Reflection of Sound and Echoes ICSE Class 10: Conditions, Numericals and SONAR

What Reflection of Sound and Echoes is about

Weightage: CISCE publishes no chapter or unit weightage for ICSE Physics. Reflection of Sound and Echoes is in the 3. Sound (i) part of the 2027 syllabus and can be asked in both Section A and Section B of the 80-mark paper.Source: CISCE ICSE Physics (Science Paper 1) syllabus, examination year 2027

Key concepts

Reflection of sound

Sound is reflected from large, hard surfaces (walls, cliffs, the sea bed), obeying the same laws as light: the angle of incidence equals the angle of reflection, and the incident and reflected sound and the normal lie in one plane.

The reflecting surface should be large compared with the wavelength of the sound. Soft, porous surfaces (curtains, carpets) absorb sound and reflect little.

Worked example

Why can a large cliff reflect your shout clearly, but a small tree cannot?

Try it first, then show the solutionSolution

1. Audible sound has wavelengths of centimetres to metres.
2. A reflector must be large compared with the wavelength: a cliff is, a thin tree is not, so the tree scatters little sound back.

Answer: The reflector must be large compared with the wavelength.

Echo and the 0.1 s condition

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

The sensation of a sound persists in our ear for about $\frac{1}{10}$ s (persistence of hearing). So the reflected sound must arrive at least 0.1 s after the original to be heard separately.

Conditions for hearing an echo:
1. Minimum distance between source and reflector so that the time gap is at least 0.1 s.
2. The reflecting surface must be large compared with the wavelength.
3. The sound must be loud enough to be heard after reflection.

Worked example

Why is no echo heard in an ordinary classroom?

Try it first, then show the solutionSolution

1. The walls are only a few metres away.
2. Reflected sound returns in much less than 0.1 s and merges with the original sound.

Answer: The walls are closer than about 17 m, so the gap is under 0.1 s.

The minimum distance for an echo

Sound travels to the reflector and back, a total distance $2d$, in time $t$:
$v=\frac{2d}{t}$, so $d=\frac{vt}{2}$.

With $t=0.1$ s and $v=340$ m/s in air, $d_{\min }=\frac{340\times 0.1}{2}=17$ m.

The minimum distance is proportional to the speed of sound in the medium: more in water (about 1450–1500 m/s) and steel, and slightly more in warm air.

Worked example

Find the minimum distance for an echo in water where sound travels at 1450 m/s.

Try it first, then show the solutionSolution

1. $d_{\min }=\frac{v\times 0.1}{2}=\frac{1450\times 0.1}{2}$.
2. $d_{\min }=72.5$ m.

Answer: 72.5 m

Echo numericals

Use $v=\frac{2d}{t}$, always counting the round trip.

• Distance of a reflector: $d=\frac{vt}{2}$.
• Speed of sound: $v=\frac{2d}{t}$.
• Time for an echo: $t=\frac{2d}{v}$.

Convert minutes to seconds and km to m first. Give the unit with the answer.

Worked example

A boy shouts and hears the echo from a hill 1.5 s later. If sound travels at 340 m/s, how far away is the hill?

Try it first, then show the solutionSolution

1. $d=\frac{vt}{2}=\frac{340\times 1.5}{2}$.
2. $d=255$ m.

Answer: 255 m

Echoes between two reflectors

A person between two cliffs at distances $d_1$ (nearer) and $d_2$ hears:

• the first echo from the nearer cliff at $t_1=\frac{2d_1}{v}$;
• the second from the farther cliff at $t_2=\frac{2d_2}{v}$;
• a third echo, of sound reflected from one cliff to the other and back, at $t_3=\frac{2(d_1+d_2)}{v}=t_1+t_2$.

The distance between the cliffs is $d_1+d_2$.

Worked example

A girl stands between two cliffs, 85 m from one and 170 m from the other. When does she hear the first and second echoes of a clap? ($v=340$ m/s)

Try it first, then show the solutionSolution

1. First (nearer cliff): $\frac{2\times 85}{340}=0.5$ s.
2. Second (farther cliff): $\frac{2\times 170}{340}=1$ s.

Answer: 0.5 s and 1 s

SONAR and ultrasonic waves

Ultrasonic waves are sound waves of frequency above 20 kHz, too high for humans to hear.

SONAR (Sound Navigation And Ranging): a ship's transmitter sends ultrasonic pulses into the water; a receiver picks up the echo from the sea bed, a submarine or a shoal of fish. From the time and the speed of sound in water, $d=\frac{vt}{2}$.

Why ultrasonic? Short wavelength, so they travel as narrow beams and are reflected even by small objects; they are not easily absorbed; they cannot be heard, so they cause no disturbance.

Ultrasonic waves are mechanical (sound) waves, not electromagnetic; do not confuse them with ultraviolet.

Worked example

A ship's SONAR receives the echo from the sea bed 0.8 s after sending the pulse. Speed of sound in sea water is 1450 m/s. Find the depth.

