IB DP Physics: Wavefronts, rays, reflection, refraction and diffraction (DP Physics)

This guide covers the first half of sub-topic C.3 (Wave phenomena) in the IB Physics guide, first assessment 2025: wavefronts and rays, behaviour at boundaries, diffraction, Snell’s law, the critical angle and total internal reflection. All of this is SL and HL. Superposition, double-slit interference, single-slit diffraction and gratings belong to the rest of C.3 and are not taught here. For quick recall use the revision notes, then try the practice questions. Find weak spots first with the free 10-minute diagnostics.

Related pages: the course hub, the printable checklist and travelling waves and the Doppler effect, which covers the wave model that this unit builds on.

What this unit covers

Guide statement	What you must be able to do	Level
C.3	Describe waves in two and three dimensions using wavefronts and rays	SL and HL
C.3	Explain reflection, refraction and transmission at a boundary	SL and HL
C.3	Describe how waves spread past an obstacle or through a gap (diffraction)	SL and HL
C.3	Draw and interpret wavefront-ray diagrams for refraction and diffraction	SL and HL
C.3	Use Snell’s law, including the speed ratio, and find a critical angle	SL and HL
C.3	Explain total internal reflection and when it happens	SL and HL

The guide says sketches and interpretations of wavefronts and rays are limited to incident, reflected and transmitted waves.

Wavefronts and rays

A wavefront is a line (in two dimensions) or surface (in three) joining points that are at the same phase of the oscillation, such as all the crests. A ray is a line drawn in the direction the wave travels, so it shows the direction of energy transfer. Wherever the wave is travelling through a uniform medium, a ray is perpendicular to the wavefronts it crosses.

Source	Wavefront shape	Rays
Point source on a water surface (2D)	Concentric circles	Straight lines pointing outwards from the source
Point source of sound or light (3D)	Concentric spheres	Straight lines pointing outwards
Source very far away, or a flat dipper	Straight lines or flat planes	Parallel lines

The distance between neighbouring wavefronts of the same type is one wavelength. Each point on a wavefront can be treated as a source of new circular wavelets, and the new wavefront is the line that touches them. This explains why straight wavefronts stay straight and why they curve at an edge.

Reflection and transmission

When a wave meets a boundary between two media, three things can happen to its energy. Some is reflected back into the first medium, some is transmitted into the second, and some may be absorbed. The frequency of the wave does not change at the boundary, because the source sets it. The speed and wavelength can change.

For reflection from a flat surface:

• the angle of incidence equals the angle of reflection;
• both angles are measured from the normal, the line at 90° to the surface;
• the incident ray, reflected ray and normal lie in one plane.

On a wavefront diagram the reflected wavefronts are the same spacing as the incident ones, because the wave stays in the same medium. The surface and the wavefront make the same angle as the ray and the normal.

Worked example 1. A ray meets a plane mirror at 35° to the mirror surface. Find the angle of incidence and the angle between the incident and reflected rays.

• Angle of incidence is measured from the normal: 90° − 35° = 55°.
• The reflected ray is also at 55° to the normal.
• The angle between the two rays is 2 × 55° = 110°.

A common slip is to give 35° as the angle of incidence. Always check what the angle is measured from.

Refraction and Snell’s law

Refraction is the change of direction of a wave when its speed changes as it crosses a boundary. If the wave meets the boundary head-on it still slows or speeds up, but it does not change direction.

Consider plane wavefronts travelling from medium 1 into medium 2 at an angle. One end of a wavefront reaches the boundary first and changes speed first. The part still in the first medium keeps going at the old speed, so the wavefront swings round.

 medium 1 (fast)      wavefronts far apart, ray at large angle to normal
 ------------------ boundary ------------------
 medium 2 (slow)      wavefronts closer together, ray nearer the normal

Take one wavefront with its ends meeting the boundary a distance AB apart along the boundary. In the time the wave in medium 1 travels one wavelength λ1, the wave in medium 2 travels λ2. The angle between a wavefront and the boundary equals the angle between the ray and the normal, so:

• sin θ1 = λ1 / AB
• sin θ2 = λ2 / AB
• dividing: sin θ1 / sin θ2 = λ1 / λ2 = v1 / v2, since the frequency is constant and v = fλ.

This is Snell’s law, which the guide gives in the form

n1 / n2 = sin θ2 / sin θ1 = v2 / v1

where n is the refractive index of each medium and θ is the angle between the ray and the normal. Cross-multiplying gives the working form n1 sin θ1 = n2 sin θ2. A higher refractive index means a slower wave. Taking medium 1 as a vacuum, the speed ratio shows that n = c / v for a medium.

What changes on refraction, and what does not:

Quantity	Changes?
Frequency	No
Speed	Yes
Wavelength	Yes, in the same ratio as the speed
Direction	Yes, unless the wave meets the boundary at 90°

Going into the medium with the higher refractive index, the ray bends towards the normal. Going into a lower refractive index, it bends away from the normal.

Worked example 2. Light of wavelength 550 nm in air meets a liquid of refractive index 1.47 at 52.0° to the normal. Take the refractive index of air as 1.00 and the speed of light in a vacuum as 3.00 × 10⁸ m s⁻¹. Find the angle of refraction, the speed in the liquid and the wavelength in the liquid.

