Reflection of Sound Waves | ICSE Class 10 Physics Notes

This note covers reflection of sound, the laws of reflection, echoes and their conditions, distance calculations, reverberation, echolocation in bats and dolphins, SONAR, and uses of reflected sound in fishing and medical imaging.

What is reflection of sound, and which laws does it obey?

How does a sound wave reach a surface?

A source is an object that produces sound. Sound is produced by vibrations, or repeated to-and-fro movements. A medium is the material through which sound travels. Solids, liquids and gases can act as media; sound cannot travel through a vacuum, where matter is absent.

In air, sound travels through alternating compressions, regions of higher air density, and rarefactions, regions of lower air density. Density means mass per unit volume. The disturbance travels through the air while its particles vibrate about their positions of rest.

Definition: Reflection of sound is the bouncing back of sound waves after they encounter a reflecting surface, such as a solid or liquid surface.

The incident sound approaches the surface; the reflected sound travels away after reflection. The point of incidence is where the incident sound meets the surface. The normal is an imaginary line perpendicular to that surface at the point of incidence.

What are the two laws of reflection?

1. The angle of incidence equals the angle of reflection. The angle of incidence is measured between the incident direction and the normal; the angle of reflection is measured between the reflected direction and the normal.
2. The incident direction, reflected direction and normal at the point of incidence lie in the same plane, meaning one flat geometrical plane contains all three.

Let i represent the angle of incidence and r the angle of reflection. The first law of reflection can then be written as:

$i = r$

Draw a straight reflecting surface and a normal at the point of incidence. Add an incident arrow towards the surface and a reflected arrow away from it. Label the angles to the normal i and r, and show that they are equal.

The normal is a reference line for measuring angles. It is not another sound path. Measuring both angles from the correct reference makes the relationship between the incoming and outgoing directions clear.

How is an echo produced, and when is it heard separately?

An echo is a repetition of sound heard after the original sound is reflected from a distant surface. Near a mountain, cliff or long corridor, a shout may return after a noticeable delay. The returning sound has travelled to the reflector and back.

What happens between producing a sound and hearing its echo?

1. A source produces sound, which travels through the surrounding medium.
2. The sound reaches a reflecting surface, such as the hard surface of a distant object.
3. The surface reflects sound, and some of the reflected sound travels back towards the listener.
4. The listener hears the returning sound separately if it arrives after a sufficient interval and is clear enough to be heard.

For two sounds to be heard separately, their arrival times must differ by at least 0.1 second. The symbol s denotes the unit second. If the interval is less than 0.1 s, we cannot clearly distinguish the sounds as separate arrivals.

This explains why sound can be reflected without producing a distinct echo. In a small room, reflected sound from the walls returns too quickly for the brain to separate it from the original sound. Reflection still occurs even though no separate repetition is heard.

Why does the nature of the reflector matter?

Echoes are stronger from hard, smooth surfaces. Soft surfaces such as curtains tend to absorb sound. Rough surfaces scatter it in different directions. With these absorbing or scattering surfaces, the echo is not heard clearly.

Absorption means that sound energy is taken up by a material. Scattering means that sound is redirected in different directions. These effects help explain why delay alone does not guarantee a clear echo.

Note: An echo needs both a suitable delay and a sufficiently clear returning sound. A calculation of the minimum distance checks the time condition; it does not establish that every surface at that distance will give a clear echo.

How is the echo-distance formula obtained?

Speed is the distance travelled per unit time. Let v be the speed of sound, t the total time from emission to reception of an echo, and d the one-way distance from the source to the reflector.

Here, emission means sending out sound and reception means detecting its return. Assume the source and receiver, the device detecting sound, are together, the reflector is stationary, and sound travels at constant speed along equal outward and return paths. These conditions give the familiar two-way distance relationship.

Derivation: Distance of a stationary reflector

1. On the outward journey, sound travels the distance d from the source to the reflector.
2. On the return journey to the same position, it travels another distance d. The total distance is therefore 2d.
3. Distance equals speed multiplied by time. Over the complete journey, the distance is vt, so $2d = vt$.
4. Divide the complete distance by two to obtain the one-way distance of the reflector.