Try it first, then show the solutionSolution

1. $d=\frac{vt}{2}=\frac{1450\times 0.8}{2}$.
2. $d=580$ m.

Answer: 580 m

Uses of echoes in nature and medicine

• Bats emit ultrasonic squeaks and locate obstacles and insects from the echoes (echolocation), so they can fly and hunt in the dark.
• Dolphins use ultrasonic clicks to find food and navigate.
• Fishermen use echo sounders (SONAR) to find shoals of fish.
• Medical field: ultrasonography (ultrasound scanning) uses echoes from internal organs to form images, e.g. of a baby in the womb or kidney stones, without harmful radiation.
• Echo methods are also used to measure the depth of the sea and the speed of sound.

Worked example

A bat emits ultrasound of frequency 50 kHz in air ($v=340$ m/s). Find its wavelength, and the distance of an insect whose echo returns in 0.05 s.

Try it first, then show the solutionSolution

1. $\lambda =\frac{v}{f}=\frac{340}{50000}=0.0068$ m $=6.8$ mm.
2. $d=\frac{340\times 0.05}{2}=8.5$ m.

Answer: 6.8 mm; 8.5 m

Physics in real life: Reflection of Sound and Echoes

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.

Topics in this chapter

• Echo Numericals (v = 2d/t)
• Conditions for an Echo and the 17 m Rule
• SONAR and Uses of Echoes

Reflection of Sound and Echoes formula sheet

Echo equation

$v=\frac{2d}{t}$

2d: to the reflector and back.

Distance of reflector

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

Convert minutes to seconds.

Minimum distance for an echo

$d_{\min }=\frac{v\times 0.1}{2}$

17 m in air at 340 m/s.

Wave equation for sound

$v=f\lambda$

Ultrasonic: f > 20 kHz.

Third echo between two cliffs

$t_3=\frac{2(d_1+d_2)}{v}=t_1+t_2$

Sound bounces off both cliffs.

Reflection of Sound and Echoes: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

How the board asks Reflection of Sound and Echoes

The question patterns to expect, and the method that works for each.

Tricks to remember Reflection of Sound and Echoes

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

• There and back again

  An echo travels twice the distance: v = 2d/t.

  Use it when: Every echo numerical.

• One-tenth, seventeen

  0.1 s of persistence → 17 m in air.

  Use it when: Conditions for echo.

• Sonic is sound, violet is light

  Ultrasonic (sound, SONAR) ≠ ultraviolet (light).

  Use it when: Naming SONAR waves.

• Faster medium, farther wall

  Minimum echo distance grows with the speed of sound.

  Use it when: Comparing media.

Reflection of Sound and Echoes: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section B style)

Read the situation, then answer every part. In the board paper this is a 10-mark question split into 3 + 3 + 4 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Reflection of Sound and Echoes

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

• ✗Mistake: Forgetting the round trip

  CISCE examiners' comments, 2025: in a submarine problem, candidates did not account for the reflection, using d = vt instead of d = vt/2.

  Wrong working

  d = 1500 × t

  Right working

  d = (1500 × t) ÷ 2

  ✓Do this: Write v = 2d/t every time.

  Usually costs: 1 to 2 marks.

• ✗Mistake: Not converting time units

  CISCE examiners' comments, 2025: candidates made errors converting 1 minute 40 seconds into seconds, and gave wrong units.

  ✓Do this: 1 min 40 s = 100 s. Convert before substituting.

  Usually costs: 1 mark.

• ✗Mistake: Wrong minimum time or ratio form

  CISCE examiners' comments, 2024: comparing minimum echo distances in two media, candidates did not use 0.1 s in both, divided speeds randomly, or did not give the answer as a ratio x : y.

  ✓Do this: d_min = v × 0.1 ÷ 2 for each medium; compare as x : y.

  Usually costs: 1 to 2 marks.

• ✗Mistake: Wrong waves for SONAR

  CISCE examiners' comments, 2024: candidates wrote ultraviolet, infrared or radio waves instead of ultrasonic waves.

  ✓Do this: SONAR uses ultrasonic (sound) waves; it works in water, where radio waves do not travel far.

  Usually costs: 1 mark.

• ✗Mistake: Ignoring the minimum distance

  CISCE examiners' comments, 2024: with two cliffs, candidates used a 10 m distance to compute an echo, not realising that a reflector closer than about 17 m in air gives no separate echo.

  ✓Do this: Check each reflector against the 17 m rule before calculating an echo time.

  Usually costs: 1 to 2 marks.

Practice questions on Reflection of Sound and Echoes

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

1. Q1 · 1 mark · MCQBoard level

   The speed of sound in air at room temperature is about:
   (a) 34 m/s
   (b) 340 m/s
   (c) 3400 m/s
   (d) $3\times 10^8$ m/s

   Hint

   Not the speed of light.

