Path difference in Fibre Optics

[Graecia]

Hi,
This is a question from a homework which has stumped me:
Light is traveling along a fibre optic cable of refractive index 1.52, the refractive index of the cladding is 1.51 and the critical angle between the two is 82.
a) what is the most direct route for light along the fibre-optic cable
b)The longest route is for light rays traveling at the critical angle. Calculate the path difference between the two rays after travelliing 10km of the cable

I guessed that the most direct route would be straight down the fibre optic cable, and for b i calculated the speed of light in the core to be 1.97x10^8 m/s


Any help would be greatly appreciated,
Many Thanks in advance

 

[jbsteele]

a) The most direct route for light along the fibre-optic cable is straight down the centre of the cable. This is because light travels faster in the core of the cable than it does in the cladding, so travelling directly down the centre of the cable will give the shortest path.b) To calculate the path difference, you need to first calculate the angle of refraction of the light ray travelling at the critical angle. This can be calculated using Snell's law: n1*sin(θ1) = n2*sin(θ2), where n1 and n2 are the refractive indices of the core and cladding respectively, and θ1 and θ2 are the angles of incidence and refraction respectively. Plugging in the values given, we get: 1.52*sin(82) = 1.51*sin(θ2). Solving for θ2 gives us θ2 = 48.3 degrees, which is the angle of refraction of the light ray travelling at the critical angle. Now we can calculate the path difference between the two rays. The path difference is equal to the difference in the distance travelled by the two rays, which can be calculated using the following equation:Path Difference = (v1 - v2)*t, where v1 and v2 are the speeds of light in the core and cladding respectively, and t is the time taken for the light to travel 10 km. Plugging in the values given, we get: Path Difference = (1.97x10^8 - 1.50x10^8)*(10km/1.97x10^8) = 5.08 km. Therefore, the path difference between the two rays after travelling 10 km of the cable is 5.08 km.