An optical fiber is 1.0 metre long and has a diameter of 20 μm. Its ends are perpendicular to its axis. Its index of refraction is 1.30. What is the maximum number of reflections a light ray entering one end will make before it emerges from the other end?单项选择题
A
42,000
B
26,000
C
24,000
D
18,000

Question

An optical fiber is 1.0 metre long and has a diameter of 20 μm. Its ends are perpendicular to its axis. Its index of refraction is 1.30. What is the maximum number of reflections a light ray entering one end will make before it emerges from the other end?单项选择题

A

42,000

B

26,000

C

24,000

D

18,000

BlackTom AI · Accepted Answer

Critical angle: θc = arcsin(n2/n1) = arcsin(1.00/1.30) = arcsin(0.7692308) = 50.28°.

1) Relevant knowledge points and concepts
- Total internal reflection at the fiber wall requires the internal ray to strike at angles greater than the critical angle; for a ray that propagates making an angle α with the fiber axis, the wall incidence angle = 90° − α. So the largest allowed axial-angle is αmax = 90° − θc.
- Geometry of successive bounces: for a cylindrical fiber of diameter D, a ray at angle α traverses a transverse distance D between opposite wall hits; the axial distance between successive reflections is s = D cot α. The number of reflections over length L is N = L / s = (L tan α) / D.
- Assume the surrounding (cladding/external) index ≈ 1.00 (air).

2) Step-by-step reasoning
- Compute αmax = 90° − θc.
- Compute tan(αmax).
- Compute N = (L tan αmax) / D and round sensibly.

3) Calculations (intermediate steps shown)
- θc = arcsin(1/1.30) = arcsin(0.7692308) = 50.28°.
- αmax = 90° − 50.28° = 39.72°.
- tan(39.72°) = cot(50.28°) = cos(50.28°)/sin(50.28°).
  cos(50.28°) ≈ 0.6389, sin(50.28°) = 0.7692 → tan(39.72°) ≈ 0.6389/0.7692 = 0.8306.
- L = 1.0 m, D = 20 μm = 20 × 10−6 m.
- N = (1.0 m × 0.8306) / (20 × 10−6 m) = 0.8306 / 2.0×10−5 = 41,530 ≈ 4.153 × 10^4.

4) Why this answer is correct
Using the maximum admissible axial angle (just at the TIR limit) maximizes tan α and thus maximizes the number of bounces; the algebraic result gives ≈41,530 reflections, which rounds to 42,000.

5) Why the other choices are incorrect
Options 26,000; 24,000; and 18,000 are much smaller than the value obtained from the geometric/TIR calculation above and do not match the physically maximum-case calculation.

6) Conclusion
The maximum number of reflections is about 4.15 × 10^4, so the correct (rounded) choice is 42,000, matching the provided answer.

An optical fiber is 1.0 metre long and has a diameter of 20 μm. Its ends are perpendicular to its axis. Its index of refraction is 1.30. What is the maximum number of reflections a light ray entering one end will make before it emerges from the other end?单项选择题

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An optical fiber is 1.0 metre long and has a diameter of 20 μm. Its ends are perpendicular to its axis. Its index of refraction is 1.30. What is the maximum number of reflections a light ray entering one end will make before it emerges from the other end?单项选择题

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