Sound bounces back when it strikes a hard surface, just as a ball rebounds off a wall. This bouncing back is the reflection of sound, and it underlies echoes, reverberation and many useful technologies.
Laws of reflection of sound. Sound obeys the same two laws as light: (1) the angle of incidence equals the angle of reflection, and (2) the incident wave, the reflected wave and the normal to the reflecting surface at the point of incidence all lie in the same plane. Hard, smooth surfaces such as walls, cliffs and metal sheets are good reflectors; soft, porous surfaces such as curtains, carpets and sponge absorb sound and reflect little.
Echo. An echo is a sound heard again after it reflects off a distant surface and returns to the listener. To hear a separate echo, the reflected sound must reach the ear at least $0.1\,\text{s}$ after the original — this is how long our brain stores a sound (the persistence of hearing). If the gap is shorter, the original and the reflection merge into one prolonged sound.
Minimum distance for an echo. In $0.1\,\text{s}$ sound travels (taking $v=344\,\text{m/s}$) a total distance of $344\times0.1=34.4\,\text{m}$. But this is the to-and-fro path, so the reflecting surface must be at least half of that away: $$d=\frac{34.4}{2}=17.2\,\text{m}.$$ So you need a wall or cliff at least about $17.2\,\text{m}$ away to hear a clear echo.
Reverberation. In a large hall, sound reflects repeatedly off the walls, floor and ceiling, so it persists even after the source stops. This prolonged, overlapping reflection is called reverberation. Too much of it makes speech unclear, so auditoriums use sound-absorbing materials — heavy curtains, carpets, padded seats and rough plaster — to control it.
SONAR. SONAR (Sound Navigation And Ranging) uses reflected ultrasound to find the depth and distance of objects underwater. A transmitter on a ship sends out an ultrasonic pulse; it travels down, reflects off the seabed or a submarine, and returns to a detector. If the speed of sound in water is $v$ and the pulse takes a total time $t$ to go down and come back, the distance $d$ to the object satisfies $$2d=v\times t,$$ because the pulse covers the depth twice. Hence $d=\dfrac{v\times t}{2}$.
Other uses of ultrasound:
- Medical imaging: ultrasonography (ultrasound scans) forms images of the heart, foetus and internal organs.
- Cleaning: ultrasonic waves dislodge dirt from delicate objects such as watches and electronic parts.
- Flaw detection: ultrasound reveals hidden cracks inside metal blocks and castings.
- Echolocation: bats and dolphins navigate and hunt by emitting ultrasound and sensing the echoes.
A person shouts towards a cliff and hears the echo $3\,\text{s}$ later. If the speed of sound in air is $340\,\text{m/s}$, how far is the cliff?
Solution- Total time for sound to go to the cliff and return: $t=3\,\text{s}$; speed $v=340\,\text{m/s}$.
- Total distance travelled $=v\times t=340\times3=1020\,\text{m}$.
- This is the to-and-fro distance, so the cliff distance is half: $d=\dfrac{1020}{2}$.
- $d=510\,\text{m}$.
Answer: The cliff is $510\,\text{m}$ away.
A SONAR pulse sent from a ship returns from the seabed after $4\,\text{s}$. If the speed of sound in sea water is $1500\,\text{m/s}$, find the depth of the sea.
Solution- Given: $t=4\,\text{s}$, $v=1500\,\text{m/s}$. Required: depth $d$.
- The pulse travels down and back, so $2d=v\times t$.
- Substitute: $2d=1500\times4=6000\,\text{m}$.
- So $d=\dfrac{6000}{2}=3000\,\text{m}$.
Answer: The depth of the sea is $3000\,\text{m}$.
Calculate the minimum distance from a reflecting wall needed to hear a distinct echo, taking the speed of sound as $344\,\text{m/s}$.
Solution- The reflected sound must arrive at least $0.1\,\text{s}$ after the original.
- In $0.1\,\text{s}$, sound travels $v\times t=344\times0.1=34.4\,\text{m}$ in total.
- This is the to-and-fro distance, so the wall must be half as far: $d=\dfrac{34.4}{2}$.
- $d=17.2\,\text{m}$.
Answer: The minimum distance for a distinct echo is about $17.2\,\text{m}$.
A submarine sends an ultrasonic signal that is reflected from a sea wreck and detected after $0.8\,\text{s}$. If the speed of sound in water is $1450\,\text{m/s}$, how far is the wreck?
Solution- Given: $t=0.8\,\text{s}$, $v=1450\,\text{m/s}$. Required: distance $d$.
- The signal travels to the wreck and back, so $2d=v\times t$.
- Substitute: $2d=1450\times0.8=1160\,\text{m}$.
- So $d=\dfrac{1160}{2}=580\,\text{m}$.
Answer: The wreck is $580\,\text{m}$ from the submarine.
Why is it difficult to hear speech clearly in a large empty hall, and how is this corrected?
Solution- In a large hall, sound reflects repeatedly off the bare hard walls, floor and ceiling.
- These overlapping reflections persist after the source stops — this is reverberation.
- The lingering sound blurs with new speech, making words hard to distinguish.
- It is reduced by covering surfaces with sound-absorbing materials such as curtains, carpets and padded seats.
Answer: Excess reverberation muddles the speech; absorbing materials (curtains, carpets, padded seats, rough plaster) cut down the reflections and restore clarity.
State the two laws of reflection of sound and give one situation where each can be observed.
Solution- First law: the angle of incidence is equal to the angle of reflection.
- Second law: the incident sound, the reflected sound and the normal at the point of incidence all lie in the same plane.
- The equal-angles law is seen in a megaphone or a curved sounding board that directs sound to listeners.
- The same-plane law is seen when an echo from a cliff returns straight back along the line you shouted.
Answer: (1) Angle of incidence equals angle of reflection (seen in a megaphone/sounding board); (2) incident sound, reflected sound and normal lie in one plane (seen in a straight cliff echo).