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An optical fibre is a thin, flexible, transparent strand of glass or plastic that guides light from one end to the other by total internal reflection.
core refractive index μ₁ > cladding refractive index μ₂
Concept
Light launched within the acceptance cone enters the core and repeatedly reflects at the core-cladding boundary. Since each incidence angle is greater than the critical angle, the ray remains trapped inside the core.
NCERT Notes
Optical fibres are based on total internal reflection.
The core must be optically denser than cladding.
Information can be transmitted using light pulses.
Optical fibre structure: core, cladding and protective jacket.
2–4. Construction, Principle and Working of Optical Fibre
Construction
The central core is made of glass or plastic with refractive index μ₁. It is surrounded by cladding of refractive index μ₂, where μ₂ is slightly smaller than μ₁. A protective jacket prevents mechanical damage.
Real-life application: fibre internet cables, medical endoscopes and sensors.
Principle
The working principle is total internal reflection. When a ray inside the denser core strikes the core-cladding boundary at an angle greater than the critical angle, it reflects completely back into the core.
sin C = μ₂ / μ₁
Working
Light pulses are launched into the core within the acceptance cone. These pulses travel in zig-zag or nearly straight paths depending on fibre type and emerge at the other end with the information signal.
Common Mistake
Students often write that the ray reflects from the outer jacket. Correctly, TIR occurs at the core-cladding boundary.
JEE Tip
Always compare the incidence angle at the core-cladding boundary with the critical angle, not the angle at the entrance face.
NEET Tip
Remember the direct formula for air: NA = √(μ₁² − μ₂²).
Total internal reflection inside the fibre core.
Numerical Aperture (Most Important)
Numerical aperture is the most exam-important quantity for optical fibre. It tells how much light the fibre can accept and guide by TIR.
Acceptance cone, acceptance angle and numerical aperture.
Acceptance Angle
The maximum angle with the fibre axis at which an incident ray can enter the fibre and still undergo total internal reflection inside the core is called the acceptance angle.
iₘₐₓ = acceptance angle
Numerical Aperture
Numerical aperture is defined as the product of refractive index of outside medium and sine of the maximum acceptance angle.
NA = μ₀ sin iₘₐₓ
Critical Angle
At the core-cladding boundary, the limiting ray is incident at the critical angle.
sin C = μ₂ / μ₁
Relation
A larger numerical aperture means a wider acceptance cone and better light-gathering ability.
larger NA ⇒ larger iₘₐₓ
Complete Derivation
Let μ₀ be the refractive index of outside medium, μ₁ of core and μ₂ of cladding.
For the limiting ray, angle inside the core is r and the incidence angle at core-cladding boundary is C.
From geometry, r = 90° − C.
By Snell's law at the entrance face: μ₀ sin i = μ₁ sin r.
Since r = 90° − C, sin r = cos C.
Therefore μ₀ sin i = μ₁ cos C.
At the critical condition, sin C = μ₂/μ₁.
So cos C = √(1 − sin²C) = √(1 − μ₂²/μ₁²).
Hence μ₀ sin i = μ₁ √(1 − μ₂²/μ₁²).
Therefore:
μ₀ sin i = √(μ₁² − μ₂²)NA = μ₀ sin iₘₐₓ = √(μ₁² − μ₂²)
For air: μ₀ = 1, so
NA = sin iₘₐₓ = √(μ₁² − μ₂²)
5–12. Fibre Diagrams, Types, Systems, Endoscopy, Advantages and Limitations
Step-index, graded-index, single-mode and multi-mode fibres.
Step-Index Fibre
Core refractive index is nearly constant and changes suddenly at the cladding boundary.
Graded-Index Fibre
Refractive index gradually decreases from the centre of the core toward the cladding, reducing pulse spreading.
Single-Mode Fibre
Thin core allows only one mode, giving very low dispersion and long-distance communication.
Multi-Mode Fibre
Thicker core allows many modes; useful for shorter distances but has greater modal dispersion.
Advantages
Low loss
Large bandwidth
No electromagnetic interference
Secure communication
Light weight
Limitations
Fragile glass core
Precise alignment required
Splicing needs skill
Initial installation can be expensive
Optical fibre communication system.
Endoscope working using fibre bundles.
Communication Systems
Electrical signals are converted into light pulses by a laser or LED. These pulses travel through the fibre and are converted back into electrical signals by a photodiode receiver.
Medical Applications and Endoscopy
Fibre bundles carry light into internal organs and return images to a camera. This allows minimally invasive diagnosis and surgery.
13. Numerical Problems
Problems cover acceptance angle, numerical aperture, critical angle, refractive index and signal transmission.
