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Question 1Physical Principles and Instrumentation

Which property of a sound wave is determined solely by the source and remains unchanged as the wave passes from one medium into another?

  1. Wavelength
  2. Propagation speed
  3. Frequency Correct answer.
  4. Acoustic impedance

Explanation

Frequency is set by the pulser/transducer and is unchanged when the wave enters a new medium. Wavelength and propagation speed both change with the medium because they are linked by the equation speed = frequency x wavelength.

Question 2Physical Principles and Instrumentation

The time required for one complete cycle of a sound wave is called the:

  1. wavelength
  2. duty factor
  3. pulse duration
  4. period Correct answer.

Explanation

The period is the duration of a single cycle and is the reciprocal of frequency. Wavelength is a distance, while duty factor and pulse duration describe pulsed-wave timing rather than a single cycle.

Question 3Physical Principles and Instrumentation

Through which of the following media does ultrasound travel fastest?

  1. Bone Correct answer.
  2. Fat
  3. Soft tissue
  4. Air

Explanation

Propagation speed increases as a medium becomes stiffer and less compressible. Bone is the stiffest of these media, so sound travels through it at roughly 4000 m/s or more, compared with about 1540 m/s in soft tissue.

Question 4Physical Principles and Instrumentation

The pulse repetition period of a pulsed-wave system equals the sum of the:

  1. pulse duration and spatial pulse length
  2. wave period and wavelength
  3. duty factor and pulse duration
  4. pulse duration and listening time Correct answer.

Explanation

The pulse repetition period is measured from the start of one pulse to the start of the next, so it includes both the transmit time (pulse duration) and the receive or listening time. It is the reciprocal of the pulse repetition frequency.

Question 5Physical Principles and Instrumentation

Which transducer component is designed to reduce reflection at the boundary between the transducer and the skin?

  1. Matching layer Correct answer.
  2. Backing layer
  3. Piezoelectric element
  4. Acoustic lens

Explanation

The matching layer has an impedance between that of the element and soft tissue, which reduces the reflection at the interface and improves transmission. Its thickness is typically one quarter of the wavelength in the matching material.

Question 6Physical Principles and Instrumentation

The piezoelectric effect refers to the:

  1. conversion of electricity into sound and back Correct answer.
  2. conversion of light energy into electrical energy
  3. conversion of sound energy into thermal energy
  4. conversion of heat energy into electrical energy

Explanation

A piezoelectric crystal changes shape when a voltage is applied (transmit) and produces a voltage when deformed by returning echoes (receive). This two-way energy conversion is the basis of all ultrasound transducers.

Question 7Physical Principles and Instrumentation

Which display mode plots echo amplitude as vertical spikes along a single line of sight?

  1. Color Doppler
  2. A-mode Correct answer.
  3. M-mode
  4. B-mode

Explanation

In A-mode (amplitude mode) the x-axis represents depth and the y-axis represents echo amplitude, producing a spike display. It was the earliest display format and is still used in some ophthalmic and echo-ranging measurements.

Question 8Physical Principles and Instrumentation

The damping (backing) material of a pulsed transducer primarily:

  1. reduces the number of cycles per pulse Correct answer.
  2. increases the emitted acoustic power
  3. improves the lateral resolution of the beam
  4. widens the beam in the elevation plane

Explanation

The backing block absorbs energy from the rear face of the crystal and mechanically damps ringing, so the pulse contains fewer cycles and has a broader bandwidth. A shorter pulse directly improves axial resolution.

Question 9Physical Principles and Instrumentation

A reflector returns an echo that is 100 times weaker than the transmitted intensity. What is this difference in decibels?

  1. 100 dB
  2. 20 dB Correct answer.
  3. 40 dB
  4. 10 dB

Explanation

Decibels are calculated as 10 log of the intensity ratio, so 10 log(100) = 20 dB. A factor of 100 in intensity therefore corresponds to a 20 dB change.

