---
title: "Acceptance angle derivation"
subject: "Propagation and Antenna"
chapter: 6
key: "semester_6/propagation_and_antenna:acceptance-angle-derivation"
source: https://www.bctengineer.com/q/semester_6/propagation_and_antenna/acceptance-angle-derivation
site: BCT Engineering
---

# Acceptance angle derivation

Propagation and Antenna · Chapter 6: Optical Fibres

- Asked in: 2070 Bhadra, 2+7 marks
- Video: [Acceptance Angle I L 9 I Optical Communication](https://www.youtube.com/watch?v=0tWGjgPgP4M) — Dopamine
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/propagation_and_antenna/acceptance-angle-derivation
- Whole chapter: https://www.bctengineer.com/md/semester_6/propagation_and_antenna/chapter-6

## Question

What is an acceptance angle? Derive an expression to calculate the acceptance angle.

## Short answer

Acceptance angle θa is the maximum input angle from the axis for guiding by total internal reflection. Derived from Snell's law at the face and TIR condition at the core-cladding boundary: sin θa = sqrt(n1^2 - n2^2)/n0, or for air θa = sin^-1 sqrt(n1^2 - n2^2).

## Answer

**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 $\theta_0$.

```text
         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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. The acceptance angle is:
   A. The angle of the cladding surface
   B. The critical angle at the core-cladding boundary
   C. The maximum angle to the axis at which a ray entering the fibre end is still guided by TIR
   D. The angle of the refracted ray inside the core
   Answer: C. Rays within the acceptance cone of half-angle theta_a are guided; rays outside refract into the cladding and are lost. (Key point: Definition of acceptance angle)

2. Why is sin(theta_1) = cos(phi) in the derivation?
   A. Because theta_1 + phi = 90 degrees
   B. Because theta_1 = phi
   C. Because n1 = n2
   D. Because the ray is parallel to the axis
   Answer: A. The boundary normal is perpendicular to the axis, so the refracted angle theta_1 and the incidence angle phi are complementary. (Key point: Geometry step)

3. For air, the acceptance angle is:
   A. sin^-1(n1 - n2)
   B. sin^-1(sqrt(n1^2 - n2^2))
   C. cos^-1(n2/n1)
   D. sin^-1(n2/n1)
   Answer: B. n0 sin(theta_a) = sqrt(n1^2 - n2^2), with n0 = 1 for air. (Key point: Result)

4. What does the acceptance angle depend on?
   A. Core diameter only
   B. The refractive indices (larger index difference widens the cone)
   C. Fibre length
   D. Source power
   Answer: B. theta_a depends only on n1, n2 (and n0), not on diameter. A larger difference widens the cone but increases modal dispersion. (Key point: Dependence of theta_a)

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