Basic 1. A fibre has μ₁ = 1.50 and μ₂ = 1.47. Find NA in air.
Given: μ₁ = 1.50, μ₂ = 1.47, μ₀ = 1
NA = √(μ₁² − μ₂²)
Substitution: NA = √(1.50² − 1.47²)
Calculation: NA = √(2.2500 − 2.1609) = √0.0891 = 0.2985
Final Answer: NA ≈ 0.30
Basic 2. If NA = 0.30 in air, find acceptance angle.
Given: NA = 0.30, μ₀ = 1
NA = sin iₘₐₓ
Substitution: sin iₘₐₓ = 0.30
Calculation: iₘₐₓ = sin⁻¹(0.30) = 17.46°
Final Answer: 17.46°
Basic 3. Core index is 1.48 and cladding index is 1.46. Find critical angle.
Given: μ₁ = 1.48, μ₂ = 1.46
sin C = μ₂/μ₁
Substitution: sin C = 1.46/1.48
Calculation: C = sin⁻¹(0.9865) = 80.57°
Final Answer: 80.57°
Medium 1. μ₁ = 1.62, μ₂ = 1.52. Find NA and acceptance angle in air.
Given: μ₁ = 1.62, μ₂ = 1.52
NA = √(μ₁² − μ₂²)
Substitution: NA = √(1.62² − 1.52²)
Calculation: NA = √(2.6244 − 2.3104) = √0.314 = 0.560
Final Answer: NA = 0.560, iₘₐₓ = sin⁻¹0.560 = 34.1°
JEE Advanced Conceptual Question. Why is single-mode fibre used for long-distance communication?
Single-mode fibre allows one mode, reducing modal dispersion and pulse broadening.
Answer: Low modal dispersion
International Curriculum Section
IB Physics
IB Physics Conceptual Question. Why does light remain trapped inside the fibre?
Because light in the denser core strikes the core-cladding boundary at angles greater than critical angle, producing total internal reflection.
Answer: TIR at core-cladding boundary
IB Physics Application Question. Name one use of optical fibres.
Examples include internet communication, medical endoscopy, sensors and defence communication.
Answer: Internet communication / endoscopy / sensors
IGCSE Physics
IGCSE Physics Conceptual Question. Why does light remain trapped inside the fibre?
Because light in the denser core strikes the core-cladding boundary at angles greater than critical angle, producing total internal reflection.
Answer: TIR at core-cladding boundary
IGCSE Physics Application Question. Name one use of optical fibres.
Examples include internet communication, medical endoscopy, sensors and defence communication.
Answer: Internet communication / endoscopy / sensors
ICSE Physics
ICSE Physics Conceptual Question. Why does light remain trapped inside the fibre?
Because light in the denser core strikes the core-cladding boundary at angles greater than critical angle, producing total internal reflection.
Answer: TIR at core-cladding boundary
ICSE Physics Application Question. Name one use of optical fibres.
Examples include internet communication, medical endoscopy, sensors and defence communication.
Answer: Internet communication / endoscopy / sensors
British Curriculum
British Curriculum Conceptual Question. Why does light remain trapped inside the fibre?
Because light in the denser core strikes the core-cladding boundary at angles greater than critical angle, producing total internal reflection.
Answer: TIR at core-cladding boundary
British Curriculum Application Question. Name one use of optical fibres.
Examples include internet communication, medical endoscopy, sensors and defence communication.
Answer: Internet communication / endoscopy / sensors
A-Level Physics
A-Level Physics Conceptual Question. Why does light remain trapped inside the fibre?
Because light in the denser core strikes the core-cladding boundary at angles greater than critical angle, producing total internal reflection.
Answer: TIR at core-cladding boundary
A-Level Physics Application Question. Name one use of optical fibres.
Examples include internet communication, medical endoscopy, sensors and defence communication.
Answer: Internet communication / endoscopy / sensors
AP Physics
AP Physics Conceptual Question. Why does light remain trapped inside the fibre?
Because light in the denser core strikes the core-cladding boundary at angles greater than critical angle, producing total internal reflection.
Answer: TIR at core-cladding boundary
AP Physics Application Question. Name one use of optical fibres.
Examples include internet communication, medical endoscopy, sensors and defence communication.
Answer: Internet communication / endoscopy / sensors
14. Previous Year Questions Section
NEET PYQs
NEET PYQs 1. State the principle of optical fibre.
Optical fibre works on total internal reflection.
Answer: Total internal reflection
NEET PYQs 2. Write the expression for numerical aperture in air.
