# Physics Of Ultrasound! Trivia Questions And Facts Quiz

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Questions: 170 | Attempts: 9,705  Settings  Are you a medical student looking for a quiz to test out what you know about the Physics of ultrasound? If so then you are in luck as the quiz below is perfect for you as it is used by students preparing for their exams too. Do give it a try and see just how much you can remember.

• 1.

### Which is not an acoustic variable?

• A.

Density

• B.

Pressure

• C.

Distance

• D.

Intensity

D. Intensity
Explanation
Intensity is not an acoustic variable because it refers to the power or energy of a sound wave per unit area. Acoustic variables, on the other hand, are physical quantities that can be directly measured or calculated from the properties of a sound wave, such as density, pressure, and distance. Intensity is derived from these variables and represents the strength or loudness of the sound wave.

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• 2.

### All of the following are true EXCEPT:

• A.

Two waves with identical frequencies must interfere constructively

• B.

Constructive interference is associated with waves that are in phase

• C.

Out of phase waves interfere destructively

• D.

Waves of different frequencies may exhibit both contructive and destructive interference at different times

A. Two waves with identical frequencies must interfere constructively
Explanation
This statement is not true because two waves with identical frequencies can interfere both constructively and destructively depending on their phase relationship. If the waves are in phase, they will interfere constructively, but if they are out of phase, they will interfere destructively.

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• 3.

### Put in decreasing order

• A.

Deca

• B.

Deci

• C.

Micro

• D.

Centi

A. Deca
B. Deci
C. Micro
D. Centi
Explanation
The given words are units of measurement that represent different magnitudes. When put in decreasing order, "deca" comes first as it represents ten times the base unit, followed by "deci" which represents one-tenth of the base unit. Next is "micro" which represents one millionth of the base unit, and finally "centi" which represents one-hundredth of the base unit. Therefore, the correct order is deca, deci, micro, centi.

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• 4.

### Which of the following sound waves is ultrasonic and least useful in diagnostic imaging?

• A.

30 KHZ

• B.

8 MHz

• C.

8,000 Hz

• D.

3,000 kHz

• E.

15 Hz

A. 30 KHZ
Explanation
Ultrasonic sound waves have a frequency greater than 20,000 Hz. Among the given options, the sound wave with a frequency of 30 KHz is ultrasonic. Diagnostic imaging typically uses sound waves with frequencies in the range of 1-20 MHz, so the sound wave with a frequency of 30 KHz would be least useful in diagnostic imaging.

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• 5.

### What is the frequency of a wave with 1msec period?

• A.

10,000 Hz

• B.

1,000 kHz

• C.

1kHz

• D.

1MHz

C. 1kHz
Explanation
The frequency of a wave is the number of complete cycles it completes in one second. A period is the time it takes for one complete cycle. In this question, the period of the wave is given as 1 millisecond (1 msec). To find the frequency, we need to invert the period by taking the reciprocal. 1 millisecond is equal to 0.001 seconds. Therefore, the frequency is 1/0.001 = 1000 Hz, which is equivalent to 1kHz.

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• 6.

### If the frequency of an US wave is doubled, what happens to the period?

• A.

Doubles

• B.

Halved

• C.

Remains the same

• D.

4 times greater

B. Halved
Explanation
When the frequency of an ultrasound wave is doubled, the period of the wave is halved. The frequency of a wave refers to the number of complete cycles of the wave that occur in one second, while the period refers to the time it takes for one complete cycle to occur. Since frequency and period are inversely related, doubling the frequency will result in halving the period. This means that the time it takes for one complete cycle of the wave to occur will be reduced by half when the frequency is doubled.

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• 7.

### The units of pulse repetition frequency are:

• A.

Per minute

• B.

Msec

• C.

Mm/us

• D.

