Measure light refraction angles at a boundary
Separate incident, refracted and reflected directions using one local normal.
Section 3.2.2 · Refraction of light
Exam scope and assessment notes
Core and Extended · 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
- Recognise a perpendicular normal.
- Distinguish approach and departure arrows.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which ray defines the refraction angle?
What you will learn
- Define normal, angle of incidence and angle of refraction.
- Measure refraction on the transmitted side of the boundary.
Name the path on each side
Refraction describes a change in light propagation when it passes between transparent regions with different speeds. Draw a normal perpendicular to the boundary at the entry point. The incidence angle i is between the approaching ray and that normal. The refraction angle r is between the transmitted ray and the normal in the second region.
Some light can also reflect on the incident side. Its angle is a reflection angle and obeys the reflection law; it is not the transmitted refraction angle. Mark each ray with an arrow and a region label before deciding which two lines form an angle.
Use the normal rather than a surface angle
A transmitted ray making 60° with a plane boundary has a refraction angle of 30°. Surface and normal angles complement to 90°. The terms remain the same for a sloping boundary: rotate the normal with the surface rather than continuing to measure from page vertical.
At a curved surface the normal is the perpendicular to the local tangent. For a circular boundary, it follows the radius through the contact point. Different contact points therefore have different normals. A centre-directed ray reaches that circular surface along its local normal.
Pause and explain
A transmitted ray makes 25° with the boundary. What is r?
Worked example
A transmitted ray makes 60° with a plane boundary. Find its refraction angle.
Show the worked solution
- Draw the perpendicular normal at the entry point.
- The boundary and normal angles total 90°.
- r = 90° − 60° = 30°.
Answer r = 30° to the normal in the transmitted region.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
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.
Find r when the transmitted ray makes 70° with a plane surface.
Give me a hint
Use the perpendicular complement.
Compare my reasoning
- Normal and surface are perpendicular.
- r = 90° − 70°.
- r = 20° in the transmitted region.
20°.
Look for these in your work
- Use the correct region.
- Identify the normal reference.
An incident ray has i = 40° and a transmitted ray has r = 25°. Find their respective surface angles.
Give me a hint
Complement each normal angle separately.
Compare my reasoning
- Incident surface angle = 90° − 40° = 50°.
- Transmitted surface angle = 90° − 25° = 65°.
- These are different surface angles on the two sides.
50° incident; 65° transmitted.
Look for these in your work
- Keep the two paths distinct.
- Do not impose i = r for refraction.
Describe the normal where a ray meets a circular block face.
Give me a hint
Use the circle's local tangent and radius.
Compare my reasoning
- Locate the incidence point.
- Draw the radius from the circle centre through it.
- That radius is perpendicular to the tangent and gives the local normal.
The radius through the contact point is the local normal.
Look for these in your work
- Use a local reference.
- Do not assume page vertical everywhere.
Common mistakes
- Measure both angles from the normal.
- Reflected and refracted rays occupy different regions.
- A circular-surface normal follows the local radius.
What can you explain now?
Is the refraction angle measured on the incident side?
Compare with the explanation
No.
It is the transmitted ray's angle to the normal in the second region.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Label incident, transmitted and reflected rays and measure refraction from a local normal.
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.