Define the critical angle for light
Identify the internal incidence angle that gives a refracted ray along the boundary.
Mapped to the named 0654 section using teaching from the corresponding single-science course. A 0654-specific review of every Core and Supplement objective is still in progress. Written practice is self-checked; it does not demonstrate supervised laboratory competence or award an official Cambridge grade.
Before you begin
- Measure angles from the normal inside the incident region.
- Describe transmission from glass to air.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which side contains the incidence angle called critical?
What you will learn
- State the meaning of critical angle.
- Distinguish the critical condition from total internal reflection.
Follow the internal incidence angle
For light travelling from a higher-index region to a lower-index region, such as glass to air, the refracted ray bends away from the normal. As the internal incidence angle increases, the refraction angle increases toward 90°. The critical angle c is the internal incidence angle for which the refracted ray is along the boundary, r = 90°.
Measure c from the normal in the higher-index region. It is not the surface angle of the incident ray and is not the 90° refracted angle itself. The two angles belong to different sides of the same interface. The phrase higher index describes the optically slower region for the light under consideration, not simply greater mass density.
Compare below, at and above
Below c there is a refracted ray into the lower-index region as well as partial internal reflection. At c the refracted direction reaches the boundary. Above c the usual school-ray model has no transmitted propagating ray into the second region and gives total internal reflection. Equality at the critical condition must not be described as an angle greater than c.
A semicircular block can isolate the flat-face condition by radial entry through the curved face. Increase the internal incidence angle and observe the emerging direction, then the disappearance of the refracted beam as the critical condition is crossed. The schematic compares conditions; it supplies no newly performed measurement of a particular block's c.
Sources for this lesson
Pause and explain
At the critical angle, what is the refraction angle?
Worked example
A supplied glass–air boundary has c = 42°. Compare internal incidence 35°, 42° and 50°.
Show the worked solution
- 35° is below c: refraction with partial reflection.
- 42° equals c: the refracted direction is along the boundary.
- 50° exceeds c: total internal reflection in the school-ray model.
Answer 35° transmits; 42° is critical; 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.
Define the critical angle for glass-to-air transmission.
Give me a hint
Identify both the incident reference and transmitted limiting direction.
Compare my reasoning
- Measure incidence inside glass from the normal.
- Increase it until refraction is along the boundary.
- That internal incidence gives r = 90° and is c.
Internal incidence angle giving a 90° refracted angle.
Look for these in your work
- Name the incident region.
- Use the normal reference.
For a supplied c = 40°, classify internal incidence 30°, 40° and 55°.
Give me a hint
Compare each incidence with the same supplied threshold.
Compare my reasoning
- 30° < c gives transmission and partial reflection.
- 40° = c gives the boundary-directed refracted ray.
- 55° > c gives total internal reflection.
Below critical, at critical, then total internal reflection.
Look for these in your work
- Distinguish equality from exceeding.
- Retain the required direction.
A sketch labels 90° on the air side as the critical angle. Correct it.
Give me a hint
The 90° transmitted angle and internal critical incidence are separate.
Compare my reasoning
- The air-side transmitted angle is r = 90°.
- The critical angle is the glass-side incidence to its normal.
- Relabel the internal incidence c.
90° is the refraction angle; c is the corresponding internal incidence angle.
Look for these in your work
- Identify both angles.
- Correct the reference and region.
Common mistakes
- Critical angle is an internal normal angle.
- At c, the refracted angle is 90°.
- The higher-index to lower-index direction is required.
What can you explain now?
Does an air-to-glass ray have the same critical-angle exit condition?
Compare with the explanation
No.
The critical condition for total internal reflection requires incidence from the higher-index side.
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Make it stick
Define c with the correct region and normal, then compare below, at and above it.
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.