MHx

A. Per minute
Explanation
The pulse repetition frequency (PRF) is a measure of the number of pulses per unit of time. In this case, the units of PRF are given as "per minute." This means that the PRF is measured in terms of the number of pulses occurring within a minute. It is a common unit used to describe the frequency at which pulses are repeated in various systems, such as radar or ultrasound.

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• 8.

### The time from the beginning of a pulse until its end is__________?

• A.

Period

• B.

Pulse duration

• C.

Pulse length

• D.

PRF

B. Pulse duration
Explanation
The time from the beginning of a pulse until its end is referred to as the pulse duration. This term specifically measures the length of time that a pulse signal remains active or "on" before it turns off. It is different from the pulse length, which typically refers to the physical length or spatial extent of the pulse. The period, on the other hand, represents the time taken for one complete cycle of a repetitive waveform, while PRF stands for Pulse Repetition Frequency, which measures the number of pulses occurring in one second.

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• 9.

### The time that a transducer is pulsing is measured at 18 seconds in one hour of total elapsed time.  What is the duty factor?

• A.

0.3

• B.

0.005

• C.

0.5Hz

• D.

0.005msec

B. 0.005
Explanation
60 seconds equal 1 minute. There are 60 minutes in 1 hour so 60 seconds X 60 minutes = 18 divided by 3600 = 0.005

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• 10.

### If a wave's amplitude is doubled, what happens to the power?

• A.

Nothing

• B.

Halved

• C.

Dpib;ed

• D.

Explanation
When a wave's amplitude is doubled, the power of the wave is quadrupled. This is because power is directly proportional to the square of the amplitude. Therefore, if the amplitude is doubled, the power will increase by a factor of four.

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• 11.

### If the level of an acoustic variable ranges from 55 to 105, what is the amplitude?

• A.

105

• B.

50

• C.

25

• D.

55

C. 25
Explanation
The amplitude of an acoustic variable refers to the maximum displacement or distance from the equilibrium position. In this case, the given range of 55 to 105 represents the maximum and minimum values of the variable. The amplitude is calculated by taking half of the difference between the maximum and minimum values. Therefore, the amplitude in this case would be (105-55)/2 = 25.

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• 12.

### If the intensity of a sound beam remains unchanged while the beam area is reduced in half, what happened to the power?

• A.

• B.

Doubled

• C.

Halved

• D.

Unchanged

C. Halved
Explanation
If the intensity of a sound beam remains unchanged while the beam area is reduced in half, the power of the sound beam would be halved. This is because power is directly proportional to intensity and area. When the beam area is reduced by half, the same amount of power is spread over a smaller area, resulting in a halving of the power.

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• 13.

### Put these intensities in decreasing order

• A.

SATP

• B.

SPTP

• C.

SATA

• D.

SPTP, SATP, SATA

• E.

SPTP SATA, SATP

D. SPTP, SATP, SATA
Explanation
The given answer suggests that the intensities should be arranged in decreasing order. Among the options provided, "SPTP" has the highest intensity, followed by "SATP", and finally "SATA" with the lowest intensity. Therefore, the correct order is "SPTP, SATP, SATA".

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• 14.

### The duty factor for continuous wave ultrasound is

• A.

10.0

• B.

1%

• C.

1.0

• D.

0.0%

C. 1.0
Explanation
The duty factor for continuous wave ultrasound refers to the percentage of time that the ultrasound wave is actually being transmitted. A duty factor of 1.0 means that the ultrasound wave is being transmitted for 100% of the time, indicating a continuous transmission without any interruptions or pauses.

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• 15.

### What is the minimum value of the SP/SA factor?

• A.

10.0

• B.

1%

• C.

1.0

• D.

0.0%

C. 1.0
Explanation
The minimum value of the SP/SA factor is 1.0. This means that the selling price (SP) is equal to the selling amount (SA). In other words, the item is being sold at its cost price without any profit. This could be due to various reasons such as promotional offers, discounts, or loss-leading strategies.

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• 16.

### Which pair of intensities has the same value for continuous wave US?

• A.

Spatial peak & spatial average

• B.