Question 10Physical Principles and Instrumentation

An echo returns 65 microseconds after transmission. Assuming a propagation speed of 1540 m/s in soft tissue, the reflector is located at a depth of approximately:

  1. 2.5 cm
  2. 5 cm Correct answer.
  3. 10 cm
  4. 13 cm

Explanation

The round-trip distance equals speed times time: 1540 m/s x 65 microseconds is about 10 cm. Because the sound traveled to the reflector and back, the depth is half of that, or about 5 cm.

Question 11Physical Principles and Instrumentation

Axial resolution is determined by the:

  1. spatial pulse length Correct answer.
  2. transducer diameter
  3. length of the focal zone
  4. beam width in the scan plane

Explanation

Two reflectors aligned along the beam can be separated only if they are farther apart than half the spatial pulse length. Shorter pulses therefore produce better axial resolution.

Question 12Physical Principles and Instrumentation

Which change would improve axial resolution?

  1. Using a broadband transducer with heavy damping Correct answer.
  2. Moving the focal zone deeper into the tissue
  3. Selecting a lower frequency imaging transducer for penetration
  4. Increasing the acoustic output power of the system

Explanation

Axial resolution depends on spatial pulse length, which equals the number of cycles multiplied by wavelength. Heavy damping and broad bandwidth shorten the pulse, and a higher frequency shortens the wavelength, both improving axial resolution.

Question 13Physical Principles and Instrumentation

Lateral resolution is best:

  1. only in the far field
  2. at the transducer face
  3. where the beam is narrowest Correct answer.
  4. throughout the entire beam

Explanation

Lateral resolution depends on beam width, so it is optimal at the focal zone where the beam is narrowest. Away from the focus the beam widens and lateral resolution degrades.

Question 14Physical Principles and Instrumentation

Elevation (slice-thickness) resolution is determined mainly by the:

  1. Receive gain
  2. Pulse duration
  3. Element height Correct answer.
  4. Doppler angle

Explanation

The dimension of the element perpendicular to the scan plane, together with a fixed acoustic lens, sets the elevation beam width and therefore the slice thickness. Thin slices require a short element height and a well-placed elevation focus.

Question 15Physical Principles and Instrumentation

Increasing the selected imaging depth on a scanner will generally:

  1. decrease the frame rate Correct answer.
  2. increase the frame rate
  3. increase the pulse repetition frequency
  4. reduce the duty factor to zero

Explanation

Deeper imaging lengthens the listening time needed for echoes to return, which increases the pulse repetition period and lowers the pulse repetition frequency. With a fixed number of lines per frame, a lower PRF means fewer frames can be produced each second.

Question 16Physical Principles and Instrumentation

Tissue harmonic imaging improves image quality primarily because harmonic signals:

  1. are stronger than the fundamental at every depth in tissue
  2. arise mainly in the strongest central part of the beam Correct answer.
  3. are generated only when a contrast agent is present
  4. have a lower frequency than the transmitted fundamental

Explanation

Harmonics arise from nonlinear propagation, which is greatest where the beam intensity is highest, so they are generated predominantly in the central beam and at depth. Because weak side-lobe and near-field echoes produce few harmonics, the harmonic image shows less clutter and better contrast.

Question 17Physical Principles and Instrumentation

A Doppler angle of 60 degrees produces a Doppler shift that is what fraction of the maximum possible shift for the same flow?

  1. Three quarters
  2. Equal to the maximum
  3. One half Correct answer.
  4. One quarter

Explanation

The Doppler equation includes the cosine of the angle between the beam and flow. The cosine of 60 degrees is 0.5, so the measured shift is half of what would be recorded at zero degrees.

Question 18Physical Principles and Instrumentation

Aliasing in spectral Doppler occurs when the Doppler shift exceeds:

  1. half the pulse repetition frequency Correct answer.
  2. the transmitted frequency
  3. the frame rate
  4. twice the pulse repetition frequency

Explanation

The Nyquist limit equals half the pulse repetition frequency, and Doppler shifts above this value are incorrectly displayed on the opposite side of the baseline. Raising the PRF, lowering the transmitted frequency, or increasing the Doppler angle can eliminate aliasing.

Question 19Physical Principles and Instrumentation

Increasing the color Doppler packet size (ensemble length) will:

  1. remove aliasing
  2. improve temporal resolution
  3. increase the color frame rate
  4. reduce the color frame rate Correct answer.