Result: $d = \frac{vt}{2}$

The factor of two represents the two journeys. It does not represent two different sound speeds or two reflecting surfaces. The source-to-reflector distance is half the distance travelled during the measured return interval.

Which units and rearrangements are useful?

SI means the International System of Units. The SI unit of distance is the metre, written m. The SI unit of time is the second. The SI unit of speed is the metre per second, written m/s or m s⁻¹.

Choose the rearrangement after identifying the unknown. In every version, t refers to the full return interval. If a question instead gives the time for the outward journey, multiply that one-way time by the speed without halving it again.

How is the minimum distance for a distinct echo calculated?

The minimum separation depends on how far sound travels during the interval needed to hear two sounds separately. For a distinct echo, use the interval of at least 0.1 s. A minimum-distance calculation uses that limiting interval as the complete outward-and-return time.

What does the familiar 17 m value mean?

Worked example 1. Take the speed of sound as 340 m s⁻¹ and the minimum interval for separate hearing as 0.1 s. Find the minimum one-way distance of a reflecting surface, with the source and listener together.

Formula: L = vt; d = L/2. Here L is the total length of the outward and return paths, v is sound speed, t is the complete travel time, and d is the one-way distance.

Substitute: L = 340 × 0.1 = 34.0 m; d = 34.0/2.

Answer: d = 17 m. Sound travels 34.0 m in total, comprising the outward journey and the return journey.

The 17 m result uses a speed of 340 m s⁻¹. It is not a fixed minimum for sound in every medium or under every condition. A different supplied speed changes the distance covered during the same hearing interval.

The speed of sound in air depends on temperature and humidity. Increasing either increases the speed. Therefore, use the speed stated in a calculation rather than replacing it with a memorised value.

How should the two distances be labelled?

Place a source and listener together opposite a wall. Label the outward path 17 m and the return path 17 m. Add arrows in opposite directions. State that the combined path is 34.0 m for a speed of 340 m s⁻¹ and an interval of 0.1 s.

The hearing interval describes the separation between arriving sounds. The distance calculation translates that interval into a path length. Keeping these two ideas separate prevents the common mistake of calling 34.0 m the minimum distance of the wall.

How are numerical problems on echoes in air solved?

Begin by deciding whether the stated interval is a return time. In a wall-echo problem with the source and listener together, an echo heard after a stated interval has completed both journeys. Multiplying speed by that interval gives the total path, not the distance of the wall.

How far away is a corridor wall?

Worked example 2. A clap in an empty corridor produces an echo after 0.5 s. The speed of sound in air is 340 m s⁻¹. Calculate the distance from the person to the reflecting wall.

Formula: d = vt/2, where d is the wall distance, v is sound speed, and t is the full return time.

Substitute: d = (340 × 0.5)/2.

Answer: d = 85 m. The wall is 85 m from the person. Halving the product accounts for the outward and return journeys.

The word after links the interval to the original clap and its returning echo. The final answer must describe a distance. Writing a numerical result without a unit or without identifying the wall distance leaves the physical meaning unclear.

How far away should an experimental reflector be placed?

Worked example 3. An experiment requires echoes to return at least 0.2 s after sound is emitted. The speed of sound is 343 m s⁻¹. Find the minimum reflector distance when emission and detection occur at the same position.

Formula: d = vt/2, with t equal to the minimum complete return time.

Substitute: d = (343 × 0.2)/2.

Answer: d = 34.3 m. For this timing requirement, the reflector must be at least 34.3 m away, assuming the echo can be detected.

This experiment specifies a longer delay than the 0.1 s interval for separate hearing. Use the stated 0.2 s requirement. Replacing it with 0.1 s would answer a different question. The required distance also uses 343 m s⁻¹, the speed supplied for this experiment.

How does reverberation differ from a distinct echo?

Reverberation is the persistence of sound after its source stops, caused by multiple reflections. In a large hall or auditorium, sound may undergo repeated reflections from the walls. Closely spaced returning sounds then prolong the sound heard inside the room.

A distinct echo is heard as a separate repetition. Reverberation is heard as a continuation of sound. Both involve reflection, but the pattern of arrival and the resulting listening experience differ.

How do arrival intervals affect what is heard?

