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IGCSE · Cambridge (CIE) · 0652

Calculate refractive index from light speeds

Define the incident-to-transmitted speed ratio and preserve which region is the reference.

Section P3.2.1 · Reflection of light

Reflection of light · Refraction of light · Cambridge IGCSE Physical Science 0652 · P3.2.1 / P3.2.2

Cross-referenced from the related Cambridge IGCSE Chemistry 0620 or Physics 0625 teaching to the 0652 syllabus. Exact 0652 Core/Supplement objective coverage and this final-series course’s examiner fit still need specialist review. Written practice is self-checked and does not predict a Cambridge grade or demonstrate supervised laboratory competence.

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.
Original air-to-material ray geometry uses supplied relative index 1.50, incidence 30° and refraction about 19.47°. The speed falls from a supplied 3.0 × 10⁸ to 2.0 × 10⁸ m/s. The transmitted ray bends toward the local normal, with stationary-boundary frequency unchanged. The drawing gives no energy fraction.Air → slower materialSupplied relative index: 1.50i = 30°r ≈ 19.5°Bends toward the normalFrequency stays unchanged
Original air-to-material ray geometry uses supplied relative index 1.50, incidence 30° and refraction about 19.47°. The speed falls from a supplied 3.0 × 10⁸ to 2.0 × 10⁸ m/s. The transmitted ray bends toward the local normal, with stationary-boundary frequency unchanged. The drawing gives no energy fraction.

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.

Sources for this lesson

Pause and explain

For c₀ = 3.0 × 10⁸ m/s and v = 2.4 × 10⁸ m/s, what is n?

Put the idea to work

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
  1. Use n = c₀/v.
  2. Cancel the common factor 10⁸ and the speed units.
  3. n = 3.0/2.0 = 1.50.

Answer n = 1.50, with no unit.

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From guided practice to a new situation

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.

1 · Guided

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
  1. n = c₀/v.
  2. n = 3.0/2.0.
  3. The speed units cancel.

n = 1.50.

Look for these in your work

  • Use the correct ratio.
  • Give a dimensionless answer.
2 · Independent

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
  1. v = c₀/n.
  2. v = 3.0 × 10⁸/1.60.
  3. 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.
3 · Transfer

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
  1. n₂ relative to 1 = v₁/v₂ = 2.4/1.8 = 4/3.
  2. The reverse ratio is v₂/v₁ = 3/4.
  3. 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.
Recall without your notes

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.

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Course scope and assessment

This page maps the 2025–2026 Cambridge IGCSE Physical Science 0652 syllabus: 38 Chemistry sections and 46 Physics sections. Every section links to original 0620/0625 teaching or one of five new 0652 lessons, with guided, independent and transfer practice and explained answers. Cambridge limits entry to Administrative Zone 3 and identifies November 2026 as the final exam series. The section crosswalk is complete as a first coverage map; a numbered-objective and Core/Supplement audit, independent subject-teacher review and cumulative assessment are still needed. Written practice does not award a Cambridge grade. Online activities cannot demonstrate supervised laboratory competence or complete Paper 5/6.

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