Explanation

A larger packet means more pulses must be transmitted along each color line, so more time is required to build each frame and the frame rate falls. The trade-off is more accurate velocity estimation.

Question 20Physical Principles and Instrumentation

Which change most improves the signal-to-noise ratio when imaging a deep structure?

  1. Increasing the acoustic output power Correct answer.
  2. Increasing the receiver gain
  3. Decreasing the pulse repetition frequency
  4. Switching to a higher frequency probe

Explanation

Signal-to-noise ratio improves only when the returning echo signal is strengthened relative to noise. Raising output power increases the transmitted intensity and therefore the echo amplitude, provided ALARA limits are respected.

Question 21Physical Principles and Instrumentation

Two media have acoustic impedances of 4 MRayl and 8 MRayl. At normal incidence, the fraction of intensity reflected at the interface is approximately:

  1. About 44%
  2. About 67%
  3. About 33%
  4. About 11% Correct answer.

Explanation

The intensity reflection coefficient equals the square of the impedance difference divided by the impedance sum. Here (8-4)/(8+4) = 0.33, and squaring gives about 0.11, so roughly 11% of the intensity is reflected.

Question 22Physical Principles and Instrumentation

A 5 MHz transducer operating in soft tissue produces a wavelength of approximately:

  1. 0.77 mm
  2. 3.1 mm
  3. 0.31 mm Correct answer.
  4. 0.15 mm

Explanation

Wavelength equals propagation speed divided by frequency, so 1540 m/s divided by 5,000,000 Hz gives about 0.000308 m, or 0.31 mm. This is a useful rule-of-thumb value for a 5 MHz probe in soft tissue.

Question 23Physical Principles and Instrumentation

Using an attenuation coefficient of 0.5 dB/cm/MHz, the attenuation of a 5 MHz beam after traveling 4 cm in soft tissue is approximately:

  1. 2 dB
  2. 20 dB
  3. 5 dB
  4. 10 dB Correct answer.

Explanation

Total attenuation equals the coefficient multiplied by frequency and path length: 0.5 x 5 x 4 = 10 dB. Note that this is a one-way attenuation figure; an echo must travel the path twice, so round-trip attenuation would be 20 dB.

Question 24Physical Principles and Instrumentation

The quality factor (Q factor) of a pulsed transducer is inversely related to the:

  1. bandwidth of the transducer Correct answer.
  2. acoustic impedance of the matching layer
  3. pulse repetition frequency
  4. wavelength in soft tissue

Explanation

Q factor is defined as the resonant frequency divided by the bandwidth, so a wide bandwidth gives a low Q factor and a short pulse. Highly damped, broadband transducers therefore have low Q values.

Question 25Physical Principles and Instrumentation

Which control amplifies returning echoes differently depending on their depth?

  1. Display dynamic range
  2. Acoustic output power
  3. Time gain compensation Correct answer.
  4. Overall receiver gain

Explanation

Time gain compensation applies progressively greater amplification to later-arriving echoes so that deeper reflectors appear equally bright despite greater attenuation. Overall gain, by contrast, amplifies all echoes by the same amount.

Question 26Physical Principles and Instrumentation

If the frequency of a continuous sound wave is doubled, the period of the wave:

  1. is unchanged
  2. is doubled
  3. increases fourfold
  4. is halved Correct answer.

Explanation

Period is the time required for one complete cycle and equals the reciprocal of frequency (T = 1/f). Doubling the frequency therefore halves the period. Period and frequency are always inversely related, independent of amplitude or intensity.

Question 27Physical Principles and Instrumentation

In soft tissue, the propagation speed of sound is determined primarily by:

  1. tissue stiffness and density
  2. the acoustic power of the beam
  3. the transmitted acoustic frequency Correct answer.
  4. the angle of incidence at a boundary

Explanation

Speed is a property of the medium, described by c = square root of (bulk modulus divided by density). Stiffer media conduct sound faster, while denser media slow it down. Frequency, power, and beam angle do not alter the propagation speed through a given medium.

