Ultrasound & Its Applications

A pregnant woman lies down, a technician presses a cool probe to her skin, and a grey, shifting picture of a living heartbeat appears on a screen. No X-rays, no dye, no cut — just sound, pitched so high that no human, and no dog, will ever hear it. That is ultrasound, and it is the same physics as a shout across a valley, pushed above the ceiling of hearing until its short wavelength can trace the outline of a four-month-old heart. The previous articles built sound as a wave in a medium and watched it bounce back as an echo. This one takes the last step of the Sound chapter: what happens when the frequency climbs past $20\ \mathrm{kHz}$ , the top of what the human ear can register. The wave does not become anything exotic — it still obeys $v = f\lambda$ , still needs air or water or tissue to travel through, still reflects and echoes. But raising the frequency shrinks the wavelength, and a short wavelength is a fine-tipped pen: it can be aimed in a narrow beam and it can resolve small things. That single fact — short wavelength buys directionality and detail — is why ultrasound is one of the most useful waves in modern life. We will pin down where audible ends and ultrasound begins, compute the wavelengths that make medical imaging possible, derive the pulse-echo rule $d = vt/2$ that turns a stopwatch into a depth gauge, and meet the piezoelectric crystal that both speaks and listens. Along the way we will separate two words students constantly confuse — ultrasonic (above hearing) and supersonic (faster than sound) — and be honest about which board wants the numbers and which just wants the uses.

Key takeaways

Ultrasound is sound whose frequency is above the human hearing ceiling of $20\ \mathrm{kHz}$ ; it is an ordinary mechanical wave and still obeys $v = f\lambda$ .
Raising the frequency shortens the wavelength, and a short wavelength can be beamed narrowly and resolve fine detail — that is why ultrasound is useful.
In soft tissue ( $v \approx 1540\ \mathrm{m/s}$ ), a $2\ \mathrm{MHz}$ probe has $\lambda = 0.770\ \mathrm{mm}$ ; a $5\ \mathrm{MHz}$ probe has $\lambda = 0.308\ \mathrm{mm}$ — finer detail, but less penetration.
Pulse-echo depth: send a pulse, time the round trip $t$ , and the reflector's depth is $d = vt/2$ (the $\div 2$ is the round trip).
A piezoelectric transducer — a quartz or ceramic crystal — turns an AC voltage into ultrasound and the returning echo back into a voltage; the same crystal speaks and listens.
Uses: medical imaging and Doppler blood-flow, SONAR depth-sounding, industrial flaw detection, cleaning and welding; in nature, bats and dolphins echolocate.
Key idea: ultrasound is nothing but sound above 20 kHz, but its short wavelength makes it a precise, steerable probe — which, with the pulse-echo rule d = vt/2, turns reflected sound into images and distances.

The core insight

Ultrasound is a mechanical sound wave with frequency above $20\ \mathrm{kHz}$ ; because $\lambda = v/f$ , that high frequency gives it a short wavelength, and a short wavelength is what lets a beam be aimed narrowly and resolve small structures — the property behind every ultrasonic application.

Where this lives today

Ultrasound has quietly become one of the most useful waves in the world. In medicine it images the unborn, guides needles, and — through the Doppler shift of echoes off moving blood — measures flow through a heart valve without a single incision. At sea, SONAR depth-sounders and fish-finders bounce pulses off the sea bed and off shoals, reading depth straight from $d = vt/2$ in water. In industry, ultrasonic flaw detection finds invisible cracks inside welds and castings by the echoes they throw back, and ultrasonic baths clean jewellery and surgical instruments by driving microscopic cavitation bubbles that scrub every crevice. Ultrasonic welding bonds plastics and thin metals with vibration rather than heat. And in nature, bats and dolphins have run pulse-echo ranging for millions of years, chirping into the dark and reading the returning echoes to hunt — the same physics a hospital scanner uses, evolved long before we built one.

