PhysicsICSE Class 10 · Reflection of Sound and Echoes

Conditions for Hearing an Echo: 0.1 s, 17 m and the Reflector (ICSE Class 10)

Q: What are the conditions for hearing an echo?

The echo must reach the ear at least 0.1 s after the original sound (a reflector at least 17 m away in air), the reflector must be large and hard, and the sound must be loud enough to be heard after reflection.

Q: Why is 17 m the minimum distance for an echo?

The ear holds a sound for about 0.1 s. Sound in air at 340 m/s travels 34 m in 0.1 s, which is 17 m to the reflector and 17 m back.

Q: Does the minimum distance for an echo change in water?

Yes. It is proportional to the speed of sound. Sound is about four to five times faster in water than in air, so the reflector must be about 75 m or more away.

Updated 15 October 2026 · Fact-checked

An echo is heard only if (1) the reflected sound reaches the ear at least 0.1 s after the original, which in air at 340 m/s needs a reflector at least 17 m away, (2) the reflector is large and hard compared with the wavelength, and (3) the sound is loud enough to be heard after reflection.

Notice the sound covers 2d, there and back: 340 m/s × 0.1 s = 34 m, so the reflector must be at least 17 m away.

Illustration drawn for QuestionBuddy.

Understand Conditions for an Echo and the 17 m Rule

Reflection of sound. Sound, like light, is reflected from surfaces and obeys the laws of reflection: the angle of incidence equals the angle of reflection, and the incident sound, the reflected sound and the normal at the point of incidence lie in the same plane. Hard, smooth surfaces (rock, concrete, metal, the sea bed) reflect sound well; soft, porous surfaces (curtains, carpets, cushions) absorb much of it.

Echo. An echo is the sound heard after reflection from a distant obstacle, separately from the original sound.

Persistence of hearing. The sensation of a sound lasts in our ear for about $\frac{1}{10}$ s (0.1 s) after the sound stops. If the reflected sound arrives within 0.1 s, it merges with the original and only makes it seem longer or louder. To be heard as a separate sound, the echo must arrive at least 0.1 s later.

Condition 1: minimum distance. In time $t$ the sound goes to the reflector and back ( $2 d$ ), so $d = \frac{v t}{2}$ . Putting $t = 0.1$ s:

$d_{m i n} = \frac{v \times 0.1}{2} .$

In air at $340$ m/s, $d_{m i n} = 17$ m. A reflector closer than this gives no distinct echo.

The medium changes the minimum distance. $d_{m i n}$ is directly proportional to the speed of sound:

Medium (speed of sound)	Minimum distance
Carbon dioxide (about 260 m/s)	about 13 m
Air at room temperature (340 m/s)	17 m
Sea water (about 1530 m/s)	about 76 m
Steel (about 5000 m/s)	about 250 m

So for two media, $d_{1} : d_{2} = v_{1} : v_{2}$ . In warmer air sound is faster, so the minimum distance is a little larger.

Condition 2: the reflector. The reflecting surface must be large compared with the wavelength of the sound, and preferably hard. Audible sound in air has wavelengths from a few centimetres up to several metres (for example, 170 Hz gives $λ = \frac{340}{170} = 2$ m), so a cliff, a large building or a hillside reflects it well but a small board does not. Ultrasonic waves have wavelengths of millimetres, which is why they are reflected even by small objects such as a fish or an insect.

Condition 3: intensity. The sound must be loud enough that, after travelling to the reflector and back (and losing energy by spreading out and by absorption at the surface), it can still be heard. A short, sharp sound (a clap, a shout, a gunshot) gives the clearest echo, because it is over before the echo returns.

Why you do not hear echoes in an ordinary room. The walls are usually closer than 17 m, so reflected sound returns in less than 0.1 s and merges with the original. In addition, furniture, curtains and people absorb sound. The many quick reflections only make the sound seem to last a little longer.