Two sounds can be distinguished when their arrival times differ by at least 0.1 s. Reverberation occurs when reflected sounds arrive with a time difference of less than 0.05 s. These statements should not be treated as interchangeable numerical conditions.

How is reverberation controlled?

Modern auditoriums and concert halls are designed to provide desirable reverberation so that speech and music can be heard clearly. Unwanted reverberation can make sound garbled, meaning unclear because sounds run together.

Sound-absorbing panels, upholstered chairs, curtains and other soft, porous surfaces reduce unwanted reverberation. Upholstered chairs have padded coverings; porous materials contain small spaces. Such materials help control reflected sound within a room.

Reflection is therefore useful in some situations and needs control in others. The aim in a hall is a suitable listening environment. The presence of reflecting walls alone does not mean the listener will hear a clear, separate echo.

How do bats and dolphins use reflected sound?

Echolocation is the ability to locate objects using reflected sound. An animal produces sound and detects its returning echoes. Those echoes provide information about nearby objects, allowing sound to help with navigation and hunting.

What is ultrasound?

Frequency is the number of complete oscillations per second. The SI unit of frequency is the hertz, written Hz. An oscillation is one complete cycle of vibration. Kilohertz, written kHz, means one thousand hertz.

Ultrasound is sound with a frequency above 20 kHz, or 20,000 Hz. The human hearing range is 20 Hz to 20,000 Hz, but this range varies from person to person and decreases with age. Humans cannot hear ultrasound, while some animals can detect it.

Derivation: Speed, wavelength and frequency of sound

The wavelength $\lambda$ is the distance between consecutive crests or consecutive troughs of the density disturbance, measured in metres. The time period $T$ is the time for one complete density oscillation, measured in seconds. Let $\nu$ denote frequency in hertz and $v$ denote sound speed.

1. In one time period, the disturbance travels one wavelength. The distance travelled is therefore $\lambda$ in the time $T$.
2. Speed is distance divided by time, giving $v = \frac{\lambda}{T}$.
3. Frequency is the number of oscillations per second. Since each oscillation takes time $T$, $\nu = \frac{1}{T}$. Substitute this reciprocal into the speed expression.

Result: $v = \lambda\nu$. The speed of sound equals its wavelength multiplied by its frequency. Using metres for wavelength and hertz for frequency gives speed in metres per second.

How do most bats locate obstacles and prey?

Bats are nocturnal, meaning active at night. They fly and search for prey in darkness. Most bats emit short bursts of ultrasonic waves. These reflect from nearby objects, and the bats sense the returning echoes to locate obstacles and prey.

1. The bat sends out a short burst of ultrasound.
2. The sound reaches a nearby object, such as prey.
3. The object reflects sound back towards the bat.
4. The bat senses the echo and uses it to determine the object's position.

The illustration shows a bat on the left and a small flying prey animal on the right. Curved wave marks between them are labelled as ultrasonic waves sent from the bat and ultrasonic waves reflected from the prey.

See Fig. 10.27 in your NCERT textbook

Dolphins also use echolocation for navigation and hunting. The shared principle is locating objects from reflected sound. The wave travels through a medium, meets an object and returns information through an echo.

How does SONAR use echoes to locate underwater objects?

SONAR stands for Sound Navigation and Ranging. Ranging means finding distance. SONAR sends ultrasonic waves into water and analyses returning reflected waves to obtain information about underwater objects, including their distance, direction and speed.

A transmitter sends out the sound; a receiver detects returning echoes. Objects such as submarines or shipwrecks can reflect the emitted waves. The time between sending a signal and receiving its echo can be used to determine the distance travelled.

What are the stages of an echo measurement?

1. Send an ultrasonic signal into the water and record its emission time.
2. The signal travels through the water until it reaches a reflecting object or the seabed.
3. The reflected signal travels back to the receiver, where its arrival is detected.
4. Measure the complete return interval and use the speed of sound in water to calculate the one-way distance.

For a simple distance calculation, the transmitter and receiver are treated as being at the same position. The outward and return distances are equal, and the speed is taken as constant. The same $d = \frac{vt}{2}$ relationship used for an air echo then applies.