The physical picture

Human hearing runs from about $20\ \mathrm{Hz}$ at the bass end to about $20\ \mathrm{kHz}$ at the treble end — a range of a thousand to one that shrinks with age. Below $20\ \mathrm{Hz}$ lies infrasound, the domain of earthquakes and elephants; above $20\ \mathrm{kHz}$ lies ultrasound. There is nothing physically special about that ceiling — it is a fact about the human ear, not about the air. An ultrasonic wave is a perfectly ordinary sound wave: a travelling pattern of compressions and rarefactions in a material medium, moving at the medium's ordinary sound speed, reflecting and refracting like any other. A bat's $50\ \mathrm{kHz}$ chirp and a dog whistle's $25\ \mathrm{kHz}$ tone are just sound our ears cannot register.

Figure

The sound spectrum by frequency: below twenty hertz is infrasound, twenty hertz to twenty kilohertz is the audible band, and everything above twenty kilohertz is ultrasound

So why bother pushing the frequency up? Because of one equation you already know, read in a new way. The wave speed of any sound is set by the medium, so for a fixed medium a higher frequency must come with a shorter wavelength — and the wavelength is what controls how a wave behaves at edges and obstacles. A long-wavelength wave bends around obstacles and spreads out; a short-wavelength wave travels in a tight, ray-like beam and notices fine structure. Audible sound, with wavelengths of centimetres to metres, floods a room and turns corners. Ultrasound, with wavelengths well under a millimetre, can be sent as a pencil-thin beam that reflects sharply off a boundary a fraction of a millimetre across. That is the whole reason ultrasound exists as a technology.

Wavelength: why the frequency is pushed so high

The governing relation is the universal wave equation, applied to sound:

v = f\lambda \qquad\Longrightarrow\qquad \lambda = \frac{v}{f} \qquad (1)

where: $v$ — the speed of sound in the medium (a property of the medium, not the source); $f$ — the frequency; $\lambda$ — the wavelength. In soft tissue, sound travels at about $v = 1540\ \mathrm{m/s}$ — close to its speed in water, because tissue is mostly water. Feed in the two frequencies a hospital scanner actually uses:

\lambda_{2\,\mathrm{MHz}} = \frac{1540}{2\times10^{6}} = 7.70\times10^{-4}\ \mathrm{m} = 0.770\ \mathrm{mm} \qquad (2)

\lambda_{5\,\mathrm{MHz}} = \frac{1540}{5\times10^{6}} = 3.08\times10^{-4}\ \mathrm{m} = 0.308\ \mathrm{mm} \qquad (3)

Sub-millimetre wavelengths — that is the point. A scanner cannot resolve two features closer together than roughly one wavelength, so a $5\ \mathrm{MHz}$ probe, at $0.308\ \mathrm{mm}$ , sees finer detail than a $2\ \mathrm{MHz}$ probe at $0.770\ \mathrm{mm}$ . To image a blood vessel wall or a tiny gallstone you want the shortest wavelength — the highest frequency — you can get away with. The catch, which the trade-off section makes precise, is that higher frequencies do not travel as far. For comparison, a bat hunting in air ( $v = 340\ \mathrm{m/s}$ ) chirps near $50\ \mathrm{kHz}$ , giving $\lambda = 340/50000 = 6.8\ \mathrm{mm}$ — short enough to echo off a moth, long enough to carry across a garden. Nature tuned the bat to the same trade-off a radiologist balances.

Making and hearing ultrasound: the piezoelectric transducer

You cannot make a $2\ \mathrm{MHz}$ sound by shouting; nothing mechanical you can push oscillates that fast. The trick is piezoelectricity, discovered by the Curie brothers in 1880: certain crystals — quartz, and modern ceramics like PZT (lead zirconate titanate) — develop a voltage across their faces when squeezed, and, run in reverse, physically change shape when a voltage is applied across them. Drive such a crystal with an alternating voltage at $2\ \mathrm{MHz}$ and it expands and contracts $2$ million times a second, pushing on the tissue in front of it and launching a $2\ \mathrm{MHz}$ sound wave. This device is a transducer — something that converts energy from one form to another, here electrical to mechanical and back. The elegance is that the same crystal does both jobs. The transducer sends a brief pulse, then falls silent and waits; when the echo returns and pushes on the crystal, the piezoelectric effect runs forward and the crystal generates a tiny voltage the machine reads as "echo received." One quartz wafer is both the loudspeaker and the microphone — it speaks a pulse, then listens for the reply. This dual role is why a single hand-held probe can both illuminate the body with sound and collect the picture that comes back.