Temporal peak & spatial peak

• C.

Pulse average & temporal average

• D.

Spatial average & temporal average

C. Pulse average & temporal average
Explanation
The pair of intensities that has the same value for continuous wave US is pulse average and temporal average. In continuous wave ultrasound, there are no pulses, so the pulse average intensity is zero. The temporal average intensity is the average intensity over time, which remains constant for continuous wave ultrasound. Therefore, both pulse average and temporal average intensities have the same value, which is non-zero for continuous wave ultrasound.

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• 17.

### The fundamental frequency of a transcucer is 2.5 MHz.  What is the second harmonic frequency?

• A.

1.25 MHz

• B.

4.5 MHz

• C.

4.8 MHz

• D.

5 cm

C. 4.8 MHz
Explanation
The second harmonic frequency is twice the fundamental frequency. Therefore, if the fundamental frequency is 2.5 MHz, the second harmonic frequency would be 2 * 2.5 MHz = 5 MHz. However, none of the options provided match this value. Therefore, the correct answer is not available.

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• 18.

### A pulse is emitted by a transducer and is traveling in soft tissue.  The go-return time, or time of flight, of a sound pulse is 130 microseconds.  What is the reflector depth?

• A.

10 cm

• B.

10 mm

• C.

10 m

• D.

130 mm

• E.

13 cm

A. 10 cm
Explanation
PRP(us) = imaging depth x 13 us/cm
130 us = ? x 13 us/cm
130 divide by 13 = 10 cm or
Depth(mm) = 1.54mm/us x go return time(us)/2
= 1.54 x 130/2
= 200.2/2
mm = 100.1 or 10 cm

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• 19.

### What is the approximate attenuation coefficient of 1MHz US in soft tissue?

• A.

0.5 dBcm

• B.

1cm

• C.

3dBcm

• D.

1dB

A. 0.5 dBcm
Explanation
The approximate attenuation coefficient of 1MHz US in soft tissue is 0.5 dBcm. Attenuation coefficient refers to the rate at which ultrasound waves decrease in intensity as they pass through a medium. In soft tissue, the attenuation coefficient at 1MHz is approximately 0.5 dB per centimeter of distance traveled. This means that for every centimeter the ultrasound waves travel in soft tissue, their intensity decreases by 0.5 dB.

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• 20.

### The relaive output of an US instrument is calibrated in dB and the operator increases the output by 60 dB.  The beam intensity is increased by which of the following?

• A.

5%

• B.

Two times

• C.

Twenty times

• D.

One hundred times

• E.

One million times

E. One million times
Explanation
3dB means 2 x bigger 6dB is 3dB + 3dB, therefore 6dB means 2 x 2 or 4 times bigger. 60dB = 10x10x10x10x10x10 or 1,000,000

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• 21.

### The more pixels per inch:

• A.

The better the temporal resolution

• B.

The better is the spatial resolution

• C.

• D.

The higher the reliability

B. The better is the spatial resolution
Explanation
The statement suggests that the higher the number of pixels per inch, the better the spatial resolution. Spatial resolution refers to the level of detail that can be captured in an image, and it is directly related to the number of pixels. Therefore, having more pixels per inch means that there are more individual dots that make up the image, resulting in a higher level of detail and better spatial resolution.

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• 22.

### If we increase the transducer diameter, the beam diameter in the far zone is:

• A.

Increased

• B.

Decreased

• C.

Unchanges

B. Decreased
Explanation
When we increase the transducer diameter, the beam diameter in the far zone decreases. This is because the beam divergence angle is inversely proportional to the transducer diameter. As the transducer diameter increases, the beam divergence angle decreases, resulting in a narrower beam diameter in the far zone. Therefore, the correct answer is decreased.

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• 23.

### If the transducer diameter increases, the lateral resolution at its smallest dimension is

• A.

Increased

• B.

Decreased

• C.

