Part 2 - Physic's
26-50
Questions and Answers
- 1.
When using the AIUM phantom, the following shoudl be kept constant for comparsion purposes
- A.
Trandsucer
- B.
DGC (depth gain compensation)
- C.
Output power
- D.
All of the above
Correct Answer
D. All of the aboveExplanation
When using the AIUM phantom, it is important to keep the transducer, DGC (depth gain compensation), and output power constant for comparison purposes. This means that these three factors should not be changed or varied during the experiment or study. Keeping them constant ensures that any differences observed in the results are not due to variations in these factors, but rather due to the specific variables being investigated.Rate this question:
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- 2.
The product of the period and the number of cycles in a pulse is
- A.
Pulse repetition frequency
- B.
Continuous wave
- C.
Pulse repetition period
- D.
Pulse duration
Correct Answer
D. Pulse durationExplanation
The product of the period and the number of cycles in a pulse refers to the total time duration of the pulse, which is known as the pulse duration. This term is commonly used in the field of electronics and signal processing to measure the length of a pulse. The pulse duration is calculated by multiplying the period (the time it takes for one cycle of the pulse) by the number of cycles in the pulse. Therefore, the correct answer is pulse duration.Rate this question:
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- 3.
As sound travels through tissue, both the intensity and amplitude are reduced. This reduction is called
- A.
Sonolucency
- B.
Progression
- C.
Attenuation
- D.
Enhancement
Correct Answer
C. AttenuationExplanation
When sound waves travel through tissue, they encounter resistance and lose energy. This causes a decrease in both the intensity (power) and amplitude (strength) of the sound waves. This phenomenon is known as attenuation. Attenuation refers to the reduction in the strength or intensity of a signal as it travels through a medium. In medical imaging, attenuation is an important concept as it helps in understanding the behavior of ultrasound waves as they pass through different tissues in the body.Rate this question:
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- 4.
When using the AIUM phantom, which of the following groups is utilized to calibrated for axial and lateral resolution?
- A.
Group A
- B.
Group B
- C.
Group C
- D.
Group D
Correct Answer
C. Group CExplanation
The AIUM phantom is used to calibrate for axial and lateral resolution in ultrasound imaging. Group C is utilized for this purpose, although the specific details of how it is used are not provided.Rate this question:
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- 5.
In order to reduce the signal dynamic range, the ultrasound system uses
- A.
Rejection
- B.
Relaxation
- C.
Elimination
- D.
Compression
Correct Answer
D. CompressionExplanation
The ultrasound system uses compression to reduce the signal dynamic range. Compression is a technique that adjusts the amplitude of the received ultrasound signals. It amplifies weaker signals and reduces the amplitude of stronger signals, thus compressing the dynamic range. This is important because ultrasound signals can vary greatly in intensity, and compressing the dynamic range allows for better visualization and analysis of the image.Rate this question:
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- 6.
How can axial resolution be improved?
- A.
By reducing the attenuation
- B.
By reducing the spatial pulse length
- C.
By reducing the bandwidth
- D.
All of the above
Correct Answer
B. By reducing the spatial pulse lengthExplanation
By reducing the spatial pulse length, the axial resolution can be improved. Axial resolution refers to the ability of an imaging system to distinguish between two closely spaced objects along the direction of the ultrasound beam. By reducing the spatial pulse length, the ultrasound waves emitted by the system become shorter in duration, allowing for better differentiation of structures along the beam path. This leads to improved axial resolution, enabling the system to provide more detailed and accurate imaging.Rate this question:
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- 7.
The piezoelectric effect describes how ________ is converted into _______ by a _______.
- A.
Electricity, image, display
- B.
Incident sound, reflected sound, boundary
- C.
Ultrasound, electricity, transducer
- D.
Ultrasound, heat, tissue
Correct Answer
C. Ultrasound, electricity, transducerExplanation
The piezoelectric effect refers to the conversion of ultrasound into electricity by a transducer. Ultrasound refers to high-frequency sound waves that are used in various applications such as medical imaging. When these ultrasound waves strike a piezoelectric material, such as a transducer, they generate an electric charge. This electric charge can then be used to create images or perform other functions in devices like ultrasound machines.Rate this question:
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- 8.
The frequencies used in diagnostic ultraound imaging
- A.
Are much higher than those in the audio range
- B.
Determine imaging depth in tissue
- C.
Determine spatial imaging resolution
- D.
All of the above
Correct Answer
D. All of the aboveExplanation
The frequencies used in diagnostic ultrasound imaging are much higher than those in the audio range, as they need to penetrate the body and provide detailed images. These higher frequencies also determine the imaging depth in tissue, as they can penetrate deeper into the body. Additionally, the frequencies used in ultrasound imaging also determine the spatial imaging resolution, as higher frequencies can provide more detailed and clearer images. Therefore, all of the given options are correct.Rate this question:
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- 9.
