Final answer:
The maximum refractive index for the cladding must be less than or equal to 1.444 to ensure total internal reflection for light traveling within 5 degrees of the fibre-axis in a glass optical fibre with a refractive index of 1.450.
Step-by-step explanation:
To determine the maximum index of refraction allowed for the cladding, we apply the concept of total internal reflection. The condition for total internal reflection is that the incident angle must be greater than the critical angle, where the critical angle θc is defined for the interface between two media of different refractive indices.
According to Snell's law, θc can be found using the equation:
n1 × sin(θc) = n2 × sin(90°)
Simplifying this gives:
sin(θc) = n2/n1
Where n1 = 1.450 (the refractive index of the core) and n2 is the refractive index of the cladding, which we want to determine. Since the light should be contained within 5 degrees of the fibre-axis, this means that the maximum angle that light makes with the surface of the core (which equals the critical angle) is 90° - 5° = 85°. Using the sine function:
sin(85°) = n2/1.450
Solving for n2, we get:
n2 = 1.450 × sin(85°)
Calculating this gives:
n2 = 1.444
Therefore, the maximum refractive index for the cladding must be less than or equal to 1.444 to ensure total internal reflection of light traveling at angles within 5 degrees of the fibre-axis.