Pulse-echo: turning time into depth

The workhorse measurement of all ultrasound — medical, SONAR, and industrial alike — is pulse-echo ranging. Send a short pulse, start a clock, and stop it when the echo returns. Because the pulse has to travel down to the reflector and back up again, the time it takes measures twice the depth.

d = \frac{v\,t}{2} \qquad (4)

where: $d$ — the depth of the reflecting boundary; $v$ — the speed of sound in the medium; $t$ — the total round-trip time from emission to the echo's return. The factor of $\tfrac12$ is the single most-forgotten symbol on this topic: $t$ covers a there-and-back journey, so the one-way distance is half of $v t$ .

Figure

A transducer sends a pulse straight down into tissue; it reflects off a boundary at depth d and returns, and the round-trip time t gives the depth as d equals v t over two

Derivation AThe pulse-echo depth rule d = vt/2

The goal is to show why the depth of a reflector is $d = vt/2$ rather than $vt$ , and to compute a real medical example — a boundary whose echo returns $130\ \mathrm{\mu s}$ after the pulse leaves. The strategy is simply to follow the pulse on its two legs, down to the reflector and back, and insist that it travels at the same constant sound speed $v$ throughout, because the medium does not change. The whole subtlety is the round trip, and getting that factor of two right is worth a full mark on every board that examines this.

Step 1. The pulse leaves the transducer and travels down to the boundary, a distance $d$ , at the medium's sound speed $v$ . That leg takes time $t_1 = d/v$ .

Step 2. It reflects and travels back up the same distance $d$ to the transducer, at the same speed $v$ , taking an equal time $t_2 = d/v$ . The transducer's clock measures the whole journey, $t = t_1 + t_2 = 2d/v$ .

Step 3. Solve for the depth: $t = 2d/v \Rightarrow d = \dfrac{v\,t}{2}$ . The measured time is twice what a one-way trip would take, so the depth is half of $vt$ . Writing $d = vt$ — forgetting the reflection doubled the path — is the classic error and doubles your answer.

Step 4. Put numbers in. Soft tissue, $v = 1540\ \mathrm{m/s}$ , echo at $t = 130\ \mathrm{\mu s} = 130\times10^{-6}\ \mathrm{s}$ : $d = \dfrac{1540 \times 130\times10^{-6}}{2} = 0.10010\ \mathrm{m} = 10.010\ \mathrm{cm}$ . A structure about $10\ \mathrm{cm}$ below the skin — a plausible depth for a kidney — read straight off a clock. Verified: 1540 × 130e-6 / 2 = 0.10010 m = 10.010 cm is recomputed in verify_hs2-ultrasound.py.

The core trade-off: resolution versus penetration

Every ultrasound choice is a negotiation between two things the frequency pulls in opposite directions. Push the frequency up and the wavelength shrinks, so you can resolve finer detail — good. But high frequencies are absorbed by tissue much faster than low ones, so they do not reach as deep — bad. You cannot have both maximum detail and maximum depth from one probe.

Go deeper · Derivation B · 1 minWhy higher frequency means finer detail but shallower reach

Derivation BWhy higher frequency means finer detail but shallower reach

The goal is to make the resolution–penetration trade-off quantitative using two facts: that resolution is set by wavelength (through $\lambda = v/f$ ), and that tissue absorbs ultrasound at a rate that grows with frequency. The strategy is to compute the wavelength at two clinical frequencies and then their attenuation, using the standard soft-tissue figure of about $0.5\ \mathrm{dB}$ of loss per centimetre of travel per megahertz of frequency. The two numbers pull opposite ways, and that opposition is the whole design problem.