How many shades of gray can a 3 bit system display?
- A.
3
- B.
6
- C.
8
- D.
10
Correct Answer
C. 8Explanation
A 3-bit system can display 2^3 = 8 different combinations. Each bit can have two possible values, either 0 or 1. Therefore, with three bits, there are 8 possible combinations (000, 001, 010, 011, 100, 101, 110, 111), which means that a 3-bit system can display 8 shades of gray.Rate this question:
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- 10.
With increased damping, which of the following is improved?
- A.
Axial resolution
- B.
Slice thickness resolution
- C.
Contrast resolution
- D.
Lateral resolution
Correct Answer
A. Axial resolutionExplanation
With increased damping, axial resolution is improved. Axial resolution refers to the ability to distinguish between two structures along the direction of the ultrasound beam. Increased damping reduces the duration of the ultrasound pulse, resulting in better axial resolution. This means that smaller structures can be accurately identified and distinguished from each other, leading to improved image quality and diagnostic capabilities.Rate this question:
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- 11.
The lateral resolution of an ultrasound system depends most upon
- A.
The transducer diameter
- B.
The transducer damping
- C.
The speed of sound in soft tissue
- D.
Memory of the display
Correct Answer
A. The transducer diameterExplanation
The lateral resolution of an ultrasound system refers to its ability to distinguish between two closely spaced objects that are side by side. It is primarily determined by the transducer diameter, as a larger diameter allows for a narrower beam width and better spatial resolution. The other options, such as transducer damping, speed of sound in soft tissue, and memory of the display, do not directly affect the lateral resolution of the ultrasound system.Rate this question:
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- 12.
In an ultrasound imaging instrument, the video display or cathode ray tub (CRT) is used as
- A.
A display
- B.
A pulser
- C.
A receiver
- D.
A memory
Correct Answer
A. A displayExplanation
The correct answer is "a display". In an ultrasound imaging instrument, the video display or cathode ray tube (CRT) is used to visually present the ultrasound images to the user. The display allows the user to view and analyze the captured ultrasound data in real-time, providing valuable information for medical diagnosis and evaluation. The CRT technology provides a high-resolution and clear image display, making it suitable for this purpose.Rate this question:
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- 13.
An important factor in the selection of a transducer for a specific application is the ultrasonic attenuation in tissue. Owing to this attenuation, a 7.5 MHz transducer should generally be used for
- A.
Imaging deep structures
- B.
Imaging superficial structures
- C.
Imaging the adult lungs
- D.
Imaging both deep and shallow structures
Correct Answer
B. Imaging superficial structuresExplanation
The ultrasonic attenuation in tissue refers to the loss of ultrasound energy as it travels through the tissue. Higher frequencies experience more attenuation than lower frequencies. Since a 7.5 MHz transducer has a higher frequency, it is more suitable for imaging superficial structures. This is because the higher frequency allows for better resolution and imaging of shallow tissues, where attenuation is less of an issue. Therefore, a 7.5 MHz transducer is generally used for imaging superficial structures.Rate this question:
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- 14.
The length of time is takes to complete a single cycle is known as
- A.
Period
- B.
Frequency
- C.
Impedance
- D.
Pulse repetition period
Correct Answer
A. PeriodExplanation
The length of time it takes to complete a single cycle is known as the period. This term is commonly used in physics and mathematics to describe the time it takes for a repeating event to occur. It is often represented by the symbol "T" and is typically measured in seconds. In the context of waves or oscillations, the period refers to the time it takes for one complete cycle to occur.Rate this question:
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- 15.
Ultrasound is defined as sound with a frequency of
- A.
Greater than 2,000 cycles per second
- B.
Less than 20,000 Hz
- C.
Greater than 20,000 Hz
- D.
Greater than 1.54 cycles per second
Correct Answer
C. Greater than 20,000 HzExplanation
Ultrasound is a type of sound that has a frequency higher than the upper limit of human hearing, which is typically around 20,000 Hz. Frequencies above this range are considered ultrasound. Therefore, the correct answer is "greater than 20,000 Hz."Rate this question:
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- 16.
Matching - wavelength formula
- A.
Propagation speed divided by frequency
- B.
Density times propagation speed
- C.
Spped at which wave travels through a medium
- D.
Power divided by area
Correct Answer
A. Propagation speed divided by frequencyExplanation
The correct answer is "propagation speed divided by frequency". This formula is used to calculate the wavelength of a wave. Wavelength is the distance between two corresponding points on a wave, such as two crests or two troughs. The propagation speed refers to how fast the wave is traveling, while the frequency refers to the number of complete cycles of the wave that occur in one second. Dividing the propagation speed by the frequency gives us the wavelength of the wave.Rate this question:
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- 17.
Matching - Impedance formula
- A.
Density times propagation speed
- B.
Material behind the rear face of the transducer
- C.
Rectangular elements in a line
- D.