The illustration shows a ship at the water surface and a submarine below it. Wave paths are labelled as ultrasonic waves sent from the ship and ultrasonic waves reflected from the submarine.

See Fig. 10.28 in your NCERT textbook

When does the result represent depth?

Depth is the vertical distance below a reference level. If a signal travels vertically down from the measuring position to the seabed and returns, the calculated one-way distance gives the depth below that position.

An echo from another underwater object gives its distance along the signal path. Read the question carefully before labelling that distance as depth. A diagram showing the direction of travel helps connect the calculated quantity to the actual arrangement.

How are SONAR distances and ocean depths calculated?

SONAR calculations use the speed of sound in water, as supplied in the question. Do not substitute the air speed from a corridor problem. The return-time principle is unchanged, but the numerical speed belongs to the medium through which the signal travels.

How far away is a reflecting underwater object?

Worked example 4. A naval SONAR signal returns through seawater after 0.90 s. Its speed is 1530 m s⁻¹. Find the distance to the reflecting object, treating the outgoing and returning paths as equal.

Formula: t₁ = t/2; d = vt₁. Here t₁ is the one-way travel time, t is the complete return time, v is sound speed, and d is the object's distance.

Substitute: t₁ = 0.90/2 = 0.45 s; d = 1530 × 0.45.

Answer: d = 688.5 m. The one-way travel time is 0.45 s.

This method halves the time before calculating distance. It gives the same result as multiplying speed by the complete time and halving the distance. Once the time has been halved, multiplying speed by that time gives the one-way distance directly.

How deep is the ocean at a measurement point?

Worked example 5. A SONAR signal sent vertically towards the ocean floor takes 4 s to return. Sound travels through seawater at 1500 m s⁻¹. Find the depth below the measuring position.

Formula: d = vt/2, where d is the depth, v is sound speed in seawater, and t is the complete return time.

Substitute: d = (1500 × 4)/2.

Answer: d = 3000 m. The measured interval includes both the downward and upward journeys.

How far down is a sunken ship?

Worked example 6. A ship receives a SONAR echo from wreckage after 5 s. Sound travels at 1525 m s⁻¹ in seawater. Treat the wreckage as vertically below the measuring position and find its depth.

Formula: d = vt/2, where d is the wreckage depth, v is sound speed, and t is the full return time.

Substitute: d = (1525 × 5)/2.

Answer: d = 3812.5 m. This is the one-way depth obtained from the supplied data.

Each problem supplies its own sound speed. Keeping those values attached to their respective questions avoids mixing assumptions from different examples.

How are reflected sounds useful in fishing and medicine?

How can fishermen use echoes?

Fishermen use echo-sounding, a method of investigating underwater objects and depths with reflected sound. A sound signal is sent into the water, and returning echoes can help locate shoals, or groups of fish. This is an application of the same principle used in SONAR.

The useful information comes from detecting the reflected signal. Where the sound speed and return interval are known, the two-way travel calculation gives the distance of the reflecting region. A signal directed towards the seabed can also be used for depth measurement.

What happens in ultrasound imaging?

Ultrasonography means making images of internal body structures using ultrasound. It allows internal organs to be imaged without surgery. In this application, ultrasonic signals enter the body and echoes return from boundaries between tissues, groups of cells that perform particular functions.

The returning signals are detected and processed to build an image. Different echo return times help indicate the positions of reflecting boundaries. Thus, the image uses information carried by reflected sound rather than a person listening for an audible repetition.

The common sequence is emission, reflection and detection. What differs is the medium, the reflecting object and the way the returning information is used. An animal senses echoes; an instrument detects and processes them for a measurement or image.

Note: The 0.1 s condition concerns hearing two sounds separately. It is not a requirement that every ultrasonic measuring instrument or imaging system must wait that long before detecting a returning signal.

For any application, name both the reflected sound and its purpose. A complete explanation connects the returning wave to the useful information: object position, underwater distance, water depth or an image of internal structures.

