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Total Internal Reflection

Key idea: O Level total internal reflection: state the conditions (denser to less dense, i greater than c), use sin c = 1/n, and explain optical fibres.

  • SEC G3 Physics 2027
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Learning objectives

  • Describe wave generation by vibrating sources, ropes and springs
  • Describe ripple-tank waves using wavefronts
  • Explain that waves transfer energy
  • Explain that wave energy transfer does not transfer matter
  • Use amplitude, frequency and wavelength to describe wave motion
  • Define and use wave speed and period and interpret wave graphs
  • Recall and apply wave speed = frequency × wavelength
  • Compare transverse and longitudinal waves and give examples
  • Explain sound production by vibration and the need for a medium
  • Describe sound using compressions and rarefactions
  • Relate sound loudness to amplitude and pitch to frequency
  • Explain reflected-sound echoes and use them to measure distance
  • Explain ultrasound use in sonar and soft-tissue scanning
  • Use the normal, angle of incidence and angle of reflection
  • Apply the law of reflection in constructions, measurements and calculations
  • Use the normal, angle of incidence and angle of refraction
  • Apply sin i divided by sin r as a constant for a fixed pair of media
  • Define refractive index as vacuum light speed divided by medium light speed
  • Explain the critical angle
  • Explain the conditions for total internal reflection
  • Apply total internal reflection to optical fibres and state advantages
  • Describe how a thin converging lens acts on a light beam
  • Define the focal length of a converging lens
  • Construct real and virtual image ray diagrams for a thin converging lens
  • Describe lens images as real or virtual, magnified or diminished, and upright or inverted

1. Definition

A. Critical angle

The critical angle, c, is the angle of incidence in the denser medium for which the angle of refraction in the less dense medium is 90°.

B. Total internal reflection (TIR)

Total internal reflection is when all the light is reflected back into the higher-index medium, with no transmitted ray in the ray model.

2. Key Ideas

A. Conditions for total internal reflection

TIR occurs only when:

  1. light travels from higher refractive index to lower refractive index (denser → less dense), and
  2. the angle of incidence in the denser medium is greater than the critical angle (i > c).

B. Critical-angle formula (from refraction)

For light going from a medium of refractive index n into air (≈ 1):

sin c = 1/n

More generally, from n₁ to n₂ where n₁ > n₂:

sin c = n₂/n₁

Refraction, critical angle, and total internal reflectionThree ray diagrams show normal incidence into a higher-index medium, the critical-angle case from higher to lower index, and total internal reflection beyond the critical angle.A: Normal incidenceair, lower nglass, higher ndirection unchangedB: At i = clower nhigher ncr = 90°C: At i > clower nhigher nitotal internal reflection
At normal incidence the ray does not bend, although its speed and wavelength still change. From higher to lower refractive index, the refracted ray reaches the boundary at i = c; total internal reflection occurs only for i > c.

3. Detailed Explanations

A. What happens as the incidence angle increases?

For light travelling from a denser medium to a less dense medium:

  • as i increases, the refracted angle r increases (bends further away from the normal)
  • at i = c, the refracted ray travels along the boundary (r = 90°)
  • for i > c, there is no transmitted ray carrying energy into the second medium in the ray model, so the light is totally reflected (TIR)

B. Using TIR in optical fibres (syllabus application)

Optical fibres guide light using repeated total internal reflections.

Structure idea:

  • core: higher refractive index
  • cladding: lower refractive index

Light rays inside the core hit the core–cladding boundary at angles greater than c, so they stay inside the core and travel long distances.

Light guided through an optical fibre by total internal reflectionA ray travels through a higher-index core surrounded by lower-index cladding. At repeated core-cladding boundaries, the incidence angle exceeds the critical angle and the ray reflects back into the core.i > chigher-index corelower-index claddingrepeated total internal reflection keeps the signal in the core
Optical fibre: light is guided by repeated total internal reflections in the higher-index core.

C. Telecommunications

In telecommunications, pulses of light carry digital information through optical fibres. Repeated total internal reflection keeps the pulses inside the higher-index core.

Compared with metal cables, optical fibres can carry more information, lose less signal energy over long distances and are not affected by electromagnetic interference.

D. Medicine

In an endoscope, one bundle of optical fibres carries light into the body and another carries an image back to the observer or camera. The fibres are thin and flexible, so doctors can view internal organs through a small opening instead of making a large incision.

E. Match each advantage to the use

UseUseful advantages
Telecommunicationshigh data capacity, low signal loss, no electromagnetic interference
Medical endoscopethin and flexible, carries light and images around bends, allows less invasive examination

4. Common Mistakes

  • Saying TIR happens when light goes from less dense to more dense (it does not).
  • Using sin c = 1/n without checking that the ray is going into air (or using the general form).
  • Thinking “critical angle” is where reflection starts (there is always some reflection; “critical angle” refers to r = 90°).

5. Exam Tips

  • Always state the two conditions for TIR (direction + i > c).
  • If the question is “find c”, start with sin c = 1/n (or n₂/n₁).
  • Keep your calculator in degree mode for sin⁻¹.
  • For optical fibre questions, use the keywords: core, cladding, higher n, repeated TIR.

