A light ray (red)enters a concentrically cladded optical fiber at its center and an angle of 29.0° to the axis of the fiber.The diameter of the fiber is 1.0 mm and its refractive index is nf = 1.35.What must be the value of ncladding to guarantee total internal reflection? (n_air = 1.00)

Question

Quizplus · Accepted Answer

To ensure total internal reflection, the refractive index of the cladding (n_cladding) must be less than the refractive index of the core (n_fiber = 1.35). The critical angle for total internal reflection is given by sin(θ_c) = n_cladding / n_fiber. For the light to be totally internally reflected, the angle of incidence inside the fiber must be greater than the critical angle. Given the angle of incidence is 29.0°, we need n_cladding < n_fiber * sin(29.0°). Calculating this gives n_cladding < 1.35 * sin(29.0°) ≈ 1.18. Thus, the correct choice is A) 1.18.

A Light Ray (Red)enters a Concentrically Cladded Optical Fiber at Its