Power divided by the beam area is equal to which quantity?

Power divided by the beam area is equal to which quantity?

• A. Probe angle
• B. Speed of sound
• C. Intensity
• D. Attenuation

Answer: (C)

In ultrasound, intensity is defined as the power flowing through a unit area, so it’s equal to the beam power divided by the cross-sectional area of the beam. This gives I = P/A, with units of watts per square meter. It tells you how much acoustic energy is delivered per unit area at that point along the beam, which is why it’s the quantity described by the question. For example, if the transducer puts out 2 watts through a beam area of 0.01 m^2, the intensity is 200 W/m^2. The other options don’t fit because they describe different ideas: the probe angle relates to direction, not energy density; the speed of sound is a property of the medium and how fast the wave travels; attenuation describes how the beam loses strength as it travels, not the instantaneous energy per area at the beam’s cross-section.

In ultrasound, intensity is defined as the power flowing through a unit area, so it’s equal to the beam power divided by the cross-sectional area of the beam. This gives I = P/A, with units of watts per square meter. It tells you how much acoustic energy is delivered per unit area at that point along the beam, which is why it’s the quantity described by the question. For example, if the transducer puts out 2 watts through a beam area of 0.01 m^2, the intensity is 200 W/m^2.

The other options don’t fit because they describe different ideas: the probe angle relates to direction, not energy density; the speed of sound is a property of the medium and how fast the wave travels; attenuation describes how the beam loses strength as it travels, not the instantaneous energy per area at the beam’s cross-section.