fizikzauglom

GRADE 9 · LIGHT PHENOMENA

Refraction and total internal reflection

Refraction law, refractive index and light speed in a medium.

28 minutes9 theory cards
Start interactive lesson

Read the full lesson theory here without an account. In the interactive lesson, the same theory is presented as cards, followed by questions with feedback and practice problems. Your progress is saved.

Understanding step by step

A spoon in a glass seems bent and the bottom seems raised. The objects have not changed; the light's route has. Why does a ray turn at a transparent boundary, and when does it fail to emerge at all?

An underwater point appears higherappearshigherSolid lines — light pathsDashes — extensions in water
The spoon remains straight: the dark outline shows its real position. Two close rays from an underwater point emerge through the water surface and reach the pupil. Their dashed backward extensions meet above the original point. This shows light paths, not the spoon's visual appearance; the rays do not cross the glass's side walls.

Different speeds

Transparency does not mean light travels through matter as it does through empty space. Interaction with a medium changes propagation speed. Define absolute refractive index as n = c/v, where c is vacuum light speed and v is its speed in the medium. Ordinary transparent glass has n > 1 and v < c. This ratio describes light of a chosen frequency: the index can differ slightly for different colours.

Example: glass with n = 1,5vacuum: cglass: v = 2c/3λ₀λ = 2λ₀/3λ = v/νFrequency ν stays the same at the boundary
Schematic glass example with n = 1,5: v = c/n = 2c/3. Plane wavefronts are shown at one instant; vacuum and glass wavelengths have ratio λ₀:λ = 3:2. Frequency is unchanged at the boundary, so lower speed means shorter wavelength. To determine speed from wavelength, frequency is also needed.

A number without a unit

If n = 1,5, then c is about one and a half times the speed in the medium: v = c/1,5. It does not mean the light is brighter by that factor. Both speeds use m/s, so n has no unit. c ≈ 3·10⁸ м/с. Greater n means lower speed for this light; at n = 1, its speed is c. The number describes speed; direction depends on how the wave crosses the boundary.

Vacuum, cMedium, v < cThe same time interval
Absolute refractive index n = c/v compares vacuum speed with speed in the medium. In the same time, light travels a shorter distance in the medium.

Direction at a boundary

An oblique wavefront does not enter the new medium all at once. One part already travels at the new speed while another still has the old speed, so the front rotates. In a medium of higher n, the ray bends towards the normal; for lower n, away from it. This is a wave explanation of refraction. If the front is parallel to the boundary, all parts change speed together: at normal incidence the ray stays straight even though speed changes.

One boundary, three casesn = 1n = 1,5n = 1n = 1,5n = 1n = 1,5towards normalaway from normalno turnA change of index changes speed
All three cases have air n = 1 above and glass n = 1,5 below. Left: oblique entry into glass bends towards the normal. Centre: the same possible path is reversed, bending away from the normal. Right: normal incidence leaves direction unchanged while speed decreases. The oblique paths have angles 45° in air and about 28.1° in glass.

Law of refraction

The quantitative angle relation is n₁ sin α = n₂ sin β. Subscript 1 refers to the incident medium, 2 to the transmitted medium; α and β are measured from the normal. Rays and normal lie in one plane. If n₁ = n₂, then β = α: no turn. If α = 0, its sine is zero and β is also zero. The law predicts an angle, not just 'towards' or 'away from' the normal.

n₁ sin α = n₂ sin βnormalair: n₁ = 1glass: n₂ = 1,5αβn sin(angle) has the same value
The diagram shows air-to-glass refraction. n₁ refers to the medium before the boundary, n₂ after it. Both angular arcs share one point: α lies between the incident ray and upper normal, β between the refracted ray and lower normal. Rays and normal lie in one plane.

Checking an angle

Predict before calculating. With incident medium and α unchanged, increasing n₂ reduces β because sin β = (n₁/n₂) sin α. The ray should move closer to the normal. Entering a lower-n₂ medium instead increases the angle, while a refracted ray exists. This check catches swapped indices or angles and explains why one large n alone, without the other medium, cannot determine the turn.

airwaterαβ
Entering water obliquely from air reduces light speed and bends the ray towards the normal. At normal incidence, direction is unchanged.

