A 2 MHz and 5 mW/cm2 ultrasound beam is traversing 10 cm of material having an attenuation of 2 dB/cm. What’s the intensity of the beam after it has passed through this material?

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Chapter17: Sound Waves
Section: Chapter Questions
Problem 17.20P: The area of a typical eardrum is about 5.00 X 10-5 m2. (a) (Calculate the average sound power...
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A 2 MHz and 5 mW/cm2 ultrasound beam is traversing 10 cm of material having an attenuation of 2 dB/cm. What’s the intensity of the beam after it has passed through this material?

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Then, why the answers are different when I use different formulas, dB= 10log (I/ I0) and I= I0 10^-ax?

The first one is 5e-5 W/cm^2 and the second one is 5e-23 W/cm^2. 

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In this question, I know the intensity of incident ultrasound and I know the attenutaion after it passes 10cm material which should be -20 dB. So, I would like to know by using the formula, dB = 10 log (I/ I0), can I directly calculate the I which is the intensity after passing the material by substituting the dB= -20 and I0 = 5 mW/cm2 given in the question?

Like -20 = 10 log( I / 5e-3), I= 5e-5 W/cm^2?

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Can we use the attenuation formula (I = I0 exp -ax) in this question? If yes, how to calculate this? If no, why?

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How do you get the I0 = 10^-12 W/m^2 ?

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