What are the acceptance angle and acceptance cone of an optical fibre?

In one line

The acceptance angle is the maximum angle to the fibre axis at which light can enter the fibre and still be guided by total internal reflection, and the cone formed by rotating this angle around the axis is the acceptance cone.

Explained simply

Think of a funnel at the mouth of the fibre. Light poured in from inside the funnel gets trapped and travels on; light coming in from outside the funnel's edge leaks away. The half-angle of this funnel is the acceptance angle, and the funnel itself is the acceptance cone.

How it works

1. A ray enters from a medium of index \(n_0\) at angle \(\theta_i\) to the axis. Snell's law at the end face: \(n_0\sin\theta_i = n_1\sin\theta_r\).
2. Inside, it hits the core-cladding boundary at \(\theta = 90° - \theta_r\).
3. For TIR, \(\theta \ge \theta_c\). The limiting case \(\theta = \theta_c\) gives the largest \(\theta_i\), called \(\theta_a\).
4. Then \(\sin\theta_r = \cos\theta_c = \sqrt{1 - n_2^2/n_1^2}\).
5. So \(n_0\sin\theta_a = n_1\sqrt{1 - n_2^2/n_1^2} = \sqrt{n_1^2 - n_2^2}\).
6. Rotating the ray at \(\theta_a\) about the axis gives a cone of full angle \(2\theta_a\): the acceptance cone.

Key formula

\(\theta_a = \sin^{-1}\left(\dfrac{\sqrt{n_1^2 - n_2^2}}{n_0}\right)\); for air \(n_0 = 1\), \(\theta_a = \sin^{-1}\sqrt{n_1^2-n_2^2}\). \(n_1\): core index, \(n_2\): cladding index, \(n_0\): index of the outside medium. Acceptance cone full angle \(= 2\theta_a\).

Quick example

\(n_1 = 1.50\), \(n_2 = 1.45\), in air: \(\sqrt{2.25 - 2.1025} = \sqrt{0.1475} = 0.384\), so \(\theta_a = \sin^{-1}(0.384) \approx 22.6°\), and the acceptance cone angle is about 45.2°. If the same fibre is dipped in water (\(n_0 = 1.33\)), \(\sin\theta_a = 0.384/1.33 = 0.289\), so \(\theta_a\) drops to about 16.8°.

Common mistakes

• Giving \(\theta_a\) as the acceptance cone angle; the cone's full angle is \(2\theta_a\).
• Forgetting to divide by \(n_0\) when the fibre is not in air.
• Measuring \(\theta_a\) from the end face instead of from the fibre axis.

How MU asks it

Asked as "Define acceptance angle and derive an expression for it" (5 marks) or as a numerical (3 to 5 marks). Draw the ray diagram with \(\theta_a\), \(\theta_r\) and \(\theta_c\) marked, show each step, and box the final result.