Question 28Physical Principles and Instrumentation

If the amplitude of a sound wave is tripled while the frequency is held constant, the intensity becomes:

  1. 27 times greater
  2. 6 times greater
  3. 9 times greater Correct answer.
  4. 3 times greater

Explanation

Intensity is proportional to the square of the pressure amplitude. Tripling the amplitude therefore multiplies intensity by 3 squared, which is 9. Frequency does not enter this relationship.

Question 29Physical Principles and Instrumentation

For specular reflection at a large, smooth interface, the angle of reflection is:

  1. smaller than the angle of incidence when speed increases
  2. equal to the angle of refraction
  3. always 90 degrees to the incident beam
  4. equal to the angle of incidence measured from the normal Correct answer.

Explanation

Specular reflection obeys the law that the angle of reflection equals the angle of incidence, both measured from the normal to the interface. The reflected beam returns on the same side of the interface as the incident beam. Refraction, by contrast, involves transmission into the second medium.

Question 30Physical Principles and Instrumentation

Absorption of ultrasound in tissue refers to the:

  1. bending of the beam where propagation speed changes
  2. conversion of acoustic energy into heat Correct answer.
  3. redirection of sound by small scatterers
  4. partial reflection at an impedance boundary

Explanation

Absorption is the portion of attenuation in which acoustic energy is converted into heat within the tissue. Together with reflection and scattering, it accounts for the total attenuation of the beam. Absorption increases with frequency and with depth traveled.

Question 31Physical Principles and Instrumentation

Which Doppler technique avoids aliasing because it does not sample the returning signal at discrete intervals?

  1. Color Doppler
  2. Continuous-wave Doppler Correct answer.
  3. Duplex spectral Doppler
  4. Pulsed-wave Doppler

Explanation

Aliasing occurs when a sampled signal exceeds the Nyquist limit imposed by the pulse repetition frequency. Continuous-wave Doppler transmits and receives simultaneously and processes the returning signal continuously, so no sampling limit applies. Pulsed and color techniques all remain subject to the Nyquist limit.

Question 32Physical Principles and Instrumentation

The wall filter in a Doppler system is used to:

  1. eliminate low-frequency echoes from stationary tissue Correct answer.
  2. remove high-velocity signals from the display
  3. increase the Doppler angle correction applied by the operator
  4. reduce the transmitted acoustic power

Explanation

Strong, low-frequency Doppler shifts arise from moving vessel walls and surrounding tissue and can obscure the flow signal. The wall filter rejects these low-frequency, high-amplitude components. Setting the filter too high can also remove genuine slow flow.

Question 33Physical Principles and Instrumentation

Increasing the overall receiver gain will:

  1. change the transmitted acoustic power
  2. amplify only echoes returning from deep structures
  3. improve axial resolution
  4. amplify all returning echoes by the same amount Correct answer.

Explanation

Overall gain is applied uniformly in the receiver, so every returning echo is boosted by the same factor. It improves brightness but does not correct for depth-dependent attenuation, which is the job of time-gain compensation. Receiver gain also cannot alter the transmitted pulse or resolution.

Question 34Physical Principles and Instrumentation

What is the wavelength of a 3 MHz beam traveling in soft tissue? Assume a propagation speed of 1540 m/s.

  1. approximately 0.51 mm Correct answer.
  2. approximately 1.5 mm
  3. approximately 4.6 mm
  4. approximately 0.31 mm

Explanation

Wavelength equals propagation speed divided by frequency. Dividing 1540 m/s by 3,000,000 Hz gives 0.000513 m, or about 0.51 mm. Wavelength shrinks as frequency rises.

Question 35Physical Principles and Instrumentation

What is the highest pulse repetition frequency that avoids range ambiguity when the deepest structure of interest is 5 cm deep? Assume c = 1540 m/s.

  1. 1,540 Hz
  2. 15,400 Hz Correct answer.
  3. 7,700 Hz
  4. 30,800 Hz

Explanation

An echo must complete a round trip of twice the depth, or 10 cm, before the next pulse is emitted. Using PRF = c/(2d) gives 1540 divided by 0.1 m, or 15,400 Hz. Higher PRFs would cause echoes from one pulse to arrive after the next pulse has been sent.

