Use the sine ratio for light refraction
Calculate index or a transmitted angle from normal-referenced incidence and refraction angles.
Section 3.2.2 · Refraction of light
Exam scope and assessment notes
Extended only (Supplement) · Physics 0972, 2026–2028. Follow the selected Light section; ray diagrams use an ideal thin-lens or plane-boundary model. Written work is self-checked. Human teacher review, supervised practical competence and cumulative assessment remain pending. Cambridge states that IGCSE 0972 has the same subject content as IGCSE 0625, with 9–1 grading. This StudyForA pathway maps the 2026–2028 syllabus only. This 9–1 graded qualification is available only in a limited number of administrative zones. Confirm that your centre can enter you for this code and exam series with Cambridge International. Cambridge has published a separate syllabus for 2029. StudyForA has not reconciled that version yet; this page supports exams in 2026, 2027 and 2028 only.
Before you begin
- Measure both angles from the local normal.
- Use sine and inverse sine in degree mode.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which operation belongs in the refraction index equation?
What you will learn
- Recall and use n = sin i / sin r.
- Use degree-mode trigonometry and the correct region order.
Use the matching speed and angle ratio
For index of the transmitted region relative to the incident region, n = sin i / sin r. The angles are measured from the same local normal on their respective sides. For air into glass, the usual air-as-reference approximation gives the glass index. A larger than one ratio corresponds to the slower transmitted region and an oblique ray closer to the normal.
For i = 30° and r = 20°, n = sin 30°/sin 20° = 0.5000/0.3420, approximately 1.46. Do not divide the angles 30/20: the relation uses their sines. State the calculator mode before obtaining numerical sine values and keep enough digits until the final answer.
Rearrange for an angle and check the branch
For a supplied relative index n = 1.50 and i = 30°, sin r = sin i/n = 1/3, so r ≈ 19.47°. The acute physical transmitted angle agrees with bending toward the normal. For a reversed region change the corresponding relative index is reciprocal; do not retain an air-to-glass ratio while relabelling the incident side.
Normal incidence gives sin i = sin r = 0, so the direct measured-angle ratio is 0/0 and cannot determine index from that one path. Speed or a suitable oblique observation is needed. If rearrangement for high-to-low transmission requests sin r greater than 1, there is no propagating refracted ray in this school model: test the critical-angle condition instead.
Pause and explain
For n = 1.50 and i = 30°, what is sin r?
Worked example
Use n = 1.50 for air-to-glass light at i = 30° to find r.
Show the worked solution
- sin r = sin i/n.
- sin r = sin 30°/1.50 = 1/3.
- In degree mode r = sin⁻¹(1/3) ≈ 19.47°.
Answer r ≈ 19.5° to the normal.
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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.
Find n from i = 30° and r = 20°.
Give me a hint
Use sine values rather than the bare degree numbers.
Compare my reasoning
- sin 30° = 0.5000 and sin 20° ≈ 0.3420.
- n = 0.5000/0.3420.
- n ≈ 1.46, dimensionless.
n ≈ 1.46.
Look for these in your work
- Use degree sines.
- Retain the reference order.
Find r for a relative index 1.50 and incidence 45°.
Give me a hint
Divide sin 45° by the index, then use inverse sine.
Compare my reasoning
- sin r = sin 45°/1.50 ≈ 0.471405.
- r = sin⁻¹(0.471405).
- r ≈ 28.13°.
r ≈ 28.1°.
Look for these in your work
- Rearrange before calculation.
- Check bending toward the normal.
A normally entering ray has i = r = 0°. A student declares n = 1 from those angles. Explain the problem.
Give me a hint
Check the denominator before interpreting a ratio.
Compare my reasoning
- Both sines are zero.
- The expression 0/0 is not a defined measured ratio.
- Use speeds or an oblique path to determine index.
The normal path gives an indeterminate ratio, not n = 1.
Look for these in your work
- Identify the undefined ratio.
- Suggest adequate evidence.
Common mistakes
- Use sines, not i/r.
- Use degree mode for supplied degree angles.
- Normal incidence cannot determine n from a 0/0 measured-angle ratio.
What can you explain now?
Can i = r = 0° alone determine n by dividing their sines?
Compare with the explanation
No.
The ratio is indeterminate; it is not evidence for n = 1.
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Make it stick
Calculate with the sine ratio, explain its reference and detect a normal-incidence limitation.
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