   Answer

   (b)

2. Q2 · 1 mark · MCQBoard level

   SONAR uses:
   (a) infrasonic waves
   (b) ultrasonic waves
   (c) ultraviolet rays
   (d) radio waves

   Hint

   Sound above 20 kHz.

   Answer

   (b)

3. Q3 · 1 mark · MCQHarder board level

   If the speed of sound in water is 1500 m/s, the minimum distance for an echo in water is:
   (a) 17 m
   (b) 75 m
   (c) 150 m
   (d) 1500 m

   Hint

   v × 0.1 ÷ 2.

   Answer

   (b)

4. Q4 · 1 mark · Assertion-reasonBoard level

   Assertion (A): Bats can fly and catch insects in complete darkness.
   Reason (R): Bats emit ultrasonic sounds and locate objects from the echoes.
   (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

   Echolocation.

   Answer

   (c)

5. Q5 · 2 marks · Short answerHarder board level

   A gun is fired and its echo is heard from a cliff after 2.5 s. Find the distance of the cliff if sound travels at 330 m/s.

   Hint

   Halve the round trip.

   Answer

   $d=\frac{330\times 2.5}{2}=412.5$ m

6. Q6 · 2 marks · Very short answerHarder board level

   Name the medical technique that uses ultrasonic echoes and give one use.

   Hint

   Scans.

   Answer

   Ultrasonography (ultrasound scanning): imaging a baby in the womb, or detecting kidney stones.

7. Q7 · 3 marks · Long answerBoard level

   (a) Define echo. (b) Show that, in air where sound travels at 340 m/s, the reflector must be at least 17 m away for an echo to be heard.

   Hint

   0.1 s and 2d.

   Answer

   (a) Sound heard after reflection from a distant obstacle, distinct from the original. (b) Gap ≥ 0.1 s; $2d=v\times 0.1=34$ m, so $d\ge 17$ m.

8. Q8 · 3 marks · Long answerHarder board level

   The minimum distances for an echo in two media A and B are x and y. Sound travels at 340 m/s in A and 1360 m/s in B. Compare x and y.

   Hint

   d_min ∝ v.

   Answer

   $x=\frac{340\times 0.1}{2}=17$ m; $y=\frac{1360\times 0.1}{2}=68$ m; $x:y=1:4$.

9. Q9 · 4 marks · Long answerHOTS / competency

   A man standing between two parallel cliffs fires a shot. He hears the first echo after 0.8 s and the second after 2 s. ($v=340$ m/s) Find the distance between the cliffs and when he hears the third echo.

   Hint

   Third echo bounces off both cliffs.

   Answer

   Distances $\frac{340\times 0.8}{2}=136$ m and $\frac{340\times 2}{2}=340$ m; the cliffs are 476 m apart. Third echo after $\frac{2\times 476}{340}=2.8$ s $=0.8+2$ s.

10. Q10 · 3 marks · Long answerHarder board level

    Explain why an echo is not heard in a classroom 10 m long, and how the 0.1 s rule explains it.

    Hint

    Find the round-trip time.

    Answer

    Round trip to the far wall is at most 20 m, taking $\frac{20}{340}\approx 0.06$ s, less than 0.1 s; the reflected sound merges with the original, so no separate echo is heard.

11. Q11 · 1 mark · MCQHOTS / competency

    On a hot day the speed of sound in air increases. The minimum distance required to hear an echo:
    (a) decreases
    (b) increases
    (c) stays the same
    (d) becomes zero

    Hint

    d_min ∝ v.

    Answer

    (b)

12. Q12 · 4 marks · Long answerHOTS / competency

    A student fires a starting pistol 165 m from a wall and hears the echo 1 s later. (a) Find the speed of sound. (b) On the same day, the echo from another cliff takes 2.4 s. How far is that cliff? (c) Why is it better to time many echoes in a row than a single one?

    Hint

    v = 2d/t.

    Answer

    (a) $v=\frac{330}{1}=330$ m/s. (b) $d=\frac{330\times 2.4}{2}=396$ m. (c) Reaction time errors are a large fraction of a single short time; timing many echoes and dividing reduces the error.

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. 1.Define echo.Show answer

   Sound heard after reflection from a distant obstacle, distinct from the original sound.

2. 2.Minimum time gap for an echo?Show answer

   0.1 s

3. 3.Minimum distance in air (v = 340 m/s)?Show answer

   17 m

4. 4.Echo formula?Show answer

   v = 2d/t

5. 5.Full form of SONAR?Show answer

   Sound Navigation And Ranging.

6. 6.Waves used in SONAR?Show answer

   Ultrasonic waves.

7. 7.Frequency of ultrasonic waves?Show answer

   Above 20 kHz.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Do you always use 2d?
• ☐Can you derive the 17 m rule?
• ☐Do you convert minutes to seconds?
• ☐Can you compare minimum distances in two media as a ratio?
• ☐Can you handle two-cliff echoes?
• ☐Can you explain why ultrasonic waves are used?

Reflection of Sound and Echoes: frequently asked questions