Unchanged

A. Increased
Explanation
When the transducer diameter increases, the lateral resolution at its smallest dimension is increased. This is because a larger transducer diameter allows for a wider beam width, resulting in a decrease in the lateral resolution. With a smaller transducer diameter, the beam width is narrower, leading to an increase in the lateral resolution. Therefore, when the transducer diameter increases, the lateral resolution at its smallest dimension is increased.

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• 24.

### If we increase the frequency the near zone length is

• A.

Increased

• B.

Decreased

• C.

Unchanged

A. Increased
Explanation
When the frequency is increased, the near zone length is increased. This is because the near zone length is inversely proportional to the frequency. As the frequency increases, the wavelength decreases, causing the near zone length to increase. This means that objects or receivers in the near zone will experience a longer distance between them, resulting in an increase in the near zone length.

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• 25.

### If the frequency is decreased, the numerical value of the radial resolution is

• A.

Increased

• B.

Decreased

• C.

Unchanged

A. Increased
Explanation
When the frequency is decreased, the numerical value of the radial resolution is increased. Radial resolution refers to the ability to distinguish between two points in a radial direction. A higher numerical value indicates a better ability to distinguish between these points. Therefore, when the frequency is decreased, the wavelength becomes longer, resulting in a decrease in the ability to distinguish between points and hence an increase in the numerical value of the radial resolution.

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• 26.

### If the spatial pulse length is 10 mm, what is the axial resolution?

• A.

0.5 cm

• B.

5 cm

• C.

10 mm

• D.

1 cm

A. 0.5 cm
Explanation
The axial resolution of an ultrasound system is determined by the spatial pulse length, which is the distance covered by one pulse. In this case, the spatial pulse length is given as 10 mm. The axial resolution is typically half of the spatial pulse length, so the axial resolution would be 0.5 cm.

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• 27.

### Which of the following is the best lateral resolution?

• A.

15 mm

• B.

6 mm

• C.

0.06 cm

• D.

2 cm

C. 0.06 cm
Explanation
The best lateral resolution is 0.06 cm. Lateral resolution refers to the ability of an imaging system to distinguish between two closely spaced objects in the horizontal direction. A smaller value indicates better resolution, as it means the system can distinguish smaller details. Therefore, 0.06 cm is the best option among the given choices as it represents the smallest value, indicating the highest level of resolution.

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• 28.

### What is the meaning of thermal index = 37

• A.

A tissue temperature may rise 3 degrees Farenheit

• B.

Tissue temperature will rise 3 degrees Celcius

• C.

Tissue temperature may rise 3 degrees Celcius

• D.

Microbubbles 3mm in diameter will burst

C. Tissue temperature may rise 3 degrees Celcius
Explanation
The thermal index of 37 indicates that the tissue temperature may rise by 3 degrees Celsius. This means that there is a possibility that the temperature of the tissue being exposed to a certain thermal source may increase by 3 degrees Celsius. It is important to note that this is just a potential increase and not a definite outcome.

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• 29.

### The interaction of microscopic bubbles and ultrasound form the basis for cavitation bioeffects.  Which of the following forms of cavitation are most likely to produce microstreaming in the intracellular fluid and shear stresses?

• A.

Stable cavitation

• B.

Normal cavitation

• C.

Transient cavitation

• D.

Inertial cavitation

• E.

Active cavitation

A. Stable cavitation
Explanation
Stable cavitation is the most likely form of cavitation to produce microstreaming in the intracellular fluid and shear stresses. Stable cavitation occurs when small bubbles oscillate in size in response to ultrasound waves. These oscillations create microstreaming, which involves the movement of fluid around the bubbles. This microstreaming can cause shear stresses on nearby cells and tissues, leading to various bioeffects. Therefore, stable cavitation is the most suitable form of cavitation for producing microstreaming and shear stresses in the intracellular fluid.

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• 30.

### Which of the following has the greatest output intensity

• A.

B-Mode

• B.

Gray scale imaging

• C.