Summary of the conditions.
1. Time gap of at least 0.1 s (distance at least $\frac{v \times 0.1}{2}$ , 17 m in air).
2. A large, hard reflecting surface (large compared with the wavelength).
3. A sound of sufficient intensity, preferably short and sharp.

Conditions for an Echo and the 17 m Rule: formulas and key facts

Minimum distance for a distinct echo: $d_{m i n} = \frac{v \times 0.1}{2}$; 17 m in air at 340 m/s.
Comparing two media: $\frac{d _{1}}{d _{2}} = \frac{v _{1}}{v _{2}}$; The time gap (0.1 s) is the same in both.
Wavelength of sound: $λ = \frac{v}{f}$; The reflector should be large compared with λ.
Three conditions for an echo: Time gap at least 0.1 s, so at least 17 m in air
Reflector large and hard compared with the wavelength
Sound loud enough, short and sharp; Learn all three; 'state two conditions' is common.

Time gap	at least 0.1 s, so at least 17 m in air
Reflector	large and hard compared with the wavelength
Sound	loud enough, short and sharp

How to solve Conditions for an Echo and the 17 m Rule questions

Questions on conditions are either explanations (why an echo is or is not heard) or minimum-distance calculations.

1For a calculation, write $d_{m i n} = \frac{v \times 0.1}{2}$ and substitute the speed in that medium.
2To compare two media, use the same 0.1 s in both and give the answer as a ratio, $d_{1} : d_{2} = v_{1} : v_{2}$ , in its simplest form.
3To judge whether an echo is heard, find $t = \frac{2 d}{v}$ and compare it with 0.1 s (or compare $d$ with $d_{m i n}$ ).
4For explanations, use the three conditions: time gap (distance), size and nature of the reflector, loudness of the sound.
5If the question mentions curtains, carpets or crowds, add that soft surfaces absorb sound and weaken the reflection.

Conditions for an Echo and the 17 m Rule in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

No echo in your bedroom

Suppose you clap loudly in your bedroom, whose walls are about 4 m apart, and then in an empty stone quarry with a rock face 40 m away. (Speed of sound 340 m/s)

Where do you hear an echo, and why?

Try it first, then show the solutionSolution

Bedroom: the nearest wall is at most 4 m away, so $t = \frac{2 \times 4}{340} \approx 0.024$ s, much less than 0.1 s.
Quarry: $t = \frac{2 \times 40}{340} \approx 0.24$ s, more than 0.1 s, and the rock face is large and hard.

Answer: Only in the quarry; in the bedroom the reflection returns too soon and merges with the clap.

Memory hook: Too close, too quick: no echo.

Echoes under water

Imagine a diver tapping a metal pipe under the sea, where sound travels at about 1530 m/s. A sunken ship's hull is 50 m away.

Will the diver hear a distinct echo from the hull?

Try it first, then show the solutionSolution

Minimum distance in sea water: $d_{m i n} = \frac{1530 \times 0.1}{2} = 76.5$ m.
The hull is only 50 m away, which is less than 76.5 m.

Answer: No: under water the reflector must be at least about 76.5 m away, so a hull 50 m away gives no distinct echo.

Memory hook: Faster medium, farther reflector needed.

Tricks to remember Conditions for an Echo and the 17 m Rule

Short memory aids. Say each one aloud once, then use it on a question the same day.

One-tenth rule
'The ear holds a sound for one-tenth of a second': the echo must wait at least 0.1 s.
Use it when: Stating persistence of hearing.
Seventeen
34 m round trip in 0.1 s means 17 m each way in air.
Use it when: Recalling the minimum distance in air.
Three Ts: Time, Target, Thunder
Time gap of 0.1 s; a large hard Target; a loud (Thunder-like) sound.
Use it when: Listing the three conditions.
Speed and distance move together
Double the speed of sound, double the minimum echo distance.
Use it when: Comparing media and temperatures.

Conditions for an Echo and the 17 m Rule: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Persistence of hearing · 1 mark · MCQBoard level

The sensation of a sound persists in the human ear for about:
(a) 0.01 s
(b) 0.1 s
(c) 1 s
(d) 10 s

Try it first, then show the solutionSolution

Persistence of hearing lasts about one-tenth of a second.