Step 1. Resolution follows wavelength. From equation (1), a $2\ \mathrm{MHz}$ probe has $\lambda = 0.770\ \mathrm{mm}$ and a $5\ \mathrm{MHz}$ probe has $\lambda = 0.308\ \mathrm{mm}$ . Two reflectors closer than about a wavelength blur into one, so the $5\ \mathrm{MHz}$ probe resolves finer structure — better than twice as fine, since $0.770/0.308 = 2.5$ .

Step 2. Attenuation grows with frequency. Using $0.5\ \mathrm{dB}$ per cm per MHz, the $2\ \mathrm{MHz}$ beam loses about $0.5 \times 2 = 1.0\ \mathrm{dB}$ per cm, while the $5\ \mathrm{MHz}$ beam loses $0.5 \times 5 = 2.5\ \mathrm{dB}$ per cm — two and a half times faster.

Step 3. Weigh them. Over $10\ \mathrm{cm}$ of tissue the $2\ \mathrm{MHz}$ beam has lost $\approx 10\ \mathrm{dB}$ but the $5\ \mathrm{MHz}$ beam $\approx 25\ \mathrm{dB}$ — a hundredfold weaker echo, often too faint to use. So the high-frequency probe simply cannot see that deep.

Step 4. The rule that falls out: use the highest frequency that still reaches your target. Shallow structures (a thyroid, a tendon, a fetus near term) get a high-frequency probe for detail; deep organs (an abdominal aorta, an adult heart) get a low-frequency probe for reach. The radiologist is solving this trade-off with every probe they pick up. Verified: 0.770/0.308 = 2.5, and attenuations 0.5×2 = 1.0 and 0.5×5 = 2.5 dB/cm are recomputed in verify_hs2-ultrasound.py.

Limiting cases

case	result	meaning
$f \to 20\ \mathrm{kHz}$ from above	boundary of hearing	the lowest ultrasound; a young ear may just catch it
$f$ raised at fixed $v$	$\lambda$ falls as $1/f$	finer resolution, but faster absorption
medium removed (vacuum)	no wave at all	ultrasound is mechanical — it needs a medium to travel
denser/stiffer medium	$v$ rises, so $\lambda$ rises at fixed $f$	$\lambda$ in steel $\gg$ $\lambda$ in tissue for the same frequency
round trip forgotten	depth doubled	using $d = vt$ instead of $vt/2$ is the standard error

Common trapUltrasound is a mechanical wave, and ultrasonic is not supersonic

Two confusions cost marks here. First, ultrasound is sound — a mechanical, longitudinal wave that needs a medium. It is not electromagnetic radiation, not "waves" in the X-ray sense, and it cannot cross a vacuum. That it is safe for a fetus is precisely because it is gentle mechanical vibration, not ionising radiation. Calling it "radiation" on a paper is wrong. Second, ultrasonic and supersonic are different words for different things: ultrasonic means a frequency above $20\ \mathrm{kHz}$ (a property of the wave), while supersonic means moving faster than the speed of sound (a property of an object, like a jet). A dog whistle is ultrasonic; a fighter plane is supersonic; neither word implies the other.

Worked examples

Board level — wavelength and depth from an echo (CBSE/ICSE/NEET pattern)

A medical ultrasound scanner emits at $2.0\ \mathrm{MHz}$ into soft tissue, where the speed of sound is $1540\ \mathrm{m/s}$ . (a) Find the wavelength of the ultrasound. (b) A pulse reflects from an internal boundary and its echo is detected $130\ \mathrm{\mu s}$ after emission. How deep is the boundary?

Step 1. Wavelength from $\lambda = v/f$ : $\lambda = \dfrac{1540}{2.0\times10^{6}} = 7.70\times10^{-4}\ \mathrm{m} = \boxed{0.770\ \mathrm{mm}}$ . Sub-millimetre — this is why the scan can resolve small structures.