• 1.00 × sin 52.0° = 1.47 × sin θ2
• sin θ2 = 0.7880 / 1.47 = 0.5361, so θ2 = 32.4°.
• v = c / n = 3.00 × 10⁸ / 1.47 = 2.04 × 10⁸ m s⁻¹.
• The frequency is unchanged, so λ2 = 550 / 1.47 = 374 nm.

The ray bends towards the normal, as expected for a higher refractive index.

Worked example 3 (a mechanical wave). Water waves in a ripple tank travel at 0.30 m s⁻¹ in a deep region and 0.21 m s⁻¹ over a shallow shelf. They have frequency 6.0 Hz and meet the shelf edge with the rays at 40° to the normal. Find the wavelengths and the angle of refraction.

• Deep region: λ1 = v / f = 0.30 / 6.0 = 0.050 m.
• Shallow region: λ2 = 0.21 / 6.0 = 0.035 m.
• sin θ2 / sin θ1 = v2 / v1, so sin θ2 = (0.21 / 0.30) × sin 40° = 0.7 × 0.6428 = 0.4500.
• θ2 = 26.7°.

On the wavefront diagram the lines are closer together in the shallow region and the rays are closer to the normal. The same equation works for any wave, which is why the guide writes it in terms of speeds as well as refractive indices.

Critical angle and total internal reflection

Now suppose a ray travels from a medium of higher refractive index n1 into one of lower refractive index n2, for example glass into air. It bends away from the normal. As the angle of incidence grows, the angle of refraction grows faster, and eventually reaches 90°, so the refracted ray runs along the boundary. The angle of incidence at which this happens is the critical angle, C.

Put θ2 = 90° in Snell’s law:

n1 sin C = n2 × 1, so sin C = n2 / n1.

If the angle of incidence is greater than C, there is no refracted ray and all the energy is reflected. This is total internal reflection. Two conditions are needed:

1. the wave travels towards a medium with a lower refractive index (slower to faster);
2. the angle of incidence exceeds C.

At angles below C a ray is partly reflected and partly transmitted. If n1 < n2, the sine would have to exceed 1, so there is no critical angle.

Worked example 4. A glass of refractive index 1.55 is in contact with a liquid of refractive index 1.36. (a) Find the critical angle. (b) A glass prism with angles 45°, 45° and 90° has a ray entering one short face along the normal and meeting the long face at 45°. Does total internal reflection occur if the prism is in air? In the liquid?

• (a) sin C = 1.36 / 1.55 = 0.8774, so C = 61.3°.
• (b) In air, sin C = 1.00 / 1.55 = 0.6452, so C = 40.2°. The angle of incidence, 45°, is greater than 40.2°, so the ray is totally internally reflected.
• In the liquid, 45° is less than 61.3°, so there is no total internal reflection. Part of the ray leaves the prism.

Diffraction

Diffraction is the way a wave fans out after meeting a gap (an aperture) or the edge of an obstacle (a body). Wavefronts bend into the region that a straight-line ray picture would call a shadow.

For a wave passing through an aperture:

• when the gap is much wider than the wavelength, the wave passes through with little change, apart from slightly curved edges and a clear shadow on either side;
• when the gap is about the same size as the wavelength, or smaller, the wave spreads out strongly on the far side, and the wavefronts become nearly circular arcs;
• the wavelength, speed and frequency are the same before and after the gap, so the spacing of wavefronts does not change. Only the shape and the spread of the wave change, and the amplitude beyond the gap is lower.

For a wave passing a body, the wave bends round the sides and fills in part of the shadow behind it. The effect is stronger when the body is not much bigger than the wavelength. A large body in a short-wavelength wave leaves a well-defined shadow.

 plane wavefronts -> | gap about one wavelength |  -> near-circular wavefronts
                       (small aperture)              spreading widely

Worked example 5. Sound of frequency 500 Hz travels at 340 m s⁻¹ in air. A radio wave of frequency 100 MHz and light of wavelength 5 × 10⁻⁷ m both pass a doorway 0.90 m wide. Compare the diffraction.

• Sound: λ = 340 / 500 = 0.68 m, comparable to the doorway, so the sound spreads into the room and can be heard around the corner.
• Radio: λ = 3.00 × 10⁸ / 1.00 × 10⁸ = 3.0 m, larger than the doorway, so it spreads strongly.
• Light: λ is about a million times smaller than the gap, so there is no noticeable spreading and a sharp shadow edge.

Common errors

• Measuring the angle from the surface instead of from the normal.
• Saying the frequency changes at a boundary. It does not; speed and wavelength do.
• Drawing refracted wavefronts with the same spacing as the incident ones.
• Saying total internal reflection can occur from glass to water at any angle. It needs an angle above C, and it needs n1 > n2.
• Writing sin C = n1 / n2. The larger index is always on the bottom.
• Saying diffraction changes the wavelength of the wave.

Official syllabus

IB Physics guide, first assessment 2025, published by the International Baccalaureate Organization: sub-topic C.3 Wave phenomena (first half of the understandings).