Calculate refractive index from light speeds
Define the incident-to-transmitted speed ratio and preserve which region is the reference.
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
- Separate speed from frequency at an interface.
- Rearrange a ratio and cancel consistent units.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which ratio defines a vacuum-reference material index?
What you will learn
- Define refractive index as a ratio of wave speeds.
- Calculate speed and index for stated transparent regions.
Identify the reference region
For the index of a material relative to vacuum, n = c₀/v, where c₀ is the vacuum light speed and v is the material light speed for the light considered. An index is a dimensionless ratio because both speeds use the same unit. Ordinary glass with supplied speed 2.0 × 10⁸ m/s and vacuum speed 3.0 × 10⁸ m/s has n = 1.50.
More generally the refractive index of the second region relative to the first is v₁/v₂. Name that order: reversing it gives the reciprocal relative index. 'Optically denser' refers to higher index and lower light speed, not a claim that mass density alone determines the speed.
Rearrange and check the interpretation
For a vacuum-reference index, v = c₀/n. With n = 1.25 and c₀ = 3.0 × 10⁸ m/s, v = 2.4 × 10⁸ m/s. A larger index gives a smaller speed at a specified frequency. At a stationary interface frequency stays unchanged; wavelength changes with speed rather than the source acquiring a new frequency.
Use the supplied reference speed and consistent units before dividing. If the incident and transmitted speeds are 2.4 × 10⁸ and 1.8 × 10⁸ m/s, the second-relative-to-first index is 4/3, about 1.33. It is not automatically the second material's vacuum-reference index. Material index can depend on frequency, so a supplied value applies to the stated light.
Pause and explain
For c₀ = 3.0 × 10⁸ m/s and v = 2.4 × 10⁸ m/s, what is n?
Worked example
Light speed is 2.0 × 10⁸ m/s in a material. Use c₀ = 3.0 × 10⁸ m/s to find its vacuum-reference index.
Show the worked solution
- Use n = c₀/v.
- Cancel the common factor 10⁸ and the speed units.
- n = 3.0/2.0 = 1.50.
Answer n = 1.50, with no unit.
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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 for v = 2.0 × 10⁸ m/s and c₀ = 3.0 × 10⁸ m/s.
Give me a hint
Keep the vacuum reference in the numerator.
Compare my reasoning
- n = c₀/v.
- n = 3.0/2.0.
- The speed units cancel.
n = 1.50.
Look for these in your work
- Use the correct ratio.
- Give a dimensionless answer.
Find material speed for n = 1.60 and c₀ = 3.0 × 10⁸ m/s.
Give me a hint
Rearrange n = c₀/v before substituting.
Compare my reasoning
- v = c₀/n.
- v = 3.0 × 10⁸/1.60.
- v = 1.875 × 10⁸ m/s.
1.875 × 10⁸ m/s, approximately 1.88 × 10⁸ m/s.
Look for these in your work
- Rearrange correctly.
- Retain the speed unit.
Speeds in regions 1 and 2 are 2.4 × 10⁸ and 1.8 × 10⁸ m/s. Find the second-relative-to-first index and its reciprocal.
Give me a hint
State the region order before forming each ratio.
Compare my reasoning
- n₂ relative to 1 = v₁/v₂ = 2.4/1.8 = 4/3.
- The reverse ratio is v₂/v₁ = 3/4.
- Neither ratio alone gives both vacuum-reference indices.
Second relative to first 4/3; reverse 3/4.
Look for these in your work
- Name both reference directions.
- Avoid an unstated vacuum assumption.
Common mistakes
- Name the index's reference region.
- Index is dimensionless.
- Higher index means lower speed for the stated light.
What can you explain now?
Is a second-relative-to-first index always a vacuum-reference index?
Compare with the explanation
No.
That occurs only when the first reference is vacuum or a stated air approximation.
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Make it stick
Define refractive index with its reference, rearrange the speed ratio and explain its units.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.