When waves hit a boundary, they can reflect or refract. Reflection keeps the wave in the same medium. Refraction sends it into a new medium, where the speed changes and the path can bend.
Use the PDF for classwork, homework or revision. It includes key ideas, activities, questions, an extend task and success-criteria proof.
A ripple in deep water moves toward shallow water at an angle. As it crosses into the shallow region, does its frequency change, its speed change, both, or neither? Predict what happens to its direction too.
Type your prediction below. You will revisit it at the end.
Write your prediction in your book. You will revisit it at the end.
Wrong: Vectors and scalars are just different ways of writing the same thing.
Right: Vectors have magnitude and direction; scalars have magnitude only. They follow different mathematical rules.
📚 Core Content
Reflection happens when a wave bounces off a boundary and stays in the original medium.
The key rule is simple: angle of incidence equals angle of reflection. Both angles are measured from the normal, which is the line perpendicular to the boundary. This applies to reflected water waves, sound, and light in the ray model. The smoothness of the boundary matters too. A smooth surface produces specular reflection, where parallel rays reflect in the same direction. A rough surface produces diffuse reflection, where rays scatter in many directions. In both cases, the law of reflection holds at every individual point on the surface.
Reflection is why we see echoes, why mirrors work, and why ocean waves bounce back from harbour walls. When a sound wave reflects off a flat wall, the angle between the incoming wave and the normal equals the angle between the reflected wave and the normal. This predictable geometry makes reflection one of the easiest wave behaviours to analyse quantitatively.
Refraction is the change in speed and direction when a wave enters a different medium.
When one side of a wavefront enters the new medium first, its speed changes before the rest of the wavefront does. That difference in speed across the wavefront makes the path bend. If the wave slows down, it bends toward the normal. If it speeds up, it bends away from the normal. This bending is not caused by the wave "choosing" a new path — it is a direct consequence of different parts of the wavefront travelling at different speeds for a brief moment.
The frequency does not change during refraction because the source is still driving the oscillation at the same rate. The particles at the boundary are forced to oscillate at the same frequency as the incoming wave, and they then become the source for the transmitted wave. If speed changes while frequency stays constant, wavelength must change too. From $v = f\lambda$, a lower speed means a shorter wavelength, and a higher speed means a longer wavelength. This is a critical checkpoint in every refraction problem.
Ray diagrams simplify the wave direction so we can focus on angle and bending.
Ray diagrams replace the continuous wavefront with a single line showing the direction of travel. This abstraction is incredibly useful because it lets us apply geometry — angles, triangles, and trigonometry — to wave problems. In a ray diagram for reflection, the incident ray and reflected ray make equal angles with the normal. In refraction, the incident ray and refracted ray are on opposite sides of the normal, and the angle changes depending on the relative speeds.
Left: reflection keeps the wave in the same medium. Right: refraction changes direction because speed changes in the new medium.
During refraction, speed and wavelength can change, but frequency stays fixed.
This is because frequency is determined by the source, not the medium. When a water wave moves from deep to shallow water, the wave crests are still being produced at the same rate, but each crest travels more slowly. Since the crests are produced at the same rate but move slower, they must be closer together — the wavelength decreases. The same logic applies in reverse: when a wave speeds up, its wavelength increases.
| Quantity | Reflection | Refraction |
|---|---|---|
| Frequency | Same | Same |
| Speed | Same medium, so same speed | May change |
| Wavelength | Usually same in same medium | Changes with speed |
| Direction | Changes by reflection rule | May bend at boundary |
When a wave tries to leave a slower medium for a faster one at a steep enough angle, it can reflect entirely instead of refracting.
This phenomenon is called total internal reflection. It occurs when a wave travels from a slower medium to a faster medium (where it would normally bend away from the normal) and the angle of incidence exceeds a certain critical angle. Above this angle, no refracted ray escapes — all the wave energy is reflected back into the original medium. This principle is the basis for optical fibres, which carry internet data and medical imaging signals over long distances with minimal loss.
For the NSW HSC syllabus, total internal reflection is most commonly discussed in the context of light, but the same physics applies to any wave. Seismic waves can undergo total internal reflection at boundaries between different rock layers, and underwater sound waves can be trapped in ocean layers where the speed gradient creates a similar effect. The critical angle depends on the ratio of wave speeds in the two media.
A few persistent errors trip up students in boundary problems. Let's address them directly.
The relationship $v = f\lambda$ is not just a formula — it is the key to predicting what happens when a wave crosses from one medium to another.