NA = √(μ₁² − μ₂²)Answer: NA = √(μ₁² − μ₂²)
NEET PYQs 3. Why should μ₁ be greater than μ₂?
For TIR, light must go from denser core to rarer cladding. Therefore μ₁ must be greater than μ₂.
Answer: To satisfy TIR condition
NEET PYQs 4. What happens when incident ray is outside acceptance cone?
It is not guided properly by the fibre and may leak out through cladding.
Answer: Ray is not guided
JEE Main PYQs
JEE Main PYQs 1. State the principle of optical fibre.
Optical fibre works on total internal reflection.
Answer: Total internal reflection
JEE Main PYQs 2. Write the expression for numerical aperture in air.
NA = √(μ₁² − μ₂²)Answer: NA = √(μ₁² − μ₂²)
JEE Main PYQs 3. Why should μ₁ be greater than μ₂?
For TIR, light must go from denser core to rarer cladding. Therefore μ₁ must be greater than μ₂.
Answer: To satisfy TIR condition
JEE Main PYQs 4. What happens when incident ray is outside acceptance cone?
It is not guided properly by the fibre and may leak out through cladding.
Answer: Ray is not guided
JEE Advanced PYQs
JEE Advanced PYQs 1. State the principle of optical fibre.
Optical fibre works on total internal reflection.
Answer: Total internal reflection
JEE Advanced PYQs 2. Write the expression for numerical aperture in air.
NA = √(μ₁² − μ₂²)Answer: NA = √(μ₁² − μ₂²)
JEE Advanced PYQs 3. Why should μ₁ be greater than μ₂?
For TIR, light must go from denser core to rarer cladding. Therefore μ₁ must be greater than μ₂.
Answer: To satisfy TIR condition
JEE Advanced PYQs 4. What happens when incident ray is outside acceptance cone?
It is not guided properly by the fibre and may leak out through cladding.
Answer: Ray is not guided
CBSE PYQs
CBSE PYQs 1. State the principle of optical fibre.
Optical fibre works on total internal reflection.
Answer: Total internal reflection
CBSE PYQs 2. Write the expression for numerical aperture in air.
NA = √(μ₁² − μ₂²)Answer: NA = √(μ₁² − μ₂²)
CBSE PYQs 3. Why should μ₁ be greater than μ₂?
For TIR, light must go from denser core to rarer cladding. Therefore μ₁ must be greater than μ₂.
Answer: To satisfy TIR condition
CBSE PYQs 4. What happens when incident ray is outside acceptance cone?
It is not guided properly by the fibre and may leak out through cladding.
Answer: Ray is not guided
Most Important Formulas Revision Sheet
Critical Anglesin C = μ₂/μ₁
Numerical ApertureNA = μ₀ sin iₘₐₓ
NA in AirNA = √(μ₁² − μ₂²)
Acceptance Angleiₘₐₓ = sin⁻¹(NA/μ₀)
Speed in Mediumv = c/μ
Time Delayt = L/v
Lossloss = attenuation × length
TIR Conditioni > C
Quick Revision Table
Point
Must Remember
Exam Use
Principle
Total internal reflection
Definition questions
Construction
Core μ₁, cladding μ₂, μ₁ > μ₂
Reasoning questions
Acceptance Angle
Maximum input angle for guided ray
Numericals
Numerical Aperture
Light-gathering ability
NEET/JEE direct formula
Single-mode
Low dispersion
Communication applications
Endoscopy
Fibre bundles carry light and image
Application questions
Common Mistakes, Exam Tips, NEET Strategy and JEE Strategy
Common Mistakes
Using μ₂/μ₁ incorrectly.
Forgetting μ₁ must be greater than μ₂.
Confusing acceptance angle with critical angle.
Writing TIR occurs at the jacket instead of core-cladding boundary.
Exam Tips
Draw core and cladding clearly.
Label μ₁ and μ₂.
For air, use NA = √(μ₁² − μ₂²).
For outside medium, use NA = μ₀ sin iₘₐₓ.
NEET Strategy
Memorise the direct formulae, definitions and applications. Most NEET questions are concept plus one-step formula based.
JEE Strategy
Practise derivation, limiting ray geometry, Snell's law at entrance face and multi-step refractive index problems.
NCERT Notes
Optical fibre is a strong application of total internal reflection in communication and medical imaging.
Real-Life Applications
Internet cables, endoscopes, aircraft sensors, defence networks, submarine cables and decorative lighting.
Need Help In Optical Fibre?
If you have doubts in Optical Fibre, Ray Optics, NEET Physics, JEE Physics, IB Physics, IGCSE Physics, AP Physics or A-Level Physics, contact Kumar Sir for one-to-one online Physics classes.
