4.3 Waves at material interfaces — Practise

Work through these on your own. Keep your working neat: the method matters as much as the answer.

Both the questions and the answers below come from a set that is checked by computer every time the book is built.

State the law of reflection for a wave at a surface.
A water wave travels at 3.0 m/s and reflects from a wall 24 m away. Calculate the time taken for the wave to travel to the wall and back.
A wave of frequency 250 Hz passes from one material into another. Its speed changes from 5.0 m/s to 8.0 m/s. The frequency does not change. Calculate the wavelength in each material.
When a wave meets a boundary, 10% of its energy is reflected and 20% is absorbed. Calculate the percentage of the energy that is transmitted.
Describe what can happen to a wave when it meets the boundary between two different materials.
Explain why the angles in a reflection diagram are measured from the normal and not from the surface.
Describe how to draw a ray diagram to show a wave reflecting from a flat surface.
A room is lined with soft, thick curtains. Explain how this reduces the loudness of an echo in the room.
A ripple tank is filled with water to a depth of 15 cm. Convert this depth to metres.
A water wave of frequency 4.0 Hz passes from shallow water into deep water. In the deep water the wavelength is 0.35 m. The frequency stays the same. Calculate the wave speed in the deep water.
A person claps and hears the echo from a wall after 0.60 s. The speed of sound in air is 330 m/s. Calculate the distance from the person to the wall.
A sound wave passes from air into water. The frequency does not change. Say which equation links the speed, the frequency and the wavelength, and explain what happens to the wavelength.

Answers and concise solutions

1. The law of reflection

The angle of reflection equals the angle of incidence.
Both angles are measured between the ray and the normal, which is the line at right angles to the surface.

2. Time for the echo

Answer: The wave travels to the wall and back, a distance of $2 \times 24 = 48\,\text{m}$ . Then $t = \dfrac{s}{v} = \dfrac{48}{3.0} = 16\,\text{s}$ .

3. Wavelengths in the two materials

Answer: The frequency stays the same. In the first material $\lambda = \dfrac{v}{f} = \dfrac{5.0}{250} = 0.020\,\text{m}$ . In the second material $\lambda = \dfrac{8.0}{250} = 0.032\,\text{m}$ . A higher speed gives a longer wavelength at the same frequency.

4. Percentage transmitted

Answer: The reflected and absorbed energy together are $10\% + 20\% = 30\%$ , so the transmitted energy is $100\% - 30\% = 70\%$ .

5. What happens at a boundary

Some of the wave can be reflected, so it bounces back from the boundary.
Some of the wave can be transmitted, so it passes through the boundary into the other material.
Some of the wave can be absorbed by the material, so its energy is transferred to the material.

6. Why the normal is used

The normal is at right angles to the surface, so it gives one fixed reference line.
The surface itself is flat and lies along the boundary, so it cannot fix the angle of a ray. Measuring from the normal gives the same angle for every point and every surface.

7. Drawing a reflection ray diagram

Draw the flat surface as a straight line, and draw the normal as a dashed line at right angles to it at the point where the ray hits.
Draw the incident ray arriving at the surface, and mark the angle between it and the normal.
Draw the reflected ray leaving the surface on the other side of the normal, with the angle of reflection equal to the angle of incidence.

8. How curtains reduce an echo

The soft curtains absorb some of the sound energy when the wave meets them.
Less of the sound is reflected back into the room, because the curtains are not a hard, smooth surface.
With less reflected energy, the echo is quieter.

9. Depth in metres

Answer: The prefix centi means one hundredth, so $15\,\text{cm} = 0.15\,\text{m}$ .

10. Wave speed in deep water

Answer: $v = f\lambda = 4.0 \times 0.35 = 1.4\,\text{m/s}$ . The frequency does not change when the wave crosses the boundary.

11. Distance to the wall

Answer: The sound travels to the wall and back, so $d = \dfrac{v t}{2} = \dfrac{330 \times 0.60}{2} = 99\,\text{m}$ .

12. Choosing the equation and predicting the wavelength

Use v = f lambda. The speed is the frequency multiplied by the wavelength.
The frequency is set by the source, so it stays the same when the wave crosses the boundary.
The speed of sound is greater in water than in air, so for the same frequency the wavelength must be longer in water.