✅ Two conditions, both needed
1. The light is travelling from a denser material towards a less dense one, such as glass to air or water to air.
2. The angle of incidence is greater than the critical angle.
Light and Sound » Total Internal Reflection and the Critical Angle
What you'll learn this session
Study time: 30 minutes
Edexcel spec: 3.21, 3.22
- What the critical angle c means
- When light is totally internally reflected, and when it is not
- How a semi-circular block shows the critical angle
- How to use sin c = 1 ÷ n to find c or n
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Unlock This CourseLight trying to leave a denser material
Remember, a ray that passes from glass or water out into air bends away from the normal. So the angle in the air is always bigger than the angle inside the material. This lesson asks a simple question: what happens as you keep making the angle inside bigger?
The table shows a ray inside glass of refractive index 1.5 meeting a flat glass-air boundary. All angles are measured from the normal.
| Angle of incidence inside the glass in ° | Angle of refraction in the air in ° |
|---|---|
| 10 | 15 |
| 20 | 31 |
| 35 | 59 |
| 40 | 75 |
| 41.8 | 90 |
| 45 | no refracted ray |
The angle in the air races ahead. At 41.8° inside the glass it reaches 90°, so the refracted ray runs along the surface. It cannot get any bigger than 90° and still be in the air. Beyond that angle no light leaves the glass at all.
Key terms:
- Critical angle (c): the angle of incidence inside the denser material for which the angle of refraction is 90°.
- Total internal reflection: when all of the light is reflected back inside the denser material and none is refracted out. It happens when the angle of incidence is greater than the critical angle.
- Denser material: in this topic, the material in which light travels more slowly. It has the higher refractive index. Glass, Perspex and water are all denser than air.
Below, at and above the critical angle
🔽 Less than c
Most of the light refracts out into the air, bending away from the normal. A little is reflected back inside, so there is also a weak reflected ray.
📐 Equal to c
The angle of refraction is 90°. The refracted ray travels along the boundary, just skimming the surface.
💡 Greater than c
There is no refracted ray. All of the light is reflected back inside. This is total internal reflection.
Each word in the name tells you something. Total: all of the light is reflected. Internal: it happens inside the denser material. Reflection: the ray obeys the law of reflection, so the angle of reflection equals the angle of incidence.
❌ Why not from air into glass?
Light going from air into glass bends towards the normal. The angle of refraction is always smaller than the angle of incidence, so it can never reach 90°. There is no critical angle in that direction.
You can see this effect in a swimming pool. A swimmer under the water who looks up at a slant sees the surface act like a mirror, because light meeting the surface from below at more than about 49° cannot get out.
The semi-circular block demonstration
You need to be able to describe this in words and sketch it.
- Place a semi-circular glass or Perspex block on paper. Draw round it and draw a normal at the centre of the flat face.
- Shine a narrow ray through the curved face, aimed at the centre of the flat face. The ray travels along a radius, so it does not bend as it enters. Any bending now happens only at the flat face.
- Start with a small angle of incidence. A bright refracted ray leaves the flat face, bending away from the normal. A faint reflected ray stays inside the block.
- Slowly increase the angle of incidence. The refracted ray swings closer to the flat face and gets dimmer. The reflected ray gets brighter.
- Stop when the refracted ray just skims along the flat face. The angle of incidence is now the critical angle. Mark the ray and measure c from the normal with a protractor.
- Increase the angle a little more. The refracted ray disappears and the reflected ray is as bright as the incident ray.
To draw it: draw the semi-circle with the flat face at the top and a dashed normal through its centre. Draw the incident ray from the curved side to the centre. For an angle above c, draw the reflected ray at an equal angle on the other side of the normal, leaving through the curved face. Draw nothing above the flat face.
The specification suggests an optional investigation like this one: finding the critical angle for Perspex-air, glass-air or water-air boundaries.
The equation: sin c = 1 ÷ n
The critical angle depends on the refractive index of the denser material:
sin c = 1 ÷ n
Here c is the critical angle and n is the refractive index. You must know and use this equation. It is not given in the exam, so learn it.
