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a = Z 2 Z 1 2 Z 1 + Z 2 2 , size 12{a= { { left (Z rSub { size 8{2} } - Z rSub { size 8{1} } right ) rSup { size 8{2} } } over { left (Z rSub { size 8{1} } +Z rSub { size 8{2} } right ) rSup { size 8{2} } } } } {}

where Z 1 size 12{Z rSub { size 8{1} } } {} and Z 2 size 12{Z rSub { size 8{2} } } {} are the acoustic impedances of the two media making up the boundary. A reflection coefficient of zero (corresponding to total transmission and no reflection) occurs when the acoustic impedances of the two media are the same. An impedance “match” (no reflection) provides an efficient coupling of sound energy from one medium to another. The image formed in an ultrasound is made by tracking reflections (as shown in [link] ) and mapping the intensity of the reflected sound waves in a two-dimensional plane.

Calculate acoustic impedance and intensity reflection coefficient: ultrasound and fat tissue

(a) Using the values for density and the speed of ultrasound given in [link] , show that the acoustic impedance of fat tissue is indeed 1.34 × 10 6 kg/(m 2 ·s) .

(b) Calculate the intensity reflection coefficient of ultrasound when going from fat to muscle tissue.

Strategy for (a)

The acoustic impedance can be calculated using Z = ρv size 12{Z= ital "pv"} {} and the values for ρ and v found in [link] .

Solution for (a)

(1) Substitute known values from [link] into Z = ρv size 12{Z= ital "pv"} {} .

Z = ρv = 925 kg /m 3 1450 m/s size 12{Z=ρv= left ("925"" kg/m" rSup { size 8{3} } right ) left ("1450"" m/s" right )} {}

(2) Calculate to find the acoustic impedance of fat tissue.

1.34 × 10 6 kg/(m 2 ·s)

This value is the same as the value given for the acoustic impedance of fat tissue.

Strategy for (b)

The intensity reflection coefficient for any boundary between two media is given by a = Z 2 Z 1 2 Z 1 + Z 2 2 size 12{a= { { left (Z rSub { size 8{2} } - Z rSub { size 8{1} } right ) rSup { size 8{2} } } over { left (Z rSub { size 8{1} } + Z rSub { size 8{2} } right ) rSup { size 8{2} } } } } {} , and the acoustic impedance of muscle is given in [link] .

Solution for (b)

Substitute known values into a = Z 2 Z 1 2 Z 1 + Z 2 2 size 12{a= { { left (Z rSub { size 8{2} } - Z rSub { size 8{1} } right ) rSup { size 8{2} } } over { left (Z rSub { size 8{1} } + Z rSub { size 8{2} } right ) rSup { size 8{2} } } } } {} to find the intensity reflection coefficient:

a = Z 2 Z 1 2 Z 1 + Z 2 2 = 1 . 34 × 10 6 kg/(m 2 · s) 1.70 × 10 6 kg/(m 2 · s) 2 1 . 70 × 10 6 kg/(m 2 · s) + 1 . 34 × 10 6 kg/(m 2 · s) 2 = 0 . 014 size 12{a= { { left (Z rSub { size 8{2} } - Z rSub { size 8{1} } right ) rSup { size 8{2} } } over { left (Z rSub { size 8{1} } +Z rSub { size 8{2} } right ) rSup { size 8{2} } } } = { { left (1 "." "34" times "10" rSup { size 8{6} } "kgm" rSup { size 8{"-2"} } s rSup { size 8{ - 1} } - 1 "." "70" times "10" rSup { size 8{6} } "kgm" rSup { size 8{"-2"} } s rSup { size 8{"-1"} } right ) rSup { size 8{2} } } over { left (1 "." "70" times "10" rSup { size 8{6} } "kgm" rSup { size 8{"-2"} } s rSup { size 8{ - 1} } +1 "." "34" times "10" rSup { size 8{6} } "kgm" rSup { size 8{"-2"} } s rSup { size 8{"-1"} } right ) rSup { size 8{2} } } } =0 "." "014"} {}

Discussion

This result means that only 1.4% of the incident intensity is reflected, with the remaining being transmitted.

The applications of ultrasound in medical diagnostics have produced untold benefits with no known risks. Diagnostic intensities are too low (about 10 2 W/m 2 size 12{"10" rSup { size 8{ - 2} } "W/m" rSup { size 8{2} } } {} ) to cause thermal damage. More significantly, ultrasound has been in use for several decades and detailed follow-up studies do not show evidence of ill effects, quite unlike the case for x-rays.

The first part of the diagram shows a rectangular shaped transducer with speaker and microphone sending spherical waves to produce echos from a fetus. The second part shows a graph of echo intensity versus time, with four sharp peaks.
(a) An ultrasound speaker doubles as a microphone. Brief bleeps are broadcast, and echoes are recorded from various depths. (b) Graph of echo intensity versus time. The time for echoes to return is directly proportional to the distance of the reflector, yielding this information noninvasively.

The most common ultrasound applications produce an image like that shown in [link] . The speaker-microphone broadcasts a directional beam, sweeping the beam across the area of interest. This is accomplished by having multiple ultrasound sources in the probe’s head, which are phased to interfere constructively in a given, adjustable direction. Echoes are measured as a function of position as well as depth. A computer constructs an image that reveals the shape and density of internal structures.

The first part of the diagram shows an ultrasound device scanning a woman’s abdomen. The second part of the diagram is an ultrasound scan report of the abdomen.
(a) An ultrasonic image is produced by sweeping the ultrasonic beam across the area of interest, in this case the woman’s abdomen. Data are recorded and analyzed in a computer, providing a two-dimensional image. (b) Ultrasound image of 12-week-old fetus. (credit: Margaret W. Carruthers, Flickr)

Questions & Answers

can someone help me with some logarithmic and exponential equations.
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ninjadapaul
20/(×-6^2)
Salomon
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ninjadapaul
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ninjadapaul
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ninjadapaul
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ninjadapaul
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The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
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At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
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Source:  OpenStax, College physics: physics of california. OpenStax CNX. Sep 30, 2013 Download for free at http://legacy.cnx.org/content/col11577/1.1
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