The limiting exit angle

Increase the incidence angle as light leaves higher n for lower n. The refracted ray bends progressively farther from the normal, reaching the boundary direction at β = 90°. This defines critical angle αкр, with sin αкр = n₂/n₁. The ratio must be less than one. If n₁ ≤ n₂, no such limiting angle for total internal reflection exists.

Glass n₁ = 1,5 to air n₂ = 1β ≈ 30,9°α = 20°β = 90°αкр ≈ 41,8°no outgoing rayα = 60°sin αкр = n₂/n₁ = 2/3Green — reflected ray
Light goes from glass n₁ = 1,5 towards air n₂ = 1. Left: at α = 20°, some light emerges at β ≈ 30,9° and some reflects. Centre: at αкр ≈ 41,8°, the limiting direction β = 90° runs along the boundary. Right: at α = 60°, no refracted ray propagates out. In all three cases, the reflection angle equals incidence.

Both conditions together

What happens at α > αкр? The law would require sin β > 1, which no direction can satisfy: there is no propagating refracted ray. In the transparent-media model, total internal reflection occurs. Both conditions are needed: n₁ > n₂ and incidence above the critical angle. Simply going from glass towards air is insufficient: at small angles, some light emerges. A field exists immediately at the boundary, but that belongs to the wave description, not an ordinary outgoing ray.

Both conditions at oncen₁ > n₂ and α > αкрair: n₂ = 1glass: n₁ = 1,560°60°60° > αкр ≈ 41,8°
Light goes from glass n₁ = 1,5 towards air n₂ = 1. The critical angle is about 41.8°, and incidence is 60°. Both conditions hold: n₁ > n₂ and α > αкр. There is no propagating refracted ray; the reflected ray makes the same 60° with the normal. At a smaller angle, some light could emerge into air.

Applying this to a fibre

An optical fibre's core and cladding have a suitable index relationship. Rays hitting their boundary at suitable angles undergo total internal reflection and are guided along the fibre. Geometry and indices both matter: not every ray is automatically trapped. Frequency stays the same at a stationary boundary; the new speed changes wavelength through v = λν. Refraction should therefore not be explained by a change in source frequency.

Total reflection at the core boundariesDashed lines — normalscore: n₁ = 1,50cladding: n₂ = 1,40α ≈ 74,7° > αкр ≈ 69°
Schematic example: core n₁ = 1,50, cladding n₂ = 1,40. The ray meets each core boundary exactly and reflects at the same angle. At all three points, the normal angle is about 74.7°, above the critical angle of about 69°. In the ideal model, the ray remains within the core. Measure the angle from the wall normal, not from the fibre axis.

From air to glass

Light goes from air, n₁ = 1, into glass, n₂ = 1,5. The incidence angle is 30°. Estimate the refraction angle, given sin 30° = 0,5.

What is knownnormalair: n₁ = 1glass: n₂ = 1,5αα = 30°β — ?sin 30° = 0,5
The incident ray makes exactly 30° with the normal and reaches the air–glass boundary. Before it, n₁ = 1; after it, n₂ = 1,5. The required β uses the lower normal. Boundary, normal and incident ray remain unchanged through all steps.

Let's look at an example · 1

First predict: n₂ = 1,5 exceeds n₁ = 1, so the ray in glass must bend towards the normal and β must be less than 30°.

First predict the directionnormalair: n₁ = 1glass: n₂ = 1,5αα = 30°0 < β < 30°n₂ > n₁ → β < α = 30°
In the same setup, glass has the higher index. The green sector marks the expected refracted direction range: between the normal and the dashed continuation of the original ray. Thus 0 < β < 30°. The refracted ray itself will be drawn after calculation.

Let's look at an example · 2

Rearrange n₁ sin α = n₂ sin β: sin β = n₁ sin α/n₂ = 1 · 0,5/1,5 = 1/3.

sin β = n₁ sin α/n₂normalair: n₁ = 1glass: n₂ = 1,5αα = 30°0 < β < 30°sin β = 1 · 0,5/1,5 = 1/3
The boundary, original 30° ray and expected-direction sector remain in place. The refraction law gives sin β = n₁ sin α/n₂ = 1 · 0,5/1,5 = 1/3. This is the sine of an angle, not the angle in degrees.

