19. A fiber has index of refraction n = 1.15, and its end surface is normal to the direction of its length. Light is incident from the air. What is the maximum incident angle so that the light can have total internal reflection inside the fiber?


(A) sin⁻¹(0.57) ≈ 35°
(B) sin⁻¹(0.66) ≈ 42°
(C) sin⁻¹(0.75) ≈ 49°
(D) sin⁻¹(0.80) ≈ 53°
(E) sin⁻¹(0.91) ≈ 65°

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