Information is transmitted over long distances using optical fibres in which a ray of light is guided along a fibre - Leaving Cert Physics - Question 12(c) - 2009

Each optical fibre is coated with glass of a lower refractive index to facilitate the principle of total internal reflection. When light travels from a denser medium (the core) to a less dense medium (the cladding), it refracts away from the normal line as it exits the fibre. This design maximizes the extent of internal reflections, ensuring that minimal light escapes into the surrounding environment. Furthermore, the lower refractive index creates a barrier for any stray light that might otherwise breach the core, ensuring efficient data transmission over long distances.

To find the speed of light as it passes through the optical fibre, we utilize the refractive index formula:

$n = \frac{c}{v}$

Where:

• $n$ is the refractive index (1.55)
• $c$ is the speed of light in vacuum ($3.0 \times 10^8 \text{ m/s}$)
• $v$ is the speed of light in the medium.

Rearranging this gives us:

$v = \frac{c}{n} = \frac{3.0 \times 10^8}{1.55} \approx 1.94 \times 10^8 \text{ m/s}$

Given that the initial power is 10 W and the power halves every 2 km, we can calculate the power after 8 km:

1. After 2 km: $P = \frac{10}{2} = 5$ W
2. After 4 km: $P = \frac{5}{2} = 2.5$ W
3. After 6 km: $P = \frac{2.5}{2} = 1.25$ W
4. After 8 km: $P = \frac{1.25}{2} = 0.625$ W

Thus, the power transmitted by the light after travelling 8 km through the fibre is approximately 0.625 W.