6. Worked Examples

Modelled example 1

Critical angle from refractive index

Core

Problem

The refractive index of glass is n = 1.50. Find the critical angle for the glass–air boundary.
Study the worked solution

  1. Select the boundary-specific relationship

    Method

    Use sin c = 1/n for light travelling from glass into air.

    Reason

    At the critical angle, the refracted ray in air is at 90°.

    Working

    sin c = 1/1.50 = 0.666…

  2. Calculate the angle

    Method

    Use inverse sine in degree mode.

    Reason

    The relationship gives the sine of the angle, not the angle itself.

    Working

    c = sin⁻¹ (0.666…) ≈ 41.8°

Guided practice 2

Does TIR occur?

About 5 min

Problem

Light travels from water (n = 1.33) to air. The angle of incidence in water is 40°. Does total internal reflection occur?

Check direction and angle threshold

Ray outcome

Hints

Hint 1: find the threshold
Use c = sin⁻¹ (1/1.33).
Hint 2: compare strictly
TIR requires i > c, not merely travel from denser to less dense.
View solution step by step

  1. Find the critical angle

    Method

    Calculate the water–air threshold.

    Reason

    The actual incidence angle must be compared with this boundary value.

    Working

    c = sin⁻¹ (1/1.33) ≈ 48.8°

  2. Apply both conditions

    Method

    State that the direction is correct but 40° < 48.8°.

    Reason

    Failure of either TIR condition means a transmitted refracted ray remains.

    Working

    No TIR; the ray refracts into air.

Common misconception 3

Finding n from critical angle

Find and correct the mistake

Learner response

A transparent material has a critical angle of 36° in air. A student writes n = sin 36° and obtains n < 1. Diagnose the rearrangement.

Use the reciprocal relationship

View solution step by step

  1. Make refractive index the subject

    Method

    Take the reciprocal of sin c.

    Reason

    The original equation is sin c = 1/n, not sin c = n.

    Working

    n = 1/(sin c)

  2. Calculate and test plausibility

    Method

    Obtain n ≈ 1.70.

    Reason

    An ordinary material’s refractive index relative to air should exceed 1.

    Working

    n = 1/(sin 36°) ≈ 1.70

Examiner practice 4

Checking TIR in glass

4 marks

Examination question

Light travels from glass (n = 1.50) to air at an incidence angle of 45° in the glass. Determine whether total internal reflection occurs and justify both conditions. [4 marks]

Calculate the threshold and test both conditions

View solution step by step

  1. Calculate the critical angle

    2 marks

    Method

    Use the glass–air critical-angle relationship.

    Reason

    The threshold depends on the refractive indices at this boundary.

    Working

    c = sin⁻¹ (1/1.50) ≈ 41.8°

  2. Check both conditions

    2 marks

    Method

    State higher-to-lower index and 45° > 41.8°.

    Reason

    TIR requires both the correct direction and incidence above the critical angle.

    Working

    Both satisfied → total internal reflection occurs.

Challenge 5

Medical endoscope

Minimal support

Application transfer

Explain how optical fibres make an endoscope useful for examining inside the body. Include the physics that keeps light inside each fibre and one benefit to the patient.

Connect the mechanism to the medical benefit

Hints

Hint 1: guide the light
Compare the refractive indices of the core and cladding.
Hint 2: state the benefit
Think about the size and flexibility of the fibre bundle.
View solution step by step

  1. Explain light guidance

    Method

    State that light undergoes repeated total internal reflection at the core–cladding boundary.

    Reason

    The core has a higher refractive index and the rays meet the boundary above the critical angle.

    Working

    Higher-index core + i > c → repeated TIR.

  2. Connect this to the endoscope

    Method

    Use fibre bundles to carry illumination into the body and an image back out.

    Reason

    The thin, flexible bundle guides light around bends.

    Working

    Small opening → less invasive examination and a smaller incision.

7. Mind Stretchers

Mind stretcher 1: Why refraction “stops”Extension

Why can’t the refracted angle r be greater than 90° at the critical angle?

Show Answer

r = 90° means the refracted ray travels along the boundary. A larger angle would require the ray to refract “back into” the denser medium, which is not possible for refraction into the less dense medium. So beyond this limit, the light reflects back into the denser medium (TIR).

Mind stretcher 2: Improving fibre performanceExtension

Why does having a cladding with a lower refractive index than the core help light stay inside the fibre?

Show Answer

It ensures the light is travelling from higher n (core) to lower n (cladding), so a critical angle exists. Many rays then meet the boundary with i > c, producing repeated total internal reflection.

8. Practice and next step

Practice Time!

Compare i, c, and the ray path in the Light & Lens Explorer, then test your understanding:

Waves Quiz  Structured Waves Practice

Continue to thin converging lenses to apply refraction to image formation.

Continue with the next resource in this course.

Course and syllabus information
Course
SEC G3 Physics
Edition
SEC G3 Physics 2027