Operating frequency divided by bandwidth
Correct Answer
A. Density times propagation speedExplanation
The impedance formula in this context refers to the calculation of the characteristic impedance of a transmission line, which is determined by the product of the density and propagation speed of the material through which the signal is being transmitted. This formula helps in understanding the behavior of the transmission line and ensuring efficient signal transfer.Rate this question:
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- 18.
Matching - Propagation speed
- A.
Speed at which wave travels through a medium
- B.
Material in front of transducer face
- C.
Material behind the rear face of the transducer
- D.
Related to time and motion
Correct Answer
A. Speed at which wave travels through a mediumExplanation
The correct answer is "speed at which wave travels through a medium". This refers to the rate at which a wave propagates or travels through a specific material or medium. It describes how quickly the wave energy is transmitted from one point to another within the medium. The propagation speed is influenced by the properties of the medium, such as its density, elasticity, and compressibility.Rate this question:
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- 19.
Matching - Spatial pulse length
- A.
Length of space over which a pulse occurs
- B.
Density times propagation speed
- C.
Power divided by area
- D.
Operating frequency divided by bandwidth
Correct Answer
A. Length of space over which a pulse occursExplanation
The spatial pulse length refers to the length of space over which a pulse occurs. This means that it measures the distance over which a pulse of energy is spread out. It is not related to density, propagation speed, power, area, operating frequency, or bandwidth. Instead, it specifically focuses on the spatial extent of the pulse.Rate this question:
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- 20.
Matching - Intensity Formula
- A.
Power divided by area
- B.
Density times propagation speed
- C.
Propagation speed divided by frequency
- D.
Operating frequency divided by bandwidth
Correct Answer
A. Power divided by areaExplanation
The intensity of a wave is determined by the amount of power it carries per unit area. This means that the power of the wave is divided by the area over which it is spread. The greater the power and the smaller the area, the higher the intensity of the wave. Therefore, the formula for intensity is power divided by area.Rate this question:
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- 21.
Matching - Linear array
- A.
Rectangular elements in a line
- B.
Picture element
- C.
Paths to and from reflector are not the same
- D.
Related to reververation artifact
Correct Answer
A. Rectangular elements in a lineExplanation
The correct answer is "Rectangular elements in a line". This answer most likely refers to the concept of a linear array in signal processing or antenna design. A linear array consists of multiple antenna elements arranged in a straight line, typically in a rectangular shape. This configuration allows for directional beamforming and improved signal reception or transmission in a specific direction. Therefore, the answer "Rectangular elements in a line" accurately describes the characteristic of a linear array.Rate this question:
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- 22.
Matching - Matching Layer
- A.
Material in front of transducer face
- B.
Material behind the rear face of the transducer
- C.
Paths to and from reflector are not the same
- D.
Picture element
Correct Answer
A. Material in front of transducer faceExplanation
The correct answer is "material in front of transducer face." In ultrasonic testing, the transducer is the device that generates and receives sound waves. The material in front of the transducer face refers to the substance or medium through which the sound waves travel before reaching the transducer. This material can affect the transmission and reception of the waves, influencing the quality and accuracy of the ultrasonic inspection.Rate this question:
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- 23.
Matching - Damping Material
- A.
Material in front of transducer face
- B.
Material behind the rear face of the transducer
- C.
Paths to and from reflector are not the same
- D.
Picture element
Correct Answer
B. Material behind the rear face of the transducerExplanation
The correct answer is "material behind the rear face of the transducer." In ultrasound imaging, the transducer emits sound waves and receives the echoes produced by the reflection of these waves. To improve image quality, it is important to minimize the reflection of sound waves from the rear face of the transducer. This is achieved by placing a damping material behind the transducer, which absorbs the sound waves and reduces their reflection.Rate this question:
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- 24.
Matching - Q factor formula
- A.
Operating frequency divided by bandwidth
- B.
Related to time and motion
- C.
Power divided by area
- D.
Occurs when doppler shift exceeds 1/2 PRF
Correct Answer
A. Operating frequency divided by bandwidthExplanation
The Q factor formula is a mathematical expression used to calculate the quality factor of a resonant circuit. It is calculated by dividing the operating frequency of the circuit by its bandwidth. The Q factor is a measure of the efficiency of the circuit, with higher values indicating a narrower bandwidth and better selectivity.Rate this question:
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- 25.
Matching - M Mode
- A.
Related to time and motion
- B.
Picture element
- C.
Rectangular elements in a line
- D.
Area where imaging cannot be performed
Correct Answer
A. Related to time and motionExplanation
The term "M Mode" is commonly used in medical imaging, specifically in ultrasound. It refers to a display mode that shows the motion of structures over time. In M Mode, a single scan line is used to create a graph-like representation of movement, allowing for the assessment of dynamic changes in tissues or organs. Therefore, the term "related to time and motion" accurately describes the concept of M Mode in medical imaging.Rate this question:
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