Answer: (b) 0.1 s

Example 2: Minimum distance in carbon dioxide · 2 marks · Short answerBoard level

Sound travels at 260 m/s in carbon dioxide. Calculate the minimum distance of a reflector from the source for a distinct echo to be heard in this gas.

Try it first, then show the solutionSolution

Time gap needed $= 0.1$ s, during which sound covers $2 d$ . (1)
$d_{m i n} = \frac{260 \times 0.1}{2} = 13$ m. (1)

Answer: 13 m

Example 3: Air against steel · 3 marks · Short answerHarder board level

Sound travels at 332 m/s in air at 0 °C and at 5000 m/s in steel. (a) How far away, at the least, must a reflector be for an echo in each? (b) Express them as a ratio in its simplest form. (c) Why is the distance so much larger in steel?

Try it first, then show the solutionSolution

(a) Air: $\frac{332 \times 0.1}{2} = 16.6$ m; steel: $\frac{5000 \times 0.1}{2} = 250$ m. (1)
(b) $16.6 : 250 = 332 : 5000 = 83 : 1250$ . (1)
(c) Sound travels about 15 times faster in steel, so it covers a much longer round trip in the same 0.1 s. (1)

Answer: (a) 16.6 m and 250 m (b) 83 : 1250 (c) $d_{m i n}$ is proportional to the speed of sound

Example 4: A clap near a wall · 3 marks · Short answerHarder board level

A boy claps 8 m from a large wall. The speed of sound is 340 m/s. (a) After how long does the reflected sound reach him? (b) Does he hear an echo? (c) What does he hear instead?

Try it first, then show the solutionSolution

(a) $t = \frac{2 \times 8}{340} \approx 0.047$ s. (1)
(b) No: 0.047 s is less than 0.1 s (and 8 m is less than 17 m). (1)
(c) The reflected sound merges with the clap, which only seems slightly louder or longer. (1)

Answer: (a) About 0.047 s (b) No (c) The reflection merges with the original sound

Example 5: Size of the reflector · 3 marks · Short answerHarder board level

(a) Find the wavelength in air of (i) a 170 Hz hum and (ii) a 40 kHz ultrasonic pulse. (Speed of sound 340 m/s) (b) Explain why a small cardboard sheet 30 cm wide gives a poor echo of the hum, while a bat's ultrasonic pulse is reflected well even by a moth.

Try it first, then show the solutionSolution

(a)(i) $λ = \frac{340}{170} = 2$ m. (ii) $λ = \frac{340}{40 000} = 0.0085$ m $= 8.5$ mm. (1)
(b) A reflector must be large compared with the wavelength. A 30 cm sheet is much smaller than 2 m, so it reflects the hum poorly. (1)
A moth is much larger than 8.5 mm, so it reflects the ultrasonic pulse well. (1)

Answer: (a) 2 m and 8.5 mm (b) The reflector must be large compared with the wavelength

Example 6: Effect of a warm day · 2 marks · Short answerHarder board level

On a winter morning the speed of sound in air is 332 m/s; on a summer afternoon it is 350 m/s. By how many metres does the minimum distance for an echo change, and in which direction?

Try it first, then show the solutionSolution

Winter: $\frac{332 \times 0.1}{2} = 16.6$ m; summer: $\frac{350 \times 0.1}{2} = 17.5$ m. (1)
It increases by $17.5 - 16.6 = 0.9$ m. (1)

Answer: It increases by 0.9 m (from 16.6 m to 17.5 m)

Example 7: Case study: the school assembly hall · 4 marks · Case-basedHOTS / competency

A school hall is 25 m long with bare concrete end walls. Sound travels at 345 m/s.
(a) A teacher claps standing against one end wall. Does she hear a distinct echo from the far wall? Show working.
(b) She now claps standing at the centre of the hall. Does she hear a distinct echo from either end wall?
(c) Find the least distance from a wall at which she could stand and still hear a distinct echo from it.
(d) Suggest one way to reduce the echo problem in the hall, giving the reason.