Glossary

• Reflection of sound — The bouncing back of sound waves after they encounter a reflecting surface.
• Incident sound — Sound travelling towards a reflecting surface before it meets that surface.
• Normal — An imaginary line perpendicular to the reflecting surface at the point of incidence.
• Angle of incidence — The angle between the incident sound direction and the normal at the reflecting surface.
• Angle of reflection — The angle between the reflected sound direction and the normal at the reflecting surface.
• Echo — A repetition heard when reflected sound returns after a sufficient interval.
• Return time — The complete interval between sending a sound signal and detecting its reflected return.
• Reverberation — Persistence of sound after its source stops, caused by multiple reflections.
• Frequency — The number of complete oscillations occurring in each second, measured in hertz.
• Ultrasound — Sound with frequency above 20,000 Hz, beyond the human audible range.
• Echolocation — Locating objects by detecting sound reflected back from those objects.
• SONAR — Sound Navigation and Ranging, using underwater ultrasonic signals and their returning reflections.
• Ultrasonography — The use of ultrasound to make images of internal body structures without surgery.

Common errors and misconceptions

• Misconception: Every reflected sound is heard as a separate echo. Correct: A distinct echo requires a sufficient arrival interval and a clear returning sound; small-room reflections may return too quickly.
• Misconception: The incidence and reflection angles are measured from the reflecting surface. Correct: Both are measured from the normal at the point of incidence.
• Misconception: Speed multiplied by the echo return time gives the distance of the wall. Correct: It gives the outward-and-return path. For equal paths, halve this distance.
• Misconception: The minimum reflector distance is 17 m under every condition. Correct: That value uses 340 m s⁻¹ and a 0.1 s minimum interval. Use the speed and timing condition supplied.
• Misconception: After halving the return time, the calculated distance must also be halved. Correct: Multiply speed by the one-way time directly; halving again would apply the same journey correction twice.
• Misconception: All bats use ultrasonic bursts for echolocation. Correct: Most bats emit such bursts; this qualification is part of the description.
• Misconception: SONAR problems should use the air speed of sound. Correct: Use the given speed in water for underwater travel.
• Misconception: An instrument cannot detect an echo returning before 0.1 s. Correct: That interval describes separate human hearing, not the detection limit of every instrument.

Exam-style questions with model answers

Q1. State the two laws of reflection of sound. [2 marks]

1. The angle of incidence equals the angle of reflection; both angles are measured from the normal at the point of incidence.
2. The incident direction, reflected direction and normal at the point of incidence lie in the same plane.

Q2. A clap produces an echo after 0.5 s. The sound speed is 340 m s⁻¹. The source and listener are together, and the outward and return distances are equal. Calculate the distance of the wall, explaining the distance used. [3 marks]

1. Let d be the wall distance, v the sound speed and t the complete echo time. Sound travels to the wall and back, covering 2d during the measured interval.
2. Use 2d = vt, so d = vt/2. Substituting gives d = (340 × 0.5)/2.
3. The wall is 85 m away. This is the one-way distance, because the measured 0.5 s includes both journeys.

Q3. Two sounds must arrive at least 0.1 s apart to be heard separately. Take sound speed as 340 m s⁻¹. For a source and listener together, calculate the minimum reflector distance and explain why 34.0 m is not that distance. [4 marks]

1. The minimum complete echo interval is 0.1 s, so use that time for the outward and return journeys together.
2. Total distance travelled equals speed multiplied by time: 340 × 0.1 = 34.0 m.
3. The sound covers equal outward and return distances, so the reflector distance is 34.0/2 = 17 m.
4. The value 34.0 m describes the whole path, not the separation from the reflector. A clear echo also needs a suitable reflecting surface and detectable returning sound.

Q4. Explain how SONAR measures the depth below a ship's measuring position. A signal travels vertically to the seabed and returns after 4 s; sound speed in the water is 1500 m s⁻¹. Assume equal outward and return paths. Include the full name and calculate the depth. [5 marks]

1. SONAR means Sound Navigation and Ranging. It uses ultrasonic waves sent into water and the echoes returning from underwater reflectors.
2. A transmitter sends the signal vertically towards the seabed. The seabed reflects the sound, and a receiver detects the returning echo.
3. The measured 4 s interval includes both journeys. If d is the depth below the measuring position, the total path length is 2d.
4. Let v be sound speed and t the return time. Using 2d = vt gives d = vt/2 = (1500 × 4)/2.
5. The depth is 3000 m. The result is a one-way vertical distance; the interval supplied was the complete downward-and-upward travel time.