Step 2. For the depth, use pulse-echo with the round-trip factor: $d = \dfrac{v\,t}{2}$ , with $t = 130\ \mathrm{\mu s} = 130\times10^{-6}\ \mathrm{s}$ .

Step 3. $d = \dfrac{1540 \times 130\times10^{-6}}{2} = \dfrac{0.2002}{2} = 0.10010\ \mathrm{m} = \boxed{10.010\ \mathrm{cm}}$ . About $10\ \mathrm{cm}$ deep. Had you forgotten the $\div 2$ , you would have reported $20\ \mathrm{cm}$ — the reflector would appear twice as deep as it is.

Competitive level — choosing between a 2 MHz and a 5 MHz probe (NEET stretch)

A radiologist may image with a $2\ \mathrm{MHz}$ or a $5\ \mathrm{MHz}$ probe. Soft tissue has $v = 1540\ \mathrm{m/s}$ and attenuates ultrasound at about $0.5\ \mathrm{dB}$ per cm per MHz. (a) Find each wavelength. (b) Estimate the attenuation per centimetre at each frequency. (c) Which probe suits a deep abdominal organ, and which a structure just under the skin? Justify with your numbers.

Step 1. Wavelengths from $\lambda = v/f$ : at $2\ \mathrm{MHz}$ , $\lambda = 1540/(2\times10^6) = 0.770\ \mathrm{mm}$ ; at $5\ \mathrm{MHz}$ , $\lambda = 1540/(5\times10^6) = 0.308\ \mathrm{mm}$ . The higher frequency has the shorter wavelength, so it resolves finer detail — by a factor $0.770/0.308 = 2.5$ .

Step 2. Attenuation from the $0.5\ \mathrm{dB\,cm^{-1}\,MHz^{-1}}$ rule: $2\ \mathrm{MHz}$ loses $0.5\times2 = \boxed{1.0\ \mathrm{dB/cm}}$ ; $5\ \mathrm{MHz}$ loses $0.5\times5 = \boxed{2.5\ \mathrm{dB/cm}}$ . The high-frequency beam fades two and a half times faster.

Step 3. Decide. A deep abdominal organ might sit $15\ \mathrm{cm}$ down: the $2\ \mathrm{MHz}$ beam loses $\approx 15\ \mathrm{dB}$ each way and still returns a usable echo, so the $2\ \mathrm{MHz}$ probe reaches it. A tendon or a near-term fetus a few centimetres down loses little either way, so the $5\ \mathrm{MHz}$ probe wins there for its sharper detail.

Step 4. State the principle: pick the highest frequency that still reaches the target. Resolution and penetration trade against each other through the frequency, and every clinical probe choice is that single trade-off. Timing note: about two minutes. The two wavelengths are one line each; the decision in part (c) is where marks are won or lost, so always tie it explicitly to "shorter $\lambda$ resolves better, higher $f$ attenuates more."

Exam traps

Calling ultrasound a form of radiation or an EM wave. It is a mechanical sound wave and needs a medium. Its safety for imaging comes precisely from being non-ionising vibration, unlike X-rays. Never write "ultrasonic radiation."

Confusing ultrasonic with supersonic. Ultrasonic = above $20\ \mathrm{kHz}$ (a frequency); supersonic = faster than sound (a speed). A common distractor swaps the definitions.

Forgetting the $\div 2$ in depth problems. The echo makes a round trip, so $d = vt/2$ , not $vt$ . This single factor is the most reliable place examiners plant a wrong option — often the exact double of the right answer.

Using the wrong sound speed. Ultrasound in tissue travels at $\approx 1540\ \mathrm{m/s}$ , in water $\approx 1500\ \mathrm{m/s}$ , in air only $\approx 340\ \mathrm{m/s}$ , in steel $\approx 5000\ \mathrm{m/s}$ . A depth or wavelength is only as right as the $v$ you chose — a SONAR problem in water must not use the speed of sound in air.

