1- An ultrasonic wave of 3 MHz with an intensity of 10 𝑚𝑊/cm2 is
incident on a flat boundary normally between two media with
acoustic impedance Z1 and Z2. Assume that the attenuation in the
two media can be neglected. The reflected power received by an
ultrasonic transducer with an aperture area of 0.1 cm2 is 0.1
𝑚𝑊.

(a) What is the transmitted intensity?

(b) If Z1 is assumed to be 1.5 × 106 kg/m2-sec, find Z2.

(c) Find reflected and transmitted pressure and show that Pi + Pr =
Pt

(d) Could Pt possibly be greater than Pi? Why?

Community Answer

I have solved for questions (a), (c) & (d)  Ans.:- Given Dato:An ultersonic wave offrequency =3MHzIntisity =10mW//cm^(2)apperture area =0.1cm^(2) is 0.1mW(a) Incident = Preflected + Prronsinitat.Now,Pincident =10mW∣cm^(2)Preffected =0.1m//N//0.1cm^(2)=1mw//cm^(2)Now, Pransmitted =10-1=9m ... See the full answer