Because the frequency $f$ is locked by the source, it acts as the constant anchor in every refraction problem. If you know the speed in each medium, you can immediately calculate the wavelength in each medium by rearranging to $\lambda = v/f$. This is why exam questions about refraction so often ask you to calculate wavelength: it tests whether you understand that $f$ stays fixed.
Students sometimes try to apply $v = f\lambda$ across the boundary as if $v$, $f$, and $\lambda$ all change simultaneously in the same equation. The correct approach is to treat each medium separately. Write $v_1 = f\lambda_1$ for the first medium and $v_2 = f\lambda_2$ for the second medium. Then equate the frequencies: $f = v_1/\lambda_1 = v_2/\lambda_2$. This gives a direct ratio: $\lambda_2/\lambda_1 = v_2/v_1$. The wavelengths scale in direct proportion to the speeds.
| Medium | Speed $v$ | Frequency $f$ | Wavelength $\lambda$ |
|---|---|---|---|
| Air | $340\ \text{m/s}$ | $500\ \text{Hz}$ | $0.68\ \text{m}$ |
| Water | $1500\ \text{m/s}$ | $500\ \text{Hz}$ | $3.0\ \text{m}$ |
| Glass | $2.0 \times 10^8\ \text{m/s}$ | $5.0 \times 10^{14}\ \text{Hz}$ | $400\ \text{nm}$ |
✏️ Worked Examples
Scenario: A wave hits a boundary with an incidence angle of 35°. Find the reflection angle.
If the 35° had been measured from the surface, you would first convert it to 55° from the normal.
Scenario: A water wave has frequency 4 Hz. In deep water its speed is 2.0 m/s. In shallow water its speed falls to 1.0 m/s. Find the wavelength in each region and state what happens to direction if the wave enters the shallow region at an angle.
If the wave entered a faster medium instead, it would bend away from the normal and the wavelength would increase.
Scenario: Light travels in an optical fibre core with speed $2.0 \times 10^8$ m/s. The surrounding cladding has speed $2.2 \times 10^8$ m/s. Explain whether total internal reflection is possible and what condition is required.
If the cladding were slower than the core, total internal reflection would be impossible — the light would always refract into the cladding.
Scenario: A sound wave of frequency 250 Hz travels from air ($v = 340$ m/s) into water ($v = 1500$ m/s). Find the wavelength in each medium and state the direction of bending if the wave enters the water at an angle.
If the wave travelled from water into air instead, it would slow down and bend toward the normal, and the wavelength would decrease to 1.36 m.
Scenario: A light wave has wavelength 600 nm in air ($v = 3.00 \times 10^8$ m/s). It enters a glass block where its speed drops to $2.00 \times 10^8$ m/s. Find the wavelength in the glass and the ratio of glass wavelength to air wavelength.
If the light entered a medium where the speed was $1.50 \times 10^8$ m/s, the new wavelength would be 300 nm — exactly half the air wavelength.
Visual Break
🏃 Activities
A wave hits a barrier at 20°, 45°, and 70° to the normal in three different cases. State the reflection angle each time.
A student says, "When a wave refracts, the frequency must change because the wavelength changes." Write a short response correcting this.
A wave bends away from the normal on entering a new medium. State whether it sped up or slowed down, and explain your reasoning.
A coin sits at the bottom of a swimming pool 2.0 m deep. Light from the coin travels from water ($v = 2.25 \times 10^8\ \text{m/s}$) into air ($v = 3.00 \times 10^8\ \text{m/s}$).
An optical fibre has a core surrounded by cladding. The refracted ray must stay inside the core to carry a signal.
A ray of light strikes a flat mirror at 55° to the surface. Draw a diagram in your book showing the incident ray, reflected ray, and normal. Calculate the angle of incidence and the angle of reflection. Explain why the angle between the incident ray and the reflected ray is 70°.
A triangular glass prism is placed in a beam of white light. The light enters one face of the prism, slows down, bends toward the normal, then exits the other face, speeding up and bending away from the normal.
A wave travels from medium X to medium Y. In medium X, its speed is 8.0 m/s and its wavelength is 4.0 m. In medium Y, its speed is 12.0 m/s.
A periscope uses two plane mirrors arranged at 45° to allow a viewer to see over obstacles.
Earlier you were asked what changes when a ripple enters shallow water at an angle.
The full answer: the wave slows down, so it bends toward the normal. Its frequency stays the same because the source still sets the oscillation rate. Because the speed drops while frequency stays constant, the wavelength becomes smaller in the shallow region.
Now revisit your prediction. What did you think would happen to speed, frequency and wavelength?
Annotate your prediction in your book with what you now understand differently.
Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?