Try it first, then show the solutionSolution

(a) $t = \frac{2 \times 25}{345} \approx 0.145$ s $> 0.1$ s: yes, a distinct echo. (1)
(b) Each wall is 12.5 m away: $t = \frac{25}{345} \approx 0.072$ s $< 0.1$ s, so no distinct echo. (1)
(c) $d_{m i n} = \frac{345 \times 0.1}{2} = 17.25$ m. (1)
(d) Hang curtains or fix soft boards on the end walls: soft, porous materials absorb sound and reflect little. (1)

Answer: (a) Yes, 0.145 s (b) No, 0.072 s (c) 17.25 m (d) Soft absorbing materials on the walls

Where students lose marks in Conditions for an Echo and the 17 m Rule

The slips examiners see most often, and what to write instead.

Mistake: Using different times in two media
The 0.1 s comes from the ear, not from the medium, so it is the same in both.
Wrong working
Taking 0.1 s in one medium and some other time in the other, or dividing the speeds at random.
Right working
$x : y = \frac{v _{1} \times 0.1}{2} : \frac{v _{2} \times 0.1}{2} = v_{1} : v_{2}$ .
Do this: Write both $d_{m i n}$ expressions with 0.1 s, then cancel.
Usually costs: Reported by CISCE examiners for 2024 Q2(vii), where answers were also not given in the ratio form x : y.
Mistake: Taking an echo from a reflector that is too close
A reflector nearer than $d_{m i n}$ gives no distinct echo, even if it is drawn in the diagram.
Wrong working
Calculating an echo time from a cliff 10 m away in air.
Right working
10 m < 17 m, so ignore that cliff; use the farther one.
Do this: Compare every distance with $d_{m i n}$ before calculating.
Usually costs: Reported by CISCE examiners for 2024 Q7(i).
Mistake: Forgetting the factor 2 in the minimum distance
In 0.1 s sound goes there and back.
Wrong working
$d_{m i n} = 340 \times 0.1 = 34$ m.
Right working
$d_{m i n} = \frac{340 \times 0.1}{2} = 17$ m.
Do this: Say 'there and back' as you write the 2.
Usually costs: The calculation mark.
Mistake: Giving only one condition
'State the conditions' expects the time gap (distance), the reflector and the loudness.
Wrong working
'The wall must be 17 m away.'
Right working
'At least 0.1 s gap (17 m in air); a large, hard reflector; a loud enough sound.'
Do this: Learn the three Ts: Time, Target, Thunder.
Usually costs: Part marks.

Exam tips

In 'state two conditions', the safest pair is the 0.1 s time gap (17 m in air) and a large reflecting surface.
For 'compare x and y' write the ratio with the medium's speeds; common factors (0.1 and 2) cancel.
The 2027 specimen paper asked whether the minimum echo distance is greater or lesser in water than in air: greater, because sound is faster in water.
Mention persistence of hearing by name when you explain the 0.1 s.