CW Doppler

• D.

Pulsed Doppler

D. Pulsed Doppler
Explanation
Pulsed Doppler has the greatest output intensity compared to B-Mode, gray scale imaging, and CW Doppler. Pulsed Doppler is a technique that uses short bursts of high-frequency sound waves to measure blood flow velocity and direction. It provides detailed information about blood flow patterns and can detect abnormalities in the circulation. The intensity of the output is higher in pulsed Doppler because it uses short bursts of sound waves, allowing for more accurate and detailed measurements of blood flow. B-Mode and gray scale imaging are used for visualizing anatomical structures, while CW Doppler measures blood flow velocity but does not provide detailed information about direction or depth.

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• 31.

### Which of the following best decribes the empirical approach to the study of bioeffects?

• A.

Exposure-response

• B.

Cause-effect

• C.

Risk-benefit

• D.

Causation-reaction

• E.

Effect-microstreaming

A. Exposure-response
Explanation
The empirical approach to the study of bioeffects involves investigating the relationship between exposure and response. This approach focuses on gathering and analyzing data to understand how different levels of exposure to a certain factor can lead to specific responses or effects. By studying this exposure-response relationship, researchers can gain insights into the potential risks or benefits associated with certain exposures and make informed decisions regarding causation and reactions in biological systems.

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• 32.

### Which of the following best describes the mechanistic approach to the study of bioeffects?

• A.

Cause-effect

• B.

Exposure-response

• C.

Risk-benefit

• D.

Causation-reaction

• E.

Effect-microstreaming

A. Cause-effect
Explanation
The mechanistic approach to the study of bioeffects involves understanding the cause-effect relationship between a certain factor or exposure and its resulting effect on biological systems. This approach focuses on identifying and explaining the specific mechanisms and pathways through which a cause leads to an effect, providing a detailed understanding of the underlying processes. It emphasizes the importance of understanding the causal relationship between a factor and its effect, rather than just observing correlations or associations.

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• 33.

### The propagation speed of US in the AIUM test object is?

• A.

1m/s

• B.

1.54 m/s

• C.

1.54 mm/us

• D.

1 km/s

C. 1.54 mm/us
Explanation
The correct answer is 1.54 mm/us. This is because the speed of ultrasound waves in a medium is determined by the density and elasticity of the medium. In the AIUM test object, the propagation speed of ultrasound is 1.54 mm/us, which indicates that it travels at a speed of 1.54 millimeters per microsecond in that object.

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• 34.

### Enhancement, multipath and side lobes result in:

• A.

Image distortion

• B.

Clearer images

• C.

Artifact

• D.

Resolution problems

C. Artifact
Explanation
Enhancement, multipath, and side lobes can result in artifacts. These artifacts are unwanted distortions or anomalies that appear in images. They can be caused by various factors such as signal interference, improper image processing, or limitations in the imaging system. Artifacts can degrade the quality and accuracy of the images, making them less reliable for analysis or diagnosis. Therefore, the presence of enhancement, multipath, and side lobes can introduce artifacts in the images.

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• 35.

### A mirror image artifact can appear along side of the true anatomy.

• A.

True

• B.

False

B. False
Explanation
A mirror image artifact cannot appear alongside the true anatomy. Mirror image artifacts occur when an object is reflected in a mirror or other reflective surface, creating a reversed image. In medical imaging, mirror image artifacts can occur when a structure is duplicated or appears in the wrong location due to technical errors or positioning. However, they do not appear alongside the true anatomy. Therefore, the statement is false.

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• 36.

### If the lines per frame is increased while the imaging depth is unchanged then:

• A.

Frame rate increases

• B.

Number of shades of gray decreases

• C.

The frame rate decreases

• D.

This cannot happen

C. The frame rate decreases
Explanation
If the lines per frame are increased while the imaging depth remains unchanged, it means that more lines are being scanned in each frame. This leads to a slower frame rate because the system takes longer to scan each line before moving on to the next frame. Therefore, the correct answer is that the frame rate decreases.

