Core, cladding and optical-fibre guidance

FREE PHYSICS REVISION · UNIT 3 · A2.1.4

Pearson BTEC Level 3 AAQ Applied Science · Issue 7 (May 2026)

What you will learn

Effect of cladding index on critical angle; core guidance and practical losses.

Specification mapping: A2.1.4

Prerequisites: A2.1.3

Physical principles

An optical fibre has a transparent core surrounded by cladding with a slightly lower refractive index. Light in the core can meet the core–cladding boundary above its critical angle and remain guided by TIR. Cladding provides a controlled optical boundary, protects the core and helps isolate it from neighbouring fibres and surface contamination.

The relevant critical angle satisfies sin C = ncladding/ncore. Since cladding index is usually much closer to the core index than air is, the core–cladding critical angle is larger than the core–air value. Raising cladding index towards core index raises C, making the TIR angular condition harder to meet. Equal indices remove this TIR-guiding boundary.

A ray diagram is an ideal model. Real fibres attenuate through absorption and scattering; excessive bending can allow radiation loss. Different paths can have different travel times, broadening a pulse in a multimode fibre. A suitable core design reduces dispersion, but cladding does not make every possible entry ray guided. Entry acceptance depends on geometry and both indices.

Scientific diagram and interpretation

Light follows a schematic zigzag path inside a higher-index core. At each core–cladding reflection incidence must exceed the relevant critical angle. Not all incident rays are guided.
Light follows a schematic zigzag path inside a higher-index core. At each core–cladding reflection incidence must exceed the relevant critical angle. Not all incident rays are guided.

Worked examples and applications

Index comparison

For ncore = 1.50, air gives C = 41.8°, whereas ncladding = 1.45 gives C = 75.2°. A core ray incident at 70° would satisfy the air case but not this cladding case.

Guidance decision

If C = 75.2° and the core ray incidence to the normal is 80°, it can undergo TIR. An angle of 10° to the boundary surface corresponds to 80° to its normal.

Common misconceptions

Cladding must have lower index than the core; a protective layer with arbitrary index is not enough.

Examination technique

State how the ratio of indices changes C before deciding whether a ray remains guided.

For an explanation, connect the physical cause to its effect. For a calculation, state the model, rearrange before substitution, retain intermediate digits, and finish with units. For a data question, refer to the actual readings and distinguish a trend from proof of a cause.

Check your understanding

Five questions. Choose an option, then Check Answer. Read the feedback and use Try Again to reset. Use Tab between controls, arrow keys within radio groups, and Enter or Space on buttons. These local practice answers are not saved.

Question 1 of 5 · A2.1.4

For TIR guidance the core index should be…

Question 2 of 5 · A2.1.4

Replacing air with cladding closer in index to the core makes C…

Question 3 of 5 · A2.1.4

For C = 75°, incidence 80° to the normal gives…

Question 4 of 5 · A2.1.4

What can excessive fibre bending cause?

Question 5 of 5 · A2.1.4

Why can long multimode fibres broaden pulses?

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