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Optical Instruments & Total Internal Reflection

Optical Fiber

Step into the high-speed telecom laboratory. Explore fiber optic cable cutaways, adjust refractive indices, launch lasers at different angles, bend the core to observe signal leakage, and inspect medical endoscope views.

Telecom Fiber Optic Laboratory ⓘ

Change incident angles and cable curvature to trace the light pulses bouncing inside the glass core.

Simulation active

Optical Fiber Telemetry

Total Internal Reflection OK
Core index ncore
1.50
Cladding index nclad
1.40
Critical Angle c
68.9°
Numerical Aperture NA
0.54
Signal / Bend Loss
100% / 0%

Physics of Optical Fibers

An optical fiber is a thin, flexible waveguide that transmits light pulses over long distances by trapping them inside a central glass core.

  • Core-Cladding Boundary: The central core has a higher refractive index (ncore) than the surrounding cladding (ncladding).
  • Total Internal Reflection (TIR): When light pulses strike the core boundary at an angle greater than the critical angle (θc), they reflect completely back into the core instead of refracting out.
  • Critical Angle Formula: The critical angle is calculated using Snell\'s Law:

Critical Angle Equation

sin θc = ncladding / ncore

Numerical Aperture (NA)

Numerical Aperture determines the light-collecting power of the fiber core. It defines the range of angles over which light launched into the fiber will be guided by TIR:

NA Formula

NA = √(ncore2 - ncladding2)

Bending Loss

If a fiber cable is bent too sharply (small bend radius), the incident angle of rays at the core boundary drops below the critical angle, causing light to refract out into the cladding and weaken the signal.

Key Applications

Beyond high-speed fiber internet, fibers are used in medical endoscopes (transmitting light and images inside the body) and industrial borescopes to inspect hard-to-reach machinery.

Step-by-Step Solved Problems

Practice calculating critical angles, numerical aperture, and light acceptance limits.

Example 1 Problem Statement

An optical fiber has a core of refractive index 1.52 and cladding of refractive index 1.45. Calculate the critical angle at the core-cladding interface and find the numerical aperture (NA) of the fiber.

View Mathematical Solution Steps
  1. Given parameters: refractive index of core ncore = 1.52, refractive index of cladding ncladding = 1.45.
  2. Recall critical angle formula: sin θc = ncladding / ncore.
  3. Substitute values: sin θc = 1.45 / 1.52 ≈ 0.9539 => θc = arcsin(0.9539) ≈ 72.5°.
  4. Recall Numerical Aperture (NA) formula: NA = √(ncore2 - ncladding2).
  5. Substitute values: NA = √(1.522 - 1.452) = √(2.3104 - 2.1025) = √0.2079 ≈ 0.456.

Final Derived Answer: Critical Angle θc = 72.5°, Numerical Aperture NA = 0.456.

Example 2 Problem Statement

A light ray is launched into an optical fiber core from air (n = 1.0). If the core refractive index is 1.60 and the cladding index is 1.40, calculate the maximum acceptance angle in air for light to be guided by total internal reflection inside the fiber.

View Mathematical Solution Steps
  1. Step 1: Calculate Numerical Aperture (NA): NA = √(1.602 - 1.402) = √(2.56 - 1.96) = √0.60 ≈ 0.775.
  2. Step 2: Recall acceptance angle relation in air: sin αmax = NA / nair = 0.775 / 1.0 = 0.775.
  3. Step 3: Solve for acceptance angle: αmax = arcsin(0.775) ≈ 50.8°.
  4. Light launched from air must enter the flat core face at an angle less than 50.8° relative to the fiber axis to propagate.

Final Derived Answer: Maximum Acceptance Angle αmax = 50.8°.

Example 3 Problem Statement

A fiber optic cable experiences bending loss. If the critical angle at the core-cladding boundary is 75°, and a sharp bend forces the ray incident angle at the boundary to drop to 70°, determine the path of the light ray.

View Mathematical Solution Steps
  1. Compare actual incident angle (70°) with critical angle (75°).
  2. Since the incident angle is less than the critical angle (70° < 75°), total internal reflection fails.
  3. The light ray refracts out of the core into the cladding layer, leading to leakage (signal loss/attenuation).

Final Derived Answer: Light refracts out of the core (leaks), causing signal loss.

Self-Check Questions

Question 1

Why is it necessary for the core refractive index of an optical fiber to be higher than that of the cladding?

Show Answer & Explanation

Total internal reflection can only occur when light travels from an optically denser medium (higher refractive index) to an optically rarer medium (lower refractive index). If the cladding had a higher index than the core, light would refract into the cladding and leak out instead of reflecting back.

Question 2

Explain the physical meaning of Numerical Aperture (NA) in fiber optics.

Show Answer & Explanation

Numerical Aperture represents the light-gathering capability of the optical fiber. It defines the size of the acceptance cone (range of angles) within which incoming light rays will be successfully trapped inside the core by total internal reflection.

Question 3

What is the difference between single-mode and multi-mode optical fibers?

Show Answer & Explanation

Single-mode fibers have a very thin core (typically 8-10 microns) allowing only a single path (mode) of light to travel, minimizing pulse dispersion and making them ideal for long-distance telecom. Multi-mode fibers have wider cores allowing multiple light paths, which is easier to couple but causes pulse dispersion, limiting them to shorter distances.

Question 4

Why does bending an optical fiber too sharply result in a loss of transmission signal?

Show Answer & Explanation

Sharp bends change the geometry of the core-cladding interface relative to the propagating light. The angle of incidence of the bouncing rays becomes smaller than the critical angle, causing light to refract through the boundary and leak out into the cladding.

Question 5

Describe how optical fibers are used in medical endoscopy.

Show Answer & Explanation

An endoscope uses two optical fiber bundles: (1) a light-guide bundle to transmit bright light from an external lamp into the body cavity to illuminate tissue, and (2) an image-guide bundle (or digital sensor at the tip) to carry the reflected light/images back to a camera display for the physician.

Question 6

Why are optical fibers completely immune to electromagnetic interference (EMI)?

Show Answer & Explanation

Optical fibers transmit signals as light waves (photons) through non-conductive glass or plastic cores, unlike copper wires which transmit electrical signals (electrons) that are vulnerable to magnetic fields and electrical noise.