Calculate critical angle and refractive index
Use n = 1/sin c for a material–air boundary and check when that special reference applies.
Section P3.2.2 · Refraction of light
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
- Define critical internal incidence.
- 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
At the critical condition, why does the limiting angle relation contain 1?
What you will learn
- Recall and use n = 1/sin c.
- Calculate a critical angle and distinguish it from the reflection angle.
Use the boundary-directed limiting ray
At the critical condition for material-to-air transmission, the refracted angle is 90° and sin 90° = 1. For a material's index relative to air, the limiting relation gives n = 1/sin c. Equivalently sin c = 1/n and c = sin⁻¹(1/n). Here c is measured internally from the normal.
A supplied material index 1.50 gives sin c = 2/3 and c ≈ 41.81°. A ray incident internally at 50° exceeds this threshold and can be totally internally reflected. A ray at 40° is below it. Rounding c too early close to a boundary can blur an angle comparison, so retain unrounded values until the conclusion.
State the reference instead of assuming it
If c = 30° for a material-to-air boundary, n = 1/sin 30° = 2.00. The index is dimensionless. The equation's unit numerator follows the air reference in the school approximation; for a different outside material, the ratio of the two indices is needed rather than inserting the incident material's vacuum index unchanged.
A critical angle requires the incident region to have higher index. An index ratio greater than 1 gives an acute c through the inverse sine. If the proposed incident-to-outside relation has no higher-to-lower change, do not force a critical angle out of an invalid physical setup. The angle formula does not supply transmitted or reflected energy below the threshold.
Sources for this lesson
Pause and explain
For n = 2.00 relative to air, what is c?
Worked example
For a supplied material–air index n = 1.50, calculate c and classify an internal ray at 50°.
Show the worked solution
- sin c = 1/1.50 = 2/3.
- c = sin⁻¹(2/3) ≈ 41.81°.
- 50° > c, so total internal reflection applies at the stated material–air boundary.
Answer c ≈ 41.8°; 50° gives total internal reflection.
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Test this topic →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.
Calculate c for n = 1.50 relative to air.
Give me a hint
Rearrange for sin c, then take inverse sine in degrees.
Compare my reasoning
- sin c = 1/1.50.
- sin c = 2/3.
- c ≈ 41.81°.
c ≈ 41.8°.
Look for these in your work
- Use the reciprocal.
- Give an angle, not its sine.
Calculate the material–air relative index when c = 40°.
Give me a hint
Evaluate the sine before taking its reciprocal.
Compare my reasoning
- sin 40° ≈ 0.642788.
- n = 1/0.642788.
- n ≈ 1.55572, dimensionless.
n ≈ 1.56.
Look for these in your work
- Use degree mode.
- Cancel any invented units.
A material has vacuum index 1.50 but the outside is another material with index 1.20. May 1/sin c be set to 1.50 unchanged?
Give me a hint
The critical condition depends on the incident-to-outside index ratio.
Compare my reasoning
- The outside reference is not air or vacuum.
- The incident-relative-to-outside ratio is 1.50/1.20 = 1.25.
- Thus sin c = 1/1.25 = 0.8 and c ≈ 53.13°.
No; use ratio 1.25, giving c ≈ 53.1° for that supplied boundary.
Look for these in your work
- Identify the outside reference.
- Apply the correct ratio.
Common mistakes
- c is measured internally from the normal.
- Use the index relative to the outside region.
- Compare i with an adequately precise c.
What can you explain now?
With c = 30°, what is the material's index relative to air?
Compare with the explanation
2.00.
n = 1/sin 30° = 1/0.5.
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Make it stick
Calculate c or n, name the outside reference and compare internal incidence with the threshold.
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.