Practice questions on Conditions for an Echo and the 17 m Rule

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The minimum distance of a reflector from the listener for an echo in air (speed 340 m/s) is:
(a) 1.7 m
(b) 17 m
(c) 34 m
(d) 170 m
Hint
Use 0.1 s and halve.
Answer
(b) 17 m
Q2 · 2 marks · Short answerBoard level
State the laws of reflection of sound.
Hint
Same as for light.
Answer
(1) The angle of incidence is equal to the angle of reflection. (2) The incident sound, the reflected sound and the normal at the point of incidence all lie in the same plane.
Q3 · 1 mark · Very short answerBoard level
Find the minimum distance for a distinct echo in sea water, where sound travels at 1530 m/s.
Hint
$d_{m i n} = \frac{v \times 0.1}{2}$ .
Answer
$\frac{1530 \times 0.1}{2} = 76.5$ m.
Q4 · 2 marks · Short answerHarder board level
Explain the term 'persistence of hearing' and how it leads to the minimum distance for an echo.
Hint
Link 0.1 s to the round trip.
Answer
The sensation of a sound remains in the ear for about 0.1 s after the sound ends. A reflected sound arriving within 0.1 s merges with the original, so for a distinct echo the round trip must take at least 0.1 s; with $2 d = v \times 0.1$ , the reflector must be at least $\frac{v \times 0.1}{2}$ away (17 m in air).
Q5 · 2 marks · Short answerHarder board level
In a certain liquid the minimum distance for a distinct echo is 60 m. Find the speed of sound in the liquid.
Hint
Rearrange $d_{m i n} = \frac{v \times 0.1}{2}$ .
Answer
$v = \frac{2 \times 60}{0.1} = 1200$ m/s.
Q6 · 2 marks · Short answerHarder board level
The minimum distances for a distinct echo are $x$ in a gas where sound travels at 320 m/s and $y$ in a liquid where it travels at 1440 m/s. Find $x : y$ and state which is larger.
Hint
Ratio of the speeds.
Answer
$x : y = \frac{320 \times 0.1}{2} : \frac{1440 \times 0.1}{2} = 16 : 72 = 2 : 9$ ; $y$ is larger because sound is faster in the liquid.
Q7 · 2 marks · Short answerHarder board level
Give two reasons why we rarely hear echoes inside a furnished living room.
Hint
Distance and surfaces.
Answer
(1) The walls are closer than 17 m, so reflected sound returns in less than 0.1 s and merges with the original. (2) Curtains, sofas, carpets and cushions absorb much of the sound, so the reflection is weak.
Q8 · 3 marks · Short answerHOTS / competency
A student finds that in a gas filling a long pipe, the minimum distance for a distinct echo is 18 m. (a) Find the speed of sound in the gas. (b) A clap is made 30 m from a closed end of the pipe. When is the echo heard? (c) Will it be distinct? Explain.
Hint
First the speed, then $t = \frac{2 d}{v}$ .
Answer
(a) $v = \frac{2 \times 18}{0.1} = 360$ m/s. (b) $t = \frac{2 \times 30}{360} \approx 0.167$ s. (c) Yes: 0.167 s is more than 0.1 s (30 m is more than 18 m), and the closed end is a hard reflector.
Q9 · 1 mark · Assertion-reasonHOTS / competency
Assertion (A): A bat can detect a small moth by echoes, but a person shouting cannot get an echo from the moth.
Reason (R): A reflector must be large compared with the wavelength of the sound, and the bat's ultrasonic waves have a much shorter wavelength than audible sound.
(a) (A) is true and (R) is false.
(b) (A) is false and (R) is true.
(c) Both (A) and (R) are true, and (R) is the correct explanation for (A).
(d) Both (A) and (R) are true, but (R) is not the correct explanation for (A).
Hint
Compare the moth's size with each wavelength.
Answer
(c) Both are true, and (R) explains (A).

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Persistence of hearing?Show answer
About 0.1 s.
2.Minimum echo distance in air at 340 m/s?Show answer
17 m.
3.Is the minimum echo distance more or less in water than in air?Show answer
More, since sound is faster in water.
4.What kind of surface gives the best echo?Show answer
A large, hard one.
5.How does $d_{m i n}$ depend on the speed of sound?Show answer
Directly proportional.

Conditions for an Echo and the 17 m Rule: frequently asked questions

What are the conditions for hearing an echo?

The echo must reach the ear at least 0.1 s after the original sound (a reflector at least 17 m away in air), the reflector must be large and hard, and the sound must be loud enough to be heard after reflection.

Why is 17 m the minimum distance for an echo?

The ear holds a sound for about 0.1 s. Sound in air at 340 m/s travels 34 m in 0.1 s, which is 17 m to the reflector and 17 m back.

Does the minimum distance for an echo change in water?

Yes. It is proportional to the speed of sound. Sound is about four to five times faster in water than in air, so the reflector must be about 75 m or more away.