin\theta_1 = \cos\phi \qquad (2)$$

Step 3: condition for TIR. The ray is totally reflected only if $\phi \ge \phi_c$, with

$$\sin\phi_c = \frac{n_2}{n_1} \qquad (3)$$

As $\theta_0$ increases, $\theta_1$ increases and $\phi$ decreases. The limiting ray has $\phi = \phi_c$ and $\theta_0 = \theta_a$.

Step 4: combine (1), (2) and (3):

$$\begin{aligned} n_0\sin\theta_a &= n_1\cos\phi_c \\ &= n_1\sqrt{1 - \sin^2\phi_c} \\ &= n_1\sqrt{1 - \frac{n_2^2}{n_1^2}} \\ &= \sqrt{n_1^2 - n_2^2} \end{aligned}$$

Step 5: acceptance angle:

$$\theta_a = \sin^{-1}\left(\frac{\sqrt{n_1^2 - n_2^2}}{n_0}\right)$$

and for air ($n_0 = 1$), $\theta_a = \sin^{-1}\sqrt{n_1^2 - n_2^2}$.

The quantity $n_0\sin\theta_a = \sqrt{n_1^2 - n_2^2}$ is the numerical aperture (NA). Using the relative index difference $\Delta = (n_1 - n_2)/n_1$, $\text{NA} \approx n_1\sqrt{2\Delta}$, so $\theta_a \approx \sin^{-1}(n_1\sqrt{2\Delta})$.

Example: $n_1 = 1.48$, $n_2 = 1.46$, in air:

$$\begin{aligned} \sin\theta_a &= \sqrt{1.48^2 - 1.46^2} = \sqrt{0.0588} = 0.2425 \\ \theta_a &= 14.03^\circ, \quad \text{full cone angle } 2\theta_a = 28.07^\circ \end{aligned}$$

Notes

• $\theta_a$ depends only on the refractive indices, not on fibre diameter.
• A larger index difference gives a wider acceptance cone (easier coupling) but more modal dispersion.