Let's look at an example · 3

An angle whose sine is 1/3 is about 19.5°. This is measured from the normal, just like the given 30°.

From sine to anglenormalair: n₁ = 1glass: n₂ = 1,5αα = 30°ββ ≈ 19,5°β = arcsin(1/3) ≈ 19,5°
At the original boundary point, draw the ray for β = arcsin(1/3) ≈ 19,5°. The arc marks this angle between the ray and lower normal. Air, glass and incidence angle 30° remain unchanged. β is measured from the normal in glass.

Let's look at an example · 4

19.5° is less than 30°: the calculation agrees with the physical prediction for slower propagation in glass.

Compare result with predictionnormalair: n₁ = 1glass: n₂ = 1,5αα = 30°ββ ≈ 19,5°19,5° < 30° — closer to the normal
Compare 19.5° and 30° in the same setup. The refracted ray is closer to the normal, consistent with n₂ > n₁. The boundary and both rays do not move during this check.

Check the result

β ≈ 19,5° from the normal. Entering glass bends the ray towards the normal, reducing the original 30° angle.

Key point · 1

n = c/v compares speeds; the ray turns because different parts of an oblique wavefront cross the boundary at different times.

Key point · 2

The refraction law relates both media and angles from the normal; at normal incidence, speed can change without a turn.

Key point · 3

Total internal reflection requires higher n to lower n and incidence above the critical angle.

What comes next

Curved lens boundaries bring rays together to form an image.

Basic Concepts

3 concepts

Refractive index

Absolute refractive index is the ratio of vacuum light speed to light speed in the medium.

A higher refractive index means lower light speed in the medium and usually stronger bending towards the normal.

Refraction of light

Light speed changes between media of different refractive indices. A ray changes direction at oblique incidence; at normal incidence it keeps its direction.

On entering an optically denser medium, a ray usually bends towards the normal.

Total internal reflection

Light is completely reflected when going from an optically denser medium towards a less dense one at an angle above the critical angle.

This is used in optical fibres and prisms; it cannot occur from an optically less dense medium into a denser one.

Physical quantities

4 quantities

Refractive index

n

The ratio of the speed of light in a vacuum to the speed in a medium.

Unit: dimensionless quantity

Speed of light

c

Speed of electromagnetic waves in vacuum.

Unit: meter per second · \frac{\mathrm{м}}{\mathrm{с}}

Speed of light in the medium

v

Speed of propagation of light in a transparent medium.

Unit: meter per second · \frac{\mathrm{м}}{\mathrm{с}}

Angle of refraction

\beta

The angle between the refracted ray and the normal to the surface.

Unit: radian · \mathrm{рад}

Lesson formulas

2 formulas

Law of refraction

n_1\sin\alpha=n_2\sin\beta

The medium's refractive index times the sine of the angle from the normal has the same value on both sides of the boundary.

Physical meaning of refractive index

n=\frac{c}{v}

Refractive index gives the factor by which vacuum light speed exceeds the speed in the medium.

Test the relationship

Make a prediction first. The button changes the stated quantity; then compare the result and explore other values.

Reflection and refraction

θотр = θ₁; n₁ sin θ₁ = n₂ sin θ₂

Plane boundary between transparent uniform media; ray optics. Line thickness does not encode power. Under total internal reflection there is no propagating transmitted ray.

At the same incidence angle, n₂ increases from 1.5 to 2. How will the refraction angle from the normal change?

Your prediction

Initial values

n₁ = 1n₂ = 1.5⊥

Calculated result

θотр
30 °
θ₂
19.47 °
Explaining the result

Angles are measured from the normal. Entering a larger n bends the ray towards the normal. Total internal reflection requires n₁ > n₂ and incidence above the critical angle.

Related formulas

The same formulas in the reference, with symbols and units.

Problems on this topic

Statements, answer options and the walkthrough.