Problem — Critical angle of an optical fibre

Problem

An optical fibre has a core of refractive index $n_1 = 1{,}50$ surrounded by air ($n_2 = 1{,}00$). Calculate the critical angle beyond which total internal reflection occurs.

Solution

Setting up. The critical angle is the one for which the refracted ray emerges at $90^\circ$; from Snell’s law $n_1\sin\theta_c = n_2\sin 90^\circ$:

$\sin\theta_c = \frac{n_2}{n_1} = \frac{1{,}00}{1{,}50} = 0{,}667$

$\theta_c = \arcsin(0{,}667) \approx \ev{41{,}8^\circ}$

For angles of incidence $> \theta_c$ the light is entirely reflected inside the core: this is the mechanism that traps the signal within the fibre.

Collegamenti

Argomenti: Optics Concetti: Refraction and Snell’s law