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PhysicsWaves and light

The critical angle formula

The critical angle formula is sin θc = n₂ ÷ n₁, where n₁ is the refractive index of the material the light is in and n₂ is the refractive index of the material it is trying to enter, with n₁ bigger than n₂. When the second material is air, with n ≈ 1, it becomes sin θc = 1 ÷ n. If light hits the boundary at more than the critical angle, it is all reflected back: total internal reflection. For fused quartz, with n = 1.46, the critical angle is sin⁻¹(1 ÷ 1.46) = 43.2°.

The critical angle equation

\sin\theta_c = \frac{n_2}{n_1} \quad (n_1 > n_2)The critical angle formula
SymbolMeaning
θcCritical angle: the angle of incidence, measured from the normal, at which the refracted ray just skims along the boundary
n₁Refractive index of the material the light starts in (the higher one)
n₂Refractive index of the material on the other side (the lower one)

AQA's A-level data booklet gives sin θc = n₂ ÷ n₁ for n₁ > n₂, and AQA's specification expects students to recall that the refractive index of air is approximately 1. That gives the shorter version for light trying to leave a material into air:

\sin\theta_c = \frac{1}{n}

Where the critical angle formula comes from

Start with Snell's law, n₁ sin θ₁ = n₂ sin θ₂. When light goes from a higher refractive index to a lower one, it bends away from the normal, so θ₂ is bigger than θ₁. As θ₁ increases, θ₂ reaches 90°: the refracted ray runs along the boundary. That value of θ₁ is the critical angle. Put θ₂ = 90° and sin 90° = 1:

n_1\sin\theta_c = n_2 \times 1 \quad \Rightarrow \quad \sin\theta_c = \frac{n_2}{n_1}
θcglassair
At θc the refracted ray just skims along the surface

Total internal reflection

Total internal reflection happens when two conditions are both met:

  1. The light is travelling from a material with a higher refractive index towards one with a lower refractive index, such as from glass into air.
  2. The angle of incidence is greater than the critical angle.

Below the critical angle, most light refracts out and a little reflects. At exactly θc, the refracted ray skims along the surface. Above it, there is no refracted ray at all: all the light reflects, obeying the law of reflection. Light going the other way, from air into glass, can never be totally internally reflected.

Critical angles of common materials

These use refractive indices from the HyperPhysics table at Georgia State University, with air taken as 1.00.

MaterialRefractive indexCritical angle into air
Ice1.3149.8°
Water at 20 °C1.3348.8°
Fused quartz1.4643.2°
Typical crown glass1.5241.1°
Heavy flint glass1.6537.3°
Diamond2.41724.4°

The higher the refractive index, the smaller the critical angle. Diamond has by far the smallest critical angle in the table, so light inside it is totally internally reflected over a much wider range of angles.

The Refraction Calculator works out critical angles, refractive indices and angles of refraction with Snell's law, and tells you when total internal reflection happens.

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Refraction Calculator

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Critical angle formula worked examples

Example 1: diamond into air

Find the critical angle for diamond, n = 2.417, surrounded by air.

\sin\theta_c = \frac{1}{2.417} = 0.4137, \quad \theta_c = 24.4°

Example 2: glass into water

Light in heavy flint glass, n = 1.65, meets a boundary with water, n = 1.33. Find the critical angle.

\sin\theta_c = \frac{n_2}{n_1} = \frac{1.33}{1.65} = 0.806, \quad \theta_c = 53.7°

It is bigger than the 37.3° for flint glass into air, because water's refractive index is closer to the glass's, so the light bends less at the boundary.

Example 3: refractive index from a critical angle

A student finds that light inside a plastic block is totally internally reflected at angles above 39.0°. Find the plastic's refractive index.

n = \frac{1}{\sin 39.0°} = 1.59

Example 4: will the light escape?

Light inside a fused quartz block, n = 1.46, hits the surface with air at an angle of incidence of 45°. Does it escape?

The critical angle is sin⁻¹(1 ÷ 1.46) = 43.2°. The angle of incidence, 45°, is bigger, so the light is totally internally reflected and does not escape.

Now try predicting what a ray does at a boundary, then press Mark to check it in the course.

Pick the picture that shows what a ray does at a boundary.

Press Mark to have your answer marked in the Light & Energy course.

Optical fibres and the cladding

An optical fibre carries light along a thin glass core by total internal reflection. The core is surrounded by cladding with a slightly lower refractive index, so the light reflects at the core-cladding boundary. AQA's A-level specification covers step-index fibres, including the function of the cladding, and material and modal dispersion.

Example 5: critical angle in a fibre

An optical fibre has a core of refractive index 1.50 and cladding of refractive index 1.46. Find the critical angle at the core-cladding boundary.

\sin\theta_c = \frac{1.46}{1.50} = 0.9733, \quad \theta_c = 76.7°

Only light meeting the boundary at over 76.7° to the normal stays in, which means rays travelling nearly along the fibre. That keeps the paths through the fibre similar in length, which reduces modal dispersion. The cladding also protects the core from scratches, which could let light escape, and stops signals passing between fibres that touch.

Common mistakes

  • Putting the bigger index on top. sin θc is always less than 1, so the smaller refractive index goes on top: n₂ ÷ n₁ with n₁ > n₂.
  • Measuring angles from the surface. All angles in refraction are measured from the normal.
  • Using the formula for light entering a denser material. There is no critical angle going from air into glass.
  • Saying total internal reflection happens at the critical angle. It happens above it. At exactly θc the ray grazes the boundary.
  • Calculator in radians. Use degrees mode for sin⁻¹.

Critical angle questions with answers

1. Find the critical angle for a material of refractive index 1.40 surrounded by air.
sin θc = 1 ÷ 1.40 = 0.714, so θc = 45.6°
2. A material has a critical angle of 45.0° in air. Find its refractive index.
n = 1 ÷ sin 45.0° = 1.41
3. Find the critical angle at a boundary from a material with n = 1.60 to one with n = 1.50.
sin θc = 1.50 ÷ 1.60 = 0.9375, so θc = 69.6°
4. Light in water (critical angle 48.8°) hits the surface at 40°. What happens?
40° is less than the critical angle, so the light refracts out into the air (with a little reflected).
5. Light travels at 2.0 × 10⁸ m/s in a material. Find its refractive index and critical angle. (c = 3.0 × 10⁸ m/s)
n = c ÷ v = 3.0 × 10⁸ ÷ 2.0 × 10⁸ = 1.5.
sin θc = 1 ÷ 1.5, so θc = 41.8°

Practise waves in a short timed paper

A-level Physics: Year 1 gives you a short timed paper every day, with 8 questions worth 25 to 35 marks across the course, so refraction and total internal reflection turn up regularly but not in every paper. The Light and Energy course covers refraction, Snell's law and total internal reflection with interactive ray diagrams.

Paper resets in

Common questions

What is the formula for the critical angle?
sin θc = n₂ ÷ n₁, where light goes from refractive index n₁ to a lower n₂. Into air, sin θc = 1 ÷ n.
What is the critical angle?
The angle of incidence that bends the refracted ray all the way to 90°, so it skims along the surface. Above it, total internal reflection happens.
What is the critical angle of glass?
About 41° for typical crown glass, n = 1.52, into air. Other glasses have different refractive indices, so different critical angles.
What is the critical angle of water?
About 48.8° into air, using n = 1.33.
Can total internal reflection happen when light goes from air into water?
No. It only happens when light travels towards a material with a lower refractive index.
Why does a higher refractive index give a smaller critical angle?
Because sin θc = 1 ÷ n. A bigger n makes sin θc smaller, so θc is smaller.

Revision cards

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