Question

Please calculate the total transmission (I2/I1) of ultrasound intensity from medium 1 to medium 2 in the following situations: (a) medium 1 directly contacts with medium 2. The acoustic impedences are noted in the figures below. (b) medium 3 is inserted between media 1 and 2. The acoustic impedence of medium 3 is 6.6 MRayls. (c) If the medium 3 is air with an impedence of 0.0004 MRayls. The attenuation in all the media is negnected.

Answer #1

a )

Accostic impedance is R = ( z_{2} - z_{1} /
z_{2} + z_{1} )^{2}

R + T = 1

T = 1 - R

T = 1 - ( z_{2} - z_{1} / z_{2} +
z_{1} )^{2}

T = 4 z_{2}z_{1} / ( z_{2} +
z_{1} )^{2}

T = 4 X 1.65 X 26.4 / ( 1.65 + 26.4 )^{2}

**T = 0.22**

b )

T_{1} = I_{3} / I_{1}

= 4 X 6.6 X 26.4 / ( 6.6 + 26.4 )^{2}

I_{3} = 0.64 I_{1}

T_{2} = 4 X 1.65 X 6.6 / ( 1.65 + 6.6 )^{2}

= 0.64

I_{2} / I_{3} = 0.64

I_{2} = 0.64 I_{3}

I_{2} = 0.64 X 0.64 X I_{1}

**I _{2} / I_{1} = 0.41**

c )

z_{3} = 0.0004 MRays

T_{1} = 4 X 4 X 10^{-4} X 26.4 / ( 4 X
10^{-4} + 26.4 )^{2}

T_{1} = 1.5 X 10^{-5}

T_{1} = I_{3} / I_{1} = 1.5 X
10^{-5}

T_{2} = 4 X 1.65 X 4 X 10^{-4} / ( 1.65 + 4 X
10^{-4} )^{2}

I_{2} / I_{3} = 9.7 X 10^{-4}

I_{2} = 9.7 X 10^{-4} X I_{3}

I_{2} = 9.7 X 10^{-4} X 1.5 X 10^{-5} X
I_{1}

**I _{2} / I_{1} = 1.46 X
10^{-8}**

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