Q5. Define echolocation, explain how most bats use it, and name another animal that uses it for navigation and hunting. [3 marks]

1. Echolocation means locating objects through reflected sound. Sound travels to an object and returns as an echo that can be detected.
2. Most bats emit short bursts of ultrasonic waves. Nearby obstacles and prey reflect those waves, and the bats sense the echoes to determine their positions in darkness.
3. Dolphins also use echolocation for navigation and hunting. In both cases, the useful information comes from reflected sound.

Q6. Distinguish a distinct echo from reverberation, and name two measures used to reduce unwanted reverberation. [3 marks]

1. A distinct echo is a separate repetition of sound. The original and returning sounds are distinguished as separate arrivals after a sufficient interval.
2. Reverberation is the persistence of sound after the source stops, caused by multiple reflections. It is experienced as prolonged sound rather than a clearly separate repetition.
3. Sound-absorbing panels and curtains reduce unwanted reverberation. They absorb sound and help prevent reflected sounds from making speech or music unclear.

Q7. Give one use of reflected sound in fishing and explain how reflected ultrasound helps produce a medical image. [3 marks]

1. Fishermen can use echo-sounding to help locate groups of fish. Sound is sent into water and returning reflections provide information about underwater reflectors.
2. In ultrasonography, ultrasonic signals enter the body and echoes return from boundaries between tissues. A receiver detects these reflected signals.
3. The echoes are processed to form an image of internal structures. Their return times help indicate the locations of reflecting boundaries, allowing internal organs to be imaged without surgery.

Q8. An experiment requires an echo to return at least 0.2 s after emission. Sound speed is 343 m s⁻¹. Emission and detection occur at the same position, with equal outward and return paths. Calculate the minimum reflector distance. [2 marks]

1. The minimum distance is d = vt/2, where v is sound speed and t is the complete return time.
2. Substitution gives d = (343 × 0.2)/2 = 34.3 m. The reflector must be at least 34.3 m away.

Key takeaways

• Sound obeys the laws of reflection: equal angles to the normal, with the incident direction, reflected direction and normal in one plane.
• A distinct echo requires at least 0.1 s between sounds heard separately, together with a clear returning reflection.
• For equal outward and return paths, reflector distance equals sound speed multiplied by the full return time, divided by two.
• The 17 m minimum reflector distance uses a speed of 340 m s⁻¹ and a hearing interval of 0.1 s.
• Reverberation prolongs sound through multiple reflections; sound-absorbing panels, upholstered chairs and curtains help reduce unwanted reverberation.
• Most bats emit ultrasonic bursts and sense the returning echoes to locate obstacles and prey; dolphins also use echolocation.
• SONAR uses underwater ultrasound and returning echoes; numerical calculations must use the sound speed supplied for the water.
• Reflected sound helps fishermen locate fish and allows medical instruments to build ultrasound images of internal body structures.

Test yourself

From which line are the incidence and reflection angles measured?

Both are measured from the normal, the perpendicular to the surface at the point of incidence.

Why can a small room reflect sound without producing a distinct echo?

Reflections can return too quickly for the brain to distinguish them from the original sound.

Why is a complete echo travel distance divided by two?

The measured path includes equal outward and return journeys, so the reflector is half that total distance away.

Why should 17 m not be used as a universal minimum echo distance?

It assumes a sound speed of 340 m s⁻¹ and a minimum interval of 0.1 s. A different speed changes the distance.

Which surfaces give stronger echoes, and what do curtains tend to do?

Hard, smooth surfaces give stronger echoes. Curtains are soft surfaces that tend to absorb sound.

What is the difference between echo time and one-way travel time?

Echo time includes travel to the reflector and back. For equal paths at constant speed, one-way time is half the echo time.

What does SONAR stand for, and what kind of sound does it use?

SONAR stands for Sound Navigation and Ranging. It sends ultrasonic waves into water and detects reflected waves.

What information do most bats obtain by sensing returning echoes?

The echoes help most bats determine the positions of nearby obstacles and prey while moving and hunting in darkness.