Explain optical fibres in telecommunications
Follow light pulses through a core and explain repeated total internal reflection at its lower-index cladding.
Section P3.2.2 · Refraction of light
Cross-referenced from the related Cambridge IGCSE Chemistry 0620 or Physics 0625 teaching to the 0652 syllabus. Exact 0652 Core/Supplement objective coverage and this final-series course’s examiner fit still need specialist review. Written practice is self-checked and does not predict a Cambridge grade or demonstrate supervised laboratory competence.
Before you begin
- State both conditions for total internal reflection.
- Measure incidence from a local boundary normal.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which index arrangement supports the stated core guidance?
What you will learn
- Describe optical-fibre light guidance.
- Connect controlled light pulses to telecommunications and realistic limits.
Guide light inside a suitable core
An optical fibre has a transparent core with a lower-index surrounding cladding. In the simple ray model, rays meeting the core–cladding boundary from inside the core above its critical angle reflect internally. Repeated reflections guide suitable light along the core. The index difference, local incidence and launch direction are all part of the explanation.
Measure each incidence angle from the normal to the local wall, not from the fibre's long axis. A ray meeting a straight horizontal wall at 15° to the axis has 75° incidence to its vertical normal. Cladding is a real second region; it is misleading to describe every fibre simply as glass surrounded by air.
Connect the path to a transmitted message
Telecommunications can encode information in controlled light pulses from a transmitter and recover it with a receiver. Fibres support high information rates and guide light through a thin flexible path. The pulses carry an encoded signal; the example does not claim that an unmodulated steady beam alone defines a complete message.
Real fibres have losses and limits from absorption, scattering, joints, bending and pulse spreading. An excessive bend can change the local boundary incidence and permit leakage rather than preserve the straight-wall drawing. Total internal reflection in the ideal model is therefore not a claim that a real communications link loses no energy or needs no repeaters or signal processing.
Sources for this lesson
Pause and explain
With c = 70°, a ray is 15° to the axis at a parallel wall. What is its incidence?
Worked example
For a supplied core–cladding critical angle 70°, a ray reaches a straight wall at 15° to the fibre axis. Test guidance there.
Show the worked solution
- The wall normal is perpendicular to the axis.
- Internal incidence = 90° − 15° = 75°.
- 75° > 70°, and the core index is higher, so that contact gives total internal reflection.
Answer 75° incidence gives total internal reflection at the stated wall.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Explain repeated light guidance in a core with lower-index cladding.
Give me a hint
Name the incident region and threshold at each wall contact.
Compare my reasoning
- The ray is inside the higher-index core.
- It meets lower-index cladding above the critical incidence.
- Repeated total internal reflection guides a suitable ray along the core.
Higher core index and above-critical wall incidence give repeated guidance.
Look for these in your work
- State both conditions.
- Use the core–cladding boundary.
With c = 70°, compare rays 10° and 25° to the axis at a straight parallel wall.
Give me a hint
Convert each axis angle to incidence at the wall normal.
Compare my reasoning
- Incidences are 80° and 65°.
- 80° > 70° satisfies the threshold.
- 65° < 70° does not give total internal reflection at that contact.
10° axis ray is above critical; 25° axis ray is below it.
Look for these in your work
- Convert the reference.
- Compare each ray separately.
Explain how a fibre carries a message and why excessive bending can impair it.
Give me a hint
Separate information encoding from boundary guidance.
Compare my reasoning
- The transmitter encodes information in controlled light pulses.
- The receiver detects the arriving signal.
- Bending can change local incidence and increase leakage or loss, so the ideal straight model has limits.
Encoded pulses are detected at the receiver; unsuitable bends can reduce guidance and signal quality.
Look for these in your work
- Identify transmitter and receiver roles.
- Explain a physical limitation.
Common mistakes
- Require a higher-index core than cladding.
- Measure incidence from the wall normal.
- Ideal reflection does not remove all real link losses.
What can you explain now?
Does a 75° angle to the fibre axis mean 75° incidence at a parallel wall?
Compare with the explanation
No.
Its incidence to the wall normal is 15° in that geometry.
After trying it yourself, choose your next review. This is your self-assessment.
Your review choice appears on Today. Sign in to sync it across devices.
Make it stick
Explain fibre guidance, angle references and how pulse transmission differs from a loss-free ideal ray sketch.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.