Question 36Physical Principles and Instrumentation

A pulsed wave has a pulse duration of 1 microsecond and a pulse repetition period of 1 millisecond. The duty factor is:

  1. 0.0001
  2. 0.01
  3. 1.0
  4. 0.001 Correct answer.

Explanation

Duty factor is the fraction of time the transducer is transmitting, calculated as pulse duration divided by pulse repetition period. Dividing 1 microsecond by 1000 microseconds gives 0.001. Duty factor is therefore always a small number for pulsed imaging.

Question 37Physical Principles and Instrumentation

A 5 MHz transducer in soft tissue emits pulses that contain 2 cycles. The spatial pulse length is approximately:

  1. 1.20 mm
  2. 0.15 mm
  3. 0.31 mm
  4. 0.62 mm Correct answer.

Explanation

Spatial pulse length equals the number of cycles in the pulse multiplied by the wavelength. At 5 MHz in soft tissue the wavelength is about 0.31 mm, so two cycles occupy about 0.62 mm. Fewer cycles and higher frequencies shorten the pulse.

Question 38Physical Principles and Instrumentation

Using an attenuation coefficient of 0.5 dB/cm/MHz, what thickness of soft tissue attenuates a 4 MHz beam by 6.9 dB?

  1. about 13.8 cm
  2. about 3.5 cm Correct answer.
  3. about 1.7 cm
  4. about 6.9 cm

Explanation

At 4 MHz the coefficient is 0.5 times 4, or 2 dB per centimeter. Dividing the 6.9 dB loss by 2 dB/cm gives about 3.5 cm, which is one half-value layer. Each additional 3.5 cm removes another 6.9 dB, halving the intensity again.

Question 39Physical Principles and Instrumentation

Refraction of an ultrasound beam at a tissue boundary occurs because:

  1. the acoustic impedance mismatch at the boundary is very large
  2. the propagation speed differs between the two media Correct answer.
  3. the second medium attenuates the beam more strongly
  4. the beam is scattered by small reflectors at the boundary

Explanation

Refraction is a change in beam direction caused by a difference in propagation speed between two media at a non-perpendicular incidence. The relationship is described by Snell's law. Scattering, reflection, and attenuation involve other physical mechanisms.

Question 40Physical Principles and Instrumentation

Which structure acts as a Rayleigh scatterer for a 5 MHz beam in soft tissue?

  1. A fetal femur
  2. The aortic wall
  3. A red blood cell Correct answer.
  4. The diaphragm

Explanation

Rayleigh scattering applies when the reflector is much smaller than the wavelength, and it increases strongly with frequency. At 5 MHz the wavelength in soft tissue is about 0.31 mm, so cells of roughly 7 micrometers behave as Rayleigh scatterers. Large structures such as the diaphragm and bone reflect specularly.

Question 41Physical Principles and Instrumentation

The rejection control on a scanner:

  1. increases the transmitted pulse amplitude
  2. amplifies deep echoes more than shallow ones
  3. shifts the focal zone toward the far field
  4. removes low-level echoes below a chosen threshold Correct answer.

Explanation

Rejection sets a minimum echo amplitude that will be displayed, discarding weaker signals such as electronic noise. Raising rejection cleans the background but can also erase genuine low-level echoes. It is a display processing function and does not change the transmitted pulse.

Question 42Physical Principles and Instrumentation

For a fixed transducer diameter, decreasing the operating frequency will:

  1. leave the far-field beam width unchanged
  2. make the far-field beam diverge more Correct answer.
  3. shorten the near field
  4. make the far-field beam diverge less

Explanation

Beam divergence is described by the sine of the divergence angle being proportional to wavelength divided by aperture. Lower frequency means longer wavelength, so the beam spreads more widely in the far field. The near-field length, which is proportional to aperture squared divided by wavelength, actually lengthens.

Question 43Physical Principles and Instrumentation

The mechanical index is calculated as:

  1. spatial peak temporal average intensity divided by 100 mW/cm squared
  2. total acoustic power divided by the beam cross-sectional area
  3. peak positive pressure multiplied by the center frequency in MHz
  4. peak negative pressure divided by the square root of frequency Correct answer.

