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Fiber Optics & Light Guide

Sobre esta simulação

[Pt] Fiber Optics & Light Guide shows how hair-thin strands of glass carry global internet traffic at the speed of light. A translucent 3D optical fiber displays its high-index inner Core (n₁), lower-index outer Cladding (n₂), entrance Acceptance Cone, and exit Photodetector. As students drag the red laser emitter or adjust the launch angle, a continuous laser ray and a stream of glowing photon pulses refract into the core face and strike the core–cladding wall at incidence angle θᵢ. As long as θᵢ stays above the critical angle θc = arcsin(n₂/n₁), 100% of the light reflects internally down the waveguide; dragging the middle bend ring too sharply drops the local incidence angle below θc and reveals cladding leakage rays.

O que os alunos podem explorar

  • [Pt] Drag the red laser emitter up and down at the entrance face and find the exact launch angle where Core incidence angle θᵢ drops below Critical angle θc, triggering red leakage rays.
  • [Pt] Drag the gold middle ring upward to bend the optical fiber and see how a tight bend causes light that was previously trapped to leak out into the cladding.
  • [Pt] Increase Core index n₁ from 1.50 to 1.80 while keeping Cladding index n₂ at 1.38 and watch the green Acceptance Cone widen as Numerical Aperture (NA) grows.
  • [Pt] Raise Cladding index n₂ closer to n₁ and observe how the Critical angle θc approaches 90°, making it harder to keep light trapped.
  • [Pt] Click the laser emitter or bend ring to toggle between lossless Total Internal Reflection and Cladding Leakage modes.

Objetivos de aprendizagem

  1. 1[Pt] Calculate the critical angle θc = arcsin(n₂ / n₁) for an interface between a high-index core (n₁) and lower-index cladding (n₂).
  2. 2[Pt] State the two physical conditions required for Total Internal Reflection: light must travel from higher to lower refractive index (n₁ exceeds n₂) and strike at θᵢ exceeding θc.
  3. 3[Pt] Relate the Numerical Aperture NA = √(n₁² − n₂²) of an optical fiber to its maximum acceptance cone angle.
  4. 4[Pt] Explain why macroscopic bends in an optical fiber cable can cause signal attenuation (bend loss).

Perguntas frequentes dos professores

[Pt] Why does light stay trapped inside a curved glass optical fiber?

[Pt] The inner glass core has a higher refractive index (n₁) than the surrounding cladding glass (n₂). When light inside the core hits the boundary at a shallow grazing angle (meaning a large angle θᵢ measured from the normal, greater than θc), Snell's law has no refracted solution and 100% of the light reflects back into the core.

[Pt] Why does an optical fiber need cladding if air already has a low refractive index (n = 1.00)?

[Pt] A bare glass fiber touching dust, moisture, scratches, or another fiber would leak light at every contact point. Surrounding the core with a clean, protective glass cladding of index n₂ guarantees a pristine reflection interface along the entire cable.

What is Numerical Aperture (NA)?

[Pt] Numerical Aperture NA = √(n₁² − n₂²) measures the light-gathering ability of the fiber. It equals sin(θ_max), where θ_max is the half-angle of the green acceptance cone at the entrance face within which entering light will successfully undergo total internal reflection.

[Pt] Why shouldn't you bend a fiber-optic cable around a sharp corner?

[Pt] At a sharp bend, the outer wall of the core curves toward the oncoming ray, reducing the angle θᵢ between the ray and the wall normal. Once θᵢ falls below the critical angle θc, light refracts out into the cladding and the signal weakens.

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