A fiber optic is made by cladding a thin fiber core of refractive index n₁ = 1.45 with a material of refractive index n₂ = 1.38. What is the maximum incident angle, Ɵ, so that the light ray is totally internally reflected inside the fiber? A) 24.1º B) 26.4º C) 28.2º D) 30.3º E) 32.7º

Question

Quizplus · Accepted Answer

The critical angle for total internal reflection is given by:
sin Ɵc = n2/n1
where n1 is the refractive index of the core and n2 is the refractive index of the cladding.
In this case, n1 = 1.45 and n2 = 1.38.
sin Ɵc = 1.38/1.45 = 0.951
Ɵc = sin-1(0.951) = 69.5º
However, this is the critical angle for total internal reflection at the boundary between the core and the cladding. To find the maximum incident angle for total internal reflection within the fiber, we need to consider the geometry of the fiber.
Assuming that the fiber has a circular cross-section, the light ray will make an angle with the normal at the boundary between the core and the cladding, as shown in the diagram below:

fib_1.png

The maximum incident angle for total internal reflection within the fiber occurs when the light ray just grazes the boundary between the core and the cladding, as shown in the diagram below:

fib_2.png

In this case, the angle of incidence is equal to the critical angle, which is given by:
sin Ɵ = n2/n1 = 1.38/1.45 = 0.951
Ɵ = sin-1(0.951) = 69.5º
However, this angle is measured with respect to the normal at the boundary between the core and the cladding. To find the angle with respect to the axis of the fiber, we need to subtract 90º from this angle. Therefore, the maximum incident angle for total internal reflection within the fiber is:
Ɵ = 90º - sin-1(0.951) = 26.4º
Therefore, the correct answer is choice B.

A Fiber Optic Is Made by Cladding a Thin Fiber