Explain internal reflection and total internal reflection
Use both threshold conditions to explain block observations and everyday optical examples.
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 the critical angle.
- Use the local reflection law.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which pair of conditions permits total internal reflection?
What you will learn
- Describe partial internal reflection and total internal reflection.
- Apply the conditions to experimental and everyday examples.
Apply both necessary conditions
Internal reflection occurs when some light travelling within a transparent region reflects at its boundary. Below the critical angle there can be both an internal reflected ray and a refracted ray. Total internal reflection requires light incident from a higher-index region toward a lower-index region and an incidence angle greater than the critical angle.
The internally reflected ray obeys i = r for reflection, using the same local normal inside the original region. A large angle by itself is insufficient: an air-to-glass incident ray does not acquire total internal reflection merely by reaching a large air-side incidence angle. At c the transmitted ray is boundary-directed rather than an above-critical case.
Explain examples with the same boundary test
With a semicircular block and radial entry, increase incidence at the glass–air flat face. Below c observe both transmitted and internally reflected paths; above c the ray stays inside by reflection. A suitable glass prism can redirect light by total internal reflection when its internal geometry gives the required boundary direction and angle.
A swimmer looking upward can see some directions through a water–air surface while other directions show internally reflected underwater surroundings. The difference follows the water-side incidence relative to its critical angle. Bright diamond paths and light guides can also involve internal reflections, but a shiny surface alone does not prove that every path satisfies total internal reflection.
Sources for this lesson
Pause and explain
A glass–air ray is exactly at c. Is it the above-critical case?
Worked example
For a supplied glass–air c = 42°, explain an internal ray at i = 50° and give its reflection angle.
Show the worked solution
- Incidence is from higher-index glass toward air.
- 50° is greater than c, so the transmitted ray is absent in the school-ray model.
- The internal reflected angle equals the incidence angle, 50°.
Answer Total internal reflection; reflected angle 50° inside glass.
Try a topic test
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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.
State both conditions for total internal reflection.
Give me a hint
One condition is the region change; the other is the angle comparison.
Compare my reasoning
- Light approaches from the higher-index region.
- The other region has lower refractive index.
- The incidence angle exceeds the corresponding critical angle.
Higher to lower index, with i > c.
Look for these in your work
- Include both conditions.
- Use a strict angle inequality.
With c = 41°, classify a glass–air ray at i = 35° and state whether a reflected ray may exist.
Give me a hint
Below the threshold is not the same as no reflection.
Compare my reasoning
- 35° < 41°.
- A refracted ray can leave the glass.
- Partial internal reflection can occur at the same time.
Below critical: refraction with possible partial internal reflection.
Look for these in your work
- Compare the angle.
- Distinguish partial from total.
Explain why a swimmer can see a reflected underwater view for some upward directions.
Give me a hint
Apply the water-to-air boundary conditions to different incidence directions.
Compare my reasoning
- The light is incident from water toward lower-index air.
- Some water-side directions exceed the critical angle.
- Those rays reflect internally rather than propagate into air in the ray model.
Above-critical water-side rays give an internally reflected underwater view.
Look for these in your work
- Name the interface direction.
- Link the observation to incidence angle.
Common mistakes
- Require higher index to lower index.
- Require i greater than c, not merely equal.
- Internal reflection below c can coexist with refraction.
What can you explain now?
Does observing an internally reflected ray alone prove total internal reflection?
Compare with the explanation
No.
Partial internal reflection can occur below c while another ray transmits.
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Make it stick
Distinguish partial and total internal reflection and test both conditions in a real example.
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.