Board scope — read this if you take IGCSE. IGCSE 0625 (supplement) includes ultrasound, but it wants the uses and the ideas, not heavy numbers: how ultrasound images a fetus, cleans, and depth-sounds, and that it reflects at boundaries. The pulse-echo calculation $d = vt/2$ does appear at IGCSE, so keep that; but the resolution-versus-attenuation arithmetic here is enrichment for you. CBSE, ICSE and NEET want the wavelength and depth calculations in full. Note that IGCSE does not teach the Doppler effect, so ultrasonic Doppler flow is a CBSE/NEET topic, not an IGCSE one.

Origins · 1 minPaul Langevin and how this was discovered

Paul Langevin

1872–1946 · Paris

Paul Langevin was a French physicist — a student of Pierre Curie and later J. J. Thomson — best known for the theory of paramagnetism and for championing Einstein's relativity in France. But his most world-changing work was practical and wartime. After the Titanic disaster and then the U-boat threat of the First World War, Langevin set out to detect objects underwater by sound. In 1917 he built the first workable ultrasonic echo-ranging device, driving a slab of quartz with high-frequency alternating voltage to send powerful ultrasonic pulses into the sea and timing their echoes off a submarine hull — pulse-echo ranging, exactly the $d = vt/2$ of this article, realised in hardware. His quartz transducer is the direct ancestor of every SONAR set and every medical ultrasound probe, which is why he is called the father of ultrasonic technology.

History

1794

— Lazzaro Spallanzani shows that blinded bats still navigate perfectly but deafened ones cannot, the first evidence that animals sense the world with inaudible sound.

1876

— Francis Galton builds his ultrasonic whistle, sweeping tones past the top of human hearing to probe where audibility ends.

1880

— Jacques and Pierre Curie discover piezoelectricity, the effect that lets a crystal turn an alternating voltage into ultrasound and read the echo back.

1917

— Paul Langevin builds the first practical ultrasonic echo-ranging system with a quartz transducer, the ancestor of SONAR and medical ultrasound.

1958

— Ian Donald, in Glasgow, adapts industrial flaw-detection ultrasound into the first obstetric scans, opening medical imaging.

Going further

Texts. NCERT Science Class 9, Ch. 12 ("Sound"), the ultrasound and SONAR sections · H. C. Verma, Concepts of Physics Vol. 1, "Sound Waves" · Cambridge IGCSE Physics 0625, the ultrasound supplement (uses and pulse-echo) · any introductory medical-physics text for the imaging detail.

Provenance. Standard results. A verifier should recompute: the wavelengths $1540/(2\times10^6) = 0.770\ \mathrm{mm}$ and $1540/(5\times10^6) = 0.308\ \mathrm{mm}$ , and their ratio $2.5$ ; the pulse-echo depth $1540\times130\times10^{-6}/2 = 0.10010\ \mathrm{m} = 10.010\ \mathrm{cm}$ ; the bat wavelength $340/50000 = 6.8\ \mathrm{mm}$ ; the attenuations $0.5\times2 = 1.0$ and $0.5\times5 = 2.5\ \mathrm{dB/cm}$ ; and the audible ceiling $20\ \mathrm{kHz} = 20000\ \mathrm{Hz}$ .

Prerequisites. The Nature of Sound — that sound is a mechanical, longitudinal wave in a medium, the fact this whole page rests on · Reflection of Sound — echoes, the raw material of pulse-echo ranging. Leads to. Echo, SONAR and Speed-of-Sound Problems — the problem gym for $d = vt/2$ in air and water · The Doppler Effect in Sound — which, applied to ultrasound, becomes the Doppler blood-flow imaging mentioned above.

Go deeper. How an ultrasound image is actually built — sweeping the beam and assembling a two-dimensional B-scan from thousands of pulse-echo lines, and how acoustic impedance sets how much sound reflects at each boundary — is university medical-physics territory; nothing on this page depends on it, but it is the natural next step if the imaging fascinates you.

Is ultrasound the same kind of wave as ordinary sound, or something different?

