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Next exam on November 14 – 15, 2026:
All concepts

Refraction and Total Internal Reflection

光的折射与全反射

The law of refraction and refractive index, the wavelength and speed expressions for refractive index, the condition for total internal reflection and the critical angle, and how optical fibres work.

6 real-exam questions · 4 sittings12 in mock papers60 practice

Formulas

  • Law of refraction (Snell's law)

    Law
    n=sin⁡θ1sin⁡θ2n=\frac{\sin\theta_1}{\sin\theta_2}

    Applies when：θ₁ is the angle in vacuum (air) and θ₂ in the medium; both measured from the normal

  • Refractive index in terms of speed

    Formula
    n=cvn=\frac{c}{v}

    Applies when：n>1, so light is slower in a medium; n varies with wavelength in the same medium

  • Critical angle for total internal reflection

    Formula
    sin⁡C=1n\sin C=\frac{1}{n}

    Applies when：Total internal reflection occurs only from denser to less dense medium at angles above C

How the real exam tests it

Questions from real CSCA papers that test this concept. Work each one out before you open the answer.

  1. CSCA Physics December 2025 Past Paper
    When light enters a medium from air, the angle of incidence i=60∘\mathrm { i } = 60 ^ { \circ } and the angle of refraction r=30∘\mathrm { r } = 30 ^ { \circ }. If the refractive index of air is n=1n = 1, then the refractive index of the medium is ( ).
    • A．2\sqrt { 2 }
    • B．3\sqrt { 3 }
    • C．5\sqrt { 5 }
    • D．1.5
    Show answer & solution

    Answer: B

    1 Snell's law: n1sin⁡i=n2sin⁡rn_1 \sin i = n_2 \sin r. 2 1×sin⁡60∘=n2×sin⁡30∘=>(3/2)=n2×(1/2)=>n2=31 \times \sin 60 ^ { \circ } = \mathrm { n } _ { 2 } \times \sin 30 ^ { \circ } = > ( \sqrt { 3 } / 2 ) = \mathrm { n } _ { 2 } \times ( 1 / 2 ) = > \mathrm { n } _ { 2 } = \sqrt { 3 }.
    See it in the full CSCA Physics December 2025 Past Paper
  2. CSCA Physics March 2026 Past Paper
    Light enters glass from air: incidence 45∘45 ^ { \circ }, refraction 30∘30 ^ { \circ }. The refractive index is
    • A．2\sqrt { 2 }
    • B．3\sqrt { 3 }
    • C．1.5
    • D．2
    Show answer & solution

    Answer: A

    n=sin⁡45∘/sin⁡30∘=(2/2)/(1/2)=2≈1.414n = \sin 45 ^ { \circ } / \sin 30 ^ { \circ } = ( \sqrt { 2 } / 2 ) / ( 1 / 2 ) = \sqrt { 2 } \approx 1.414. Different from Q16: different angles yield different indices. Answer: A.
    Take the full CSCA Physics March 2026 Past Paper — timed & graded
  3. CSCA Physics April 2026 Past Paper
    When light passes from an optically denser medium to a less dense medium with an angle of incidence greater than the critical angle, what occurs is
    • A．total internal reflection
    • B．refraction
    • C．reflection
    • D．scattering
    Show answer & solution

    Answer: A

    When the angle of incidence exceeds the critical angle in this geometry, total internal reflection occurs and all light returns to the denser medium.
    Take the full CSCA Physics April 2026 Past Paper — timed & graded

Real-exam evidence

Recorded past-paper appearances are listed below; online verification is labelled separately.

6 questions across 4 sittings1 of 4 sittings available to verify free online
  • CSCA Physics April 2026 Past Paper1 questionreading page not yet public
  • CSCA Physics March 2026 Past Paper2 questionsreading page not yet public
  • CSCA Physics January 2026 Past Paper2 questionsreading page not yet public
  • CSCA Physics December 2025 Past Paper1 questionView question 20 and answer free (full paper)

12 questions across 6 mock papers — our own papers, so a coverage signal rather than real-exam evidence

60 related practice questions availablePractise free: Geometric Optics

Sitting evidence comes from the CSCA past papers on record; questions whose original paper is missing are excluded, and sprint papers are deliberately not counted.