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 fibre is coated with glass of lower refractive index to facilitate total internal reflection. This allows the ray of light to stay within the core of the fibre, minimizing loss of light as it travels. The difference in refractive indices ensures that the light does not escape into the coating but is instead reflected back into the core, maintaining signal integrity over long distances.

The speed of light in a medium can be calculated using the formula:

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

where:

• $c$ is the speed of light in vacuum ( extit{approximately } 3.0 \times 10^8 , \text{m s}^{-1})
• $n$ is the refractive index of the medium (1.55 for the fibre)

Thus, substituting the values:

$v = \frac{3.0 \times 10^8}{1.55} \approx 1.94 \times 10^8 \, \text{m s}^{-1}$

To find the power transmitted after 8 km, we note that the power reduces by half every 2 km.

• At 2 km, power = 10 W / 2 = 5 W
• At 4 km, power = 5 W / 2 = 2.5 W
• At 6 km, power = 2.5 W / 2 = 1.25 W
• At 8 km, power = 1.25 W / 2 = 0.625 W

Thus, the power being transmitted by the light after it has travelled 8 km through the fibre is 0.625 W.