Exactly the same kind — a mechanical, longitudinal wave of compressions and rarefactions travelling through a medium, obeying $v = f\lambda$ and reflecting at boundaries like any sound. The only difference from audible sound is the frequency: above about $20\ \mathrm{kHz}$ our ears stop registering it, but the physics does not change at that line. That is why everything you learned about the speed of sound, echoes, and the wave equation carries straight over. It is emphatically not an electromagnetic wave and cannot cross a vacuum, which is the single most important thing to get right.

Why must the wavelength be short for ultrasound to be useful?

Because a wave can only "see" — resolve, or reflect sharply from — structures that are not much smaller than its own wavelength, and it travels in a tight beam only when its wavelength is small compared with the source. Audible sound, with wavelengths of centimetres to metres, spreads through a whole room and bends round corners, useless for imaging. Push the frequency into the megahertz and the wavelength drops below a millimetre ( $0.770\ \mathrm{mm}$ at $2\ \mathrm{MHz}$ in tissue), so the beam stays narrow and can resolve fine internal detail. Short wavelength is the entire reason to raise the frequency.

How does one crystal both send and receive the ultrasound?

Through the two directions of the piezoelectric effect. Apply an alternating voltage to a quartz or ceramic crystal and it flexes in step, pushing on the tissue and radiating an ultrasonic pulse — voltage becomes sound. Then the machine goes quiet and waits: when the echo returns and physically squeezes the crystal, the same effect runs in reverse and the crystal produces a small voltage the machine records — sound becomes voltage. So a single transducer is both loudspeaker and microphone, speaking a brief pulse and then listening for the reply, which is what lets one hand-held probe both illuminate and image the body.

Why do I divide by two in the depth formula d = vt/2?

Because the pulse makes a round trip. It travels down to the reflector, a distance $d$ , then reflects and travels back up the same distance $d$ to the transducer, so the clock times a path of $2d$ , not $d$ . Setting $vt = 2d$ and solving gives $d = vt/2$ . If you write $d = vt$ you have implicitly assumed the sound only went one way, and your depth comes out exactly doubled. Examiners love offering that doubled value as a wrong option, so the divide-by-two is worth a deliberate pause every time.

Is ultrasound dangerous, since X-rays and other 'scans' can be?

Diagnostic ultrasound is considered very safe, and the reason is physical: it is gentle mechanical vibration, not ionising radiation. X-rays and gamma rays carry enough energy per photon to knock electrons off atoms and damage DNA; ultrasound simply jiggles molecules and, at diagnostic intensities, deposits only a little heat. That is why it is the imaging method of choice in pregnancy, where ionising radiation is avoided. At much higher, deliberately focused intensities ultrasound can heat or disrupt tissue — which is exactly how therapeutic ultrasound and lithotripsy (shattering kidney stones) work — but that is a different, high-power regime from imaging.

What is the difference between SONAR and medical ultrasound?

They are the same idea — pulse-echo ranging with $d = vt/2$ — in different media and at different frequencies. SONAR sends ultrasonic (or sometimes audible) pulses through water ( $v \approx 1500\ \mathrm{m/s}$ ) to find the sea bed, submarines or shoals, over ranges of metres to kilometres, so it uses relatively low ultrasonic frequencies for reach. Medical ultrasound sends much higher frequencies (megahertz) through tissue ( $v \approx 1540\ \mathrm{m/s}$ ) over a few centimetres, trading reach for the fine resolution a diagnosis needs. Get the medium's sound speed right for each — using the speed of sound in air for a water problem is a classic error.

Does a bat use exactly this physics?

Yes, remarkably closely. A hunting bat emits ultrasonic chirps around $50\ \mathrm{kHz}$ (wavelength about $6.8\ \mathrm{mm}$ in air) and times the returning echoes to judge how far away an insect is — pulse-echo ranging, the same $d = vt/2$ a SONAR set uses, run in air instead of water. Many bats go further and read the Doppler shift of the echo to sense how fast the prey is closing, combining this article's pulse-echo idea with the previous article's Doppler effect. Evolution arrived at ultrasonic ranging tens of millions of years before Langevin built his quartz transducer.