Explanation

The mechanical index estimates the likelihood of cavitation and is defined as the peak rarefactional (negative) pressure in megapascals divided by the square root of the center frequency in megahertz. Higher peak negative pressure raises the index, while higher frequency lowers it. Intensity-based quantities relate to thermal risk instead.

Question 44Physical Principles and Instrumentation

The matching layer of a pulsed transducer serves to:

  1. damp crystal vibration to shorten the pulse
  2. electrically isolate the crystal from the patient
  3. focus the beam at a selected depth
  4. reduce the impedance mismatch between crystal and tissue Correct answer.

Explanation

The matching layer has an impedance intermediate between the crystal and soft tissue, improving transmission of sound across that interface. This reduces reflection losses at the face of the transducer. Damping and focusing are handled by other components.

Question 45Physical Principles and Instrumentation

Compression of the echo signal before display:

  1. widens the dynamic range of the signal
  2. selectively amplifies echoes according to their depth
  3. has no effect on the displayed gray scale
  4. narrows the range of echo amplitudes displayed Correct answer.

Explanation

Echo amplitudes span a range far wider than a display can show, so compression logarithmically reduces that range. Weak echoes are boosted relative to strong ones, allowing both to appear in the same gray-scale image. Depth-selective amplification is a separate function.

Question 46Physical Principles and Instrumentation

A piezoelectric crystal propagates sound at 4000 m/s. To resonate at 5 MHz, the crystal thickness should be approximately:

  1. 1.6 mm
  2. 0.4 mm Correct answer.
  3. 0.2 mm
  4. 0.8 mm

Explanation

A crystal resonates when its thickness equals one half-wavelength of sound in the crystal material. The wavelength is 4000 m/s divided by 5,000,000 Hz, or 0.8 mm, so the half-wavelength thickness is 0.4 mm. Higher frequencies therefore require thinner crystals.

Question 47Physical Principles and Instrumentation

An unfocused 5 MHz transducer has an element radius of 1 cm. Its near-field length in soft tissue is approximately:

  1. 0.32 cm
  2. 3.2 cm
  3. 32 cm Correct answer.
  4. 8 cm

Explanation

Near-field length equals the square of the element radius divided by the wavelength. The wavelength at 5 MHz is about 0.031 cm, so 1 cm squared divided by 0.031 cm gives roughly 32 cm. A larger aperture or higher frequency lengthens the near field.

Question 48Physical Principles and Instrumentation

A scanner calibrated for 1540 m/s displays a reflector at 10 cm depth in a medium where sound actually travels at 1450 m/s. The true depth of the reflector is approximately:

  1. 11.0 cm
  2. 9.4 cm Correct answer.
  3. 10.6 cm
  4. 8.5 cm

Explanation

The scanner converts echo time to distance using an assumed speed, so depth is proportional to the speed actually traveled. Multiplying 10 cm by 1450/1540 gives about 9.4 cm. When the true speed is lower than the assumed speed, the machine overestimates depth.

Question 49Physical Principles and Instrumentation

A 4 MHz pulse in soft tissue contains 3 cycles. The axial resolution is approximately:

  1. 0.96 mm
  2. 1.2 mm
  3. 0.39 mm
  4. 0.58 mm Correct answer.

Explanation

Axial resolution equals half the spatial pulse length. At 4 MHz the wavelength is about 0.385 mm, so three cycles occupy about 1.16 mm, and half of that is roughly 0.58 mm. Fewer cycles per pulse improve axial resolution.

Question 50Physical Principles and Instrumentation

If the maximum pulse repetition frequency is 9,625 Hz and each frame contains 100 lines, the highest possible frame rate is:

  1. 192 Hz
  2. 96 Hz Correct answer.
  3. 770 Hz
  4. 48 Hz

Explanation

The time available per frame is shared among all lines, so frame rate equals PRF divided by the number of lines per frame. Dividing 9,625 Hz by 100 gives about 96 frames per second. Increasing line density or depth lowers the achievable frame rate.