It comes straight from the equation in Refractive Index. At the critical angle, the angle in the air is 90° and the angle in the material is c. So n = sin 90° ÷ sin c. Since sin 90° = 1, n = 1 ÷ sin c, which rearranges to sin c = 1 ÷ n.
| You want | Use | Calculator |
|---|---|---|
| c from n | sin c = 1 ÷ n | Work out 1 ÷ n, then press sin−1 |
| n from c | n = 1 ÷ sin c | Work out sin c, then divide 1 by it |
Worked example 1: finding c
Question: Water in a fish tank has a refractive index of 1.33. Calculate the critical angle for the water-air boundary.
sin c = 1 ÷ n
sin c = 1 ÷ 1.33 = 0.752
c = sin−1(0.752) = 48.8°, which is 49° to 2 significant figures.
Worked example 2: finding n
Question: The critical angle for a clear plastic is 40°. Calculate its refractive index.
n = 1 ÷ sin c
n = 1 ÷ sin 40° = 1 ÷ 0.643
n = 1.56. There is no unit.
Worked example 3: what happens to the ray?
Question: A ray inside a Perspex block (n = 1.49) meets a Perspex-air boundary at an angle of incidence of 60°. Explain what happens.
sin c = 1 ÷ 1.49 = 0.671, so c = 42.2°.
60° is greater than the critical angle, so the ray is totally internally reflected. The angle of reflection is 60°.
Comparing materials
| Material | Refractive index n | 1 ÷ n | Critical angle c in ° |
|---|---|---|---|
| Water | 1.33 | 0.752 | 48.8 |
| Perspex | 1.49 | 0.671 | 42.2 |
| Glass | 1.5 | 0.667 | 41.8 |
| Diamond | 2.4 | 0.417 | 24.6 |
The pattern: the bigger the refractive index, the smaller the critical angle. Diamond has a very small critical angle, so light inside it is totally internally reflected very easily.
Common mistakes
- Writing sin c = n. For glass that gives sin c = 1.5 and your calculator shows an error, because a sine is never bigger than 1. It is 1 ÷ n.
- Stopping at sin c. The value 0.667 is not an angle. Press sin−1 to get c.
- Saying the light is totally internally reflected at the critical angle. At c the ray skims the surface. Total internal reflection needs an angle greater than c.
- Measuring the angle from the surface. The critical angle is measured from the normal.
- Forgetting the direction. It only happens from the denser material towards the less dense one.
Exam-style question
A student shines a ray of light through the curved face of a semi-circular block of clear plastic. The ray meets the centre of the flat face from inside the block.
(a) State what is meant by the critical angle. [2 marks]
(b) The student measures the critical angle of the plastic as 43°. Calculate the refractive index of the plastic. [3 marks]
(c) The student increases the angle of incidence at the flat face to 50°. Describe what happens to the ray at the flat face. [2 marks]
(d) The plastic block is replaced with a glass block of refractive index 1.60. Calculate the critical angle of this glass. [3 marks]
(e) Explain why total internal reflection cannot happen when a ray travels from air into the block. [2 marks]
Model answer
(a) The angle of incidence in the denser material (1) for which the angle of refraction is 90° (1).
(b) n = 1 ÷ sin c (1). n = 1 ÷ sin 43° = 1 ÷ 0.682 (1). n = 1.47 (1).
(c) All of the light is reflected back inside the block and none is refracted out, which is total internal reflection (1). The angle of reflection is 50° (1).
(d) sin c = 1 ÷ n (1). sin c = 1 ÷ 1.60 = 0.625 (1). c = 38.7°, or 39° (1).
(e) Light going from air into the block bends towards the normal, so the angle of refraction is always smaller than the angle of incidence (1). The angle of refraction can never reach 90°, so there is no critical angle (1).
Exam tip
In parts (b) and (d), check your answer makes sense before moving on. A refractive index must be bigger than 1, and a critical angle must be less than 90°. If you get n = 0.68 in part (b), you have used sin c instead of 1 ÷ sin c.