✅ Check Your Understanding
1. The angle of reflection is measured from:
2. A wave slows down on entering a new medium. It bends:
3. Which quantity stays constant during refraction?
4. A wave has incidence angle 28° to the normal. The reflection angle is:
5. If a wave enters a slower medium and frequency stays constant, the wavelength:
6. A wave bends away from the normal on entering a new medium. This means it has:
7. Explain the difference between reflection and refraction of waves. 3 MARKS
8. A wave of frequency 6 Hz travels at 3.0 m/s in medium A and 1.5 m/s in medium B. Calculate the wavelength in each medium. 3 MARKS
9. Explain why a wave bending toward the normal must be entering a slower medium, and why the frequency does not change. 4 MARKS
1. C — all reflection angles are measured from the normal.
2. A — slowing down means bending toward the normal.
3. D — frequency stays fixed during refraction.
4. B — law of reflection gives the same angle from the normal.
5. C — lower speed with same frequency means shorter wavelength.
6. D — bending away from the normal means the wave entered a faster medium.
Q7 (3 marks): Reflection is when a wave bounces off a boundary and remains in the original medium. The angle of incidence equals the angle of reflection, both measured from the normal. Refraction is when a wave crosses into a different medium and changes speed, which may cause its direction to bend toward or away from the normal. During refraction, the frequency stays constant because it is determined by the source, while the speed and wavelength may change according to $v = f\lambda$.
Q8 (3 marks): In medium A, $\lambda = v/f = 3.0/6 = 0.50\ \text{m}$. In medium B, $\lambda = 1.5/6 = 0.25\ \text{m}$. The frequency remains 6 Hz in both media because frequency is a property of the source and does not change when a wave crosses a boundary.
Q9 (4 marks): A wave bends toward the normal when it enters a slower medium because the part of the wavefront that enters the new medium first slows down before the rest of the wavefront does. This difference in speed across the wavefront causes the whole wavefront to pivot toward the normal. Frequency does not change because the particles at the boundary are forced to oscillate at the same rate as the incoming wave; they then become the source for the transmitted wave. Since $v = f\lambda$ and $f$ is constant, a decrease in $v$ must be accompanied by a decrease in $\lambda$.
Activity 4: (1) The light speeds up as it leaves the water because $3.00 \times 10^8 > 2.25 \times 10^8$ m/s. (2) It bends away from the normal when speeding up. (3) The coin appears shallower because light rays from the coin refract away from the normal as they exit the water. Our brain interprets these rays as having travelled in straight lines, so the apparent position of the coin is closer to the surface than its actual position.
Activity 5: (1) The core must have the lower wave speed (higher refractive index) compared with the cladding. (2) At sharp bends, the angle of incidence inside the core can fall below the critical angle, allowing some light to refract out into the cladding rather than reflecting back into the core. (3) One advantage is much lower signal loss over long distances because light experiences less absorption and interference compared with electrical signals in copper cables.
Activity 6: Angle measured from surface = 55°, so angle of incidence from normal = 90° − 55° = 35°. By the law of reflection, angle of reflection = 35°. The angle between the incident ray and the reflected ray is the sum of the angle of incidence and the angle of reflection = 35° + 35° = 70°.
Activity 7: (1) Light bends toward the normal on entering glass because glass is an optically denser medium where light travels more slowly. The part of the wavefront that enters first slows down, causing the whole wavefront to pivot toward the normal. (2) Different colours have slightly different wavelengths and therefore different speeds in glass. Violet light slows down more than red light, so it bends more sharply. (3) Spectroscopes use prisms to separate starlight into its component colours. By analysing which wavelengths are present or absent, astronomers can determine the chemical composition, temperature, and motion of distant stars.
Activity 8: (1) Frequency $f = v_X/\lambda_X = 8.0/4.0 = 2.0$ Hz. (2) Wavelength in Y: $\lambda_Y = v_Y/f = 12.0/2.0 = 6.0$ m. (3) The wave bends away from the normal because it is speeding up in medium Y. The part of the wavefront that enters Y first travels faster, causing the wavefront to pivot away from the normal. (4) Ratio = $\lambda_Y/\lambda_X = 6.0/4.0 = 1.5$.
Activity 9: (1) The light ray from the object hits the top mirror at 45°, reflects down to the bottom mirror, reflects again at 45°, and enters the eye. (2) The image is not laterally inverted because the two reflections cancel out the lateral inversion that a single mirror would produce. (3) Other applications include medical endoscopes for viewing inside the body, armoured vehicle periscopes for driver safety, and some types of optical surveying equipment.
Face the boss using your knowledge of how waves reflect and refract. Pool: lessons 1–5.
Tick when you have finished the activities and checked the answers.