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• 37.

### The more pixels per inch:

• A.

The better the temporal resolution

• B.

The better the spatial resolution

• C.

• D.

The higher the reliability

B. The better the spatial resolution
Explanation
The statement suggests that as the number of pixels per inch increases, the spatial resolution improves. Spatial resolution refers to the level of detail that can be observed in an image, and it is directly influenced by the number of pixels. Therefore, a higher number of pixels per inch would result in a clearer and more detailed image, leading to better spatial resolution.

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• 38.

### If we increase the transducer diameter, the beam diameter in the far zone is

• A.

Increased

• B.

Decreased

• C.

Unchanges

B. Decreased
Explanation
When the transducer diameter is increased, the beam diameter in the far zone is decreased. This is because the transducer acts as the source of the ultrasound beam, and a larger transducer diameter means that the beam will spread out less as it propagates in the far zone. As a result, the beam becomes narrower and the diameter decreases.

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• 39.

### If the transducer diameter increases, the lateral resolution at its smalles dimension is

• A.

Increased

• B.

Decreased

• C.

Unchanged

A. Increased
Explanation
When the transducer diameter increases, the lateral resolution at its smallest dimension is increased. This is because a larger transducer diameter allows for a narrower beam width, resulting in improved lateral resolution. A narrower beam width means that the ultrasound waves are more focused and can better distinguish small structures or details in the imaging area. Therefore, increasing the transducer diameter enhances the ability to resolve fine details and improves the lateral resolution.

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• 40.

### If we increase the frequency the near zone length is

• A.

Increased

• B.

Decreased

• C.

Unchanges

A. Increased
Explanation
Increasing the frequency of a wave causes the near zone length to increase. The near zone is the region close to the source where the wave is still undergoing changes and has not fully developed into a far-field wave. As the frequency increases, the wavelength decreases, resulting in a shorter distance for the wave to fully develop. This causes the near zone length to increase as the wave takes longer to reach the far-field.

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• 41.

### If the frequency is decreased, the numberical value of the radial resolution is

• A.

Increased

• B.

Decreased

• C.

Unchanges

A. Increased
Explanation
When the frequency is decreased, the numerical value of the radial resolution is increased. Radial resolution refers to the ability to distinguish between two closely spaced objects in an imaging system. A higher numerical value indicates a better ability to distinguish between these objects, resulting in improved resolution. By decreasing the frequency, the wavelength of the signal is increased, leading to a larger distance between the peaks and troughs of the wave. This increased distance allows for better differentiation between closely spaced objects, hence increasing the radial resolution.

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• 42.

### If the spatial pulse length is 10mm, what is the axial resolution?

• A.

0.5 cm

• B.

5 cm

• C.

10 mm

• D.

1 cm

A. 0.5 cm
Explanation
The axial resolution is a measure of the ability of an ultrasound system to distinguish between two closely spaced objects along the direction of the ultrasound beam. It is determined by the spatial pulse length, which is the distance that an ultrasound pulse occupies in the axial direction. In this case, the spatial pulse length is given as 10mm. The axial resolution is typically half of the spatial pulse length, so the axial resolution would be 0.5 cm.

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• 43.

### Which of the following is the best lateral resolution?

• A.

15 mm

• B.

6 mm

• C.

0.06 cm

• D.

2 cm

C. 0.06 cm
Explanation
The best lateral resolution is 0.06 cm. Lateral resolution refers to the ability of an imaging system to distinguish between two closely spaced objects in the horizontal direction. A smaller value indicates better resolution, as it means the system can differentiate between smaller distances. Therefore, 0.06 cm is the best option among the given choices as it provides the highest level of detail and accuracy in distinguishing between closely spaced objects.

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• 44.

### What are the units of longitudinal resolution?

• A.

Hertz

• B.

Rayls

• C.

M/s

• D.

Feet

D. Feet
Explanation
The units of longitudinal resolution are feet. Longitudinal resolution refers to the ability to distinguish between two closely spaced objects in the direction of propagation. In this context, feet is a unit of length that is commonly used to measure distances, making it an appropriate unit for expressing the resolution in this case. Hertz, Rayls, and m/s are units that are typically used to measure frequency, impedance, and velocity, respectively, and are not applicable to longitudinal resolution.

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• 45.

### The range equation relates distant from the reflector to __________ and __________.

• A.

Time-of-flight, distance

• B.

Frequency, wavelength

• C.

Time of flight, propagation speed

• D.

Propagation speed, density

C. Time of flight, propagation speed
Explanation
The range equation relates the distance from the reflector to the time of flight and propagation speed. The time of flight refers to the time it takes for the wave to travel from the source to the reflector and back, while the propagation speed is the speed at which the wave travels through the medium. By knowing the time of flight and propagation speed, we can calculate the distance from the reflector.

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• 46.

### Doppler shifts always occur if the source and observer are in motion relative to each other and the angle between the motion and the sound beam is not 90 degrees.

• A.

True

• B.

False

A. True
Explanation
The Doppler effect is a phenomenon that occurs when there is relative motion between a source of sound waves and an observer. It causes a shift in the frequency of the sound waves heard by the observer. This shift in frequency, known as the Doppler shift, can be either an increase or decrease depending on the relative motion. However, for the Doppler shift to occur, it is necessary that the angle between the motion of the source and observer and the sound beam is not 90 degrees. Therefore, the statement that Doppler shifts always occur if the source and observer are in motion relative to each other and the angle between the motion and the sound beam is not 90 degrees is true.

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• 47.

### What is the most typical Doppler shift measured clinically?

• A.

3.5 MHz

• B.

3,500,000 Hz

• C.

2 kHz

• D.

1,000 kHz

• E.

20,000 Hz

C. 2 kHz
Explanation
The most typical Doppler shift measured clinically is 2 kHz. Doppler shift is a change in frequency that occurs when there is relative motion between a source of waves and an observer. In clinical settings, Doppler ultrasound is commonly used to measure blood flow in the body. The frequency shift of the reflected sound waves from moving blood cells is typically in the range of a few kilohertz. Therefore, 2 kHz is the most typical Doppler shift measured clinically.

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• 48.

### Assuming a constant frequency, what happens if the diameter of an unfocused circular transducer is increased?

• A.

The distance to the far field is reduced

• B.

The beam width is the near field is reduced

• C.

The beam width in the near field is increased

• D.

The US wavelength is increased

• E.

The sensitivity is reduced

C. The beam width in the near field is increased
Explanation
When the diameter of an unfocused circular transducer is increased, the beam width in the near field is increased. This means that the beam becomes wider, resulting in a larger area of coverage. The near field refers to the region close to the transducer where the beam is still converging. By increasing the diameter, the beam spreads out more, leading to a wider beam width in the near field.

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• 49.

### Which one of the following sets of properties of a test phantom is most relevant when assessing depth calibration accuracy?

• A.

Reflector spacing and reflection coefficient

• B.

Attenuation and speed of the US in the medium

• C.

Reflector spacing and US attenuation in the medium

• D.

Reflector reflection coefficient and US attenuation in the medium

• E.

Reflector spacing and propagation speed

E. Reflector spacing and propagation speed
Explanation
When assessing depth calibration accuracy, the most relevant properties of a test phantom are reflector spacing and propagation speed. Reflector spacing refers to the distance between the reflectors in the phantom, which is important for accurately measuring depth. Propagation speed refers to the speed at which ultrasound waves travel through the medium, which affects the accuracy of depth measurements. By considering both reflector spacing and propagation speed, one can assess the accuracy of depth calibration in ultrasound imaging.

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• 50.

### Doppler shifts are always created when the source and receiver are in motion